problem solving in velocity

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Physics Problems with Solutions

Velocity and speed: solutions to problems.

Solutions to the problems on velocity and speed of moving objects. More tutorials can be found in this website.

A man walks 7 km in 2 hours and 2 km in 1 hour in the same direction. a) What is the man's average speed for the whole journey? b) What is the man's average velocity for the whole journey? Solution to Problem 1: a)

You start walking from a point on a circular field of radius 0.5 km and 1 hour later you are at the same point. a) What is your average speed for the whole journey? b) What is your average velocity for the whole journey? Solution to Problem 3: a) If you walk around a circular field and come back to the same point, you have covered a distance equal to the circumference of the circle.

John drove South 120 km at 60 km/h and then East 150 km at 50 km/h. Determine a) the average speed for the whole journey? b) the magnitude of the average velocity for the whole journey? Solution to Problem 4: a)

If I can walk at an average speed of 5 km/h, how many miles I can walk in two hours? Solution to Problem 5: distance = (average speed) * (time) = 5 km/h * 2 hours = 10 km using the rate of conversion 0.62 miles per km, the distance in miles is given by distance = 10 km * 0.62 miles/km = 6.2 miles

A train travels along a straight line at a constant speed of 60 mi/h for a distance d and then another distance equal to 2d in the same direction at a constant speed of 80 mi/h. a)What is the average speed of the train for the whole journey?

A car travels 22 km south, 12 km west, and 14 km north in half an hour. a) What is the average speed of the car? b) What is the final displacement of the car? c) What is the average velocity of the car?

Solution to Problem 7: a)

Solution to Problem 8: a)

More References and links

• Velocity and Speed: Tutorials with Examples
• Velocity and Speed: Problems with Solutions
• Acceleration: Tutorials with Examples
• Uniform Acceleration Motion: Problems with Solutions
• Uniform Acceleration Motion: Equations with Explanations

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Solved Speed, Velocity, and Acceleration Problems

Simple problems on speed, velocity, and acceleration with descriptive answers are presented for the AP Physics 1 exam and college students. In each solution, you can find a brief tutorial. 

Speed and velocity Problems: 

Problem (1): What is the speed of a rocket that travels $8000\,{\rm m}$ in $13\,{\rm s}$?

Solution : Speed is defined in physics  as the total distance divided by the elapsed time,  so the rocket's speed is \[\text{speed}=\frac{8000}{13}=615.38\,{\rm m/s}\]

Problem (2): How long will it take if you travel $400\,{\rm km}$ with an average speed of $100\,{\rm m/s}$?

Solution : Average speed is the ratio of the total distance to the total time. Thus, the elapsed time is \begin{align*} t&=\frac{\text{total distance}}{\text{average speed}}\\ \\ &=\frac{400\times 10^{3}\,{\rm m}}{100\,{\rm m/s}}\\ \\ &=4000\,{\rm s}\end{align*} To convert it to hours, it must be divided by $3600\,{\rm s}$ which gives $t=1.11\,{\rm h}$.

Problem (3): A person walks $100\,{\rm m}$ in $5$ minutes, then $200\,{\rm m}$ in $7$ minutes, and finally $50\,{\rm m}$ in $4$ minutes. Find its average speed. 

Solution : First find its total distance traveled ($D$) by summing all distances in each section, which gets $D=100+200+50=350\,{\rm m}$. Now, by definition of average speed, divide it by the total time elapsed $T=5+7+4=16$ minutes.

But keep in mind that since the distance is in SI units, so the time traveled must also be in SI units, which is $\rm s$. Therefore, we have\begin{align*}\text{average speed}&=\frac{\text{total distance} }{\text{total time} }\\ \\ &=\frac{350\,{\rm m}}{16\times 60\,{\rm s}}\\ \\&=0.36\,{\rm m/s}\end{align*}

Problem (4): A person walks $750\,{\rm m}$ due north, then $250\,{\rm m}$ due east. If the entire walk takes $12$ minutes, find the person's average velocity. 

Solution : Average velocity , $\bar{v}=\frac{\Delta x}{\Delta t}$, is displacement divided by the elapsed time. Displacement is also a vector that obeys the addition vector rules. Thus, in this velocity problem, add each displacement to get the total displacement . 

In the first part, displacement is $\Delta x_1=750\,\hat{j}$ (due north) and in the second part $\Delta x_2=250\,\hat{i}$ (due east). The total displacement vector is $\Delta x=\Delta x_1+\Delta x_2=750\,\hat{i}+250\,\hat{j}$ with magnitude of  \begin{align*}|\Delta x|&=\sqrt{(750)^{2}+(250)^{2}}\\ \\&=790.5\,{\rm m}\end{align*} In addition, the total elapsed time is $t=12\times 60$ seconds. Therefore, the magnitude of the average velocity is \[\bar{v}=\frac{790.5}{12\times 60}=1.09\,{\rm m/s}\]

Problem (5): An object moves along a straight line. First, it travels at a velocity of $12\,{\rm m/s}$ for $5\,{\rm s}$ and then continues in the same direction with $20\,{\rm m/s}$ for $3\,{\rm s}$. What is its average speed?

Solution: Average velocity is displacement divided by elapsed time, i.e., $\bar{v}\equiv \frac{\Delta x_{tot}}{\Delta t_{tot}}$.

Here, the object goes through two stages with two different displacements, so add them to find the total displacement. Thus,\[\bar{v}=\frac{x_1 + x_2}{t_1 +t_2}\] Again, to find the displacement, we use the same equation as the average velocity formula, i.e., $x=vt$. Thus, displacements are obtained as $x_1=v_1\,t_1=12\times 5=60\,{\rm m}$ and $x_2=v_2\,t_2=20\times 3=60\,{\rm m}$. Therefore, we have \begin{align*} \bar{v}&=\frac{x_1+x_2}{t_1+t_2}\\ \\&=\frac{60+60}{5+3}\\ \\&=\boxed{15\,{\rm m/s}}\end{align*}

Problem (6): A plane flies the distance between two cities in $1$ hour and $30$ minutes with a velocity of $900\,{\rm km/h}$. Another plane covers that distance at $600\,{\rm km/h}$. What is the flight time of the second plane?

Solution: first find the distance between two cities using the average velocity formula $\bar{v}=\frac{\Delta x}{\Delta t}$ as below \begin{align*} x&=vt\\&=900\times 1.5\\&=1350\,{\rm km}\end{align*} where we wrote one hour and a half minutes as $1.5\,\rm h$. Now use again the same kinematic equation above to find the time required for another plane \begin{align*} t&=\frac xv\\ \\ &=\frac{1350\,\rm km}{600\,\rm km/h}\\ \\&=2.25\,{\rm h}\end{align*} Thus, the time for the second plane is $2$ hours and $0.25$ of an hour, which converts to minutes as $2$ hours and ($0.25\times 60=15$) minutes.

Problem (7): To reach a park located south of his jogging path, Henry runs along a 15-kilometer route. If he completes the journey in 1.5 hours, determine his speed and velocity.

Solution:  Henry travels his route to the park without changing direction along a straight line. Therefore, the total distance traveled in one direction equals the displacement, i.e, \[\text{distance traveled}=\Delta x=15\,\rm km\]Velocity is displacement divided by the time of travel \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{15\,\rm km}{1.5\,\rm h} \\\\ &=\boxed{10\,\rm km/h}\end{align*} and by definition, its average speed is \begin{align*} \text{speed}&=\frac{\text{distance covered}}{\text{time interval}}\\\\&=\frac{15\,\rm km}{1.5\,\rm h}\\\\&=\boxed{10\,\rm km/h}\end{align*} Thus, Henry's velocity is $10\,\rm km/h$ to the south, and its speed is $10\,\rm km/h$. As you can see, speed is simply a positive number, with units but velocity specifies the direction in which the object is moving. 

Problem (8): In 15 seconds, a football player covers the distance from his team's goal line to the opposing team's goal line and back to the midway point of the field having 100-yard-length. Find, (a) his average speed, and (b) the magnitude of the average velocity.

Solution:  The total length of the football field is $100$ yards or in meters, $L=91.44\,\rm m$. Going from one goal's line to the other and back to the midpoint of the field takes $15\,\rm s$ and covers a distance of $D=100+50=150\,\rm yd$. 

Distance divided by the time of travel gets the average speed, \[\text{speed}=\frac{150\times 0.91}{15}=9.1\,\rm m/s\] To find the average velocity, we must find the displacement of the player between the initial and final points. 

The initial point is her own goal line and her final position is the midpoint of the field, so she has displaced a distance of $\Delta x=50\,\rm yd$ or $\Delta x=50\times 0.91=45.5\,\rm m$. Therefore, her velocity is calculated as follows \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time elapsed}} \\\\ &=\frac{45.5\,\rm m}{15\,\rm s} \\\\&=\boxed{3.03\quad \rm m/s}\end{align*} Contrary to the previous problem, here the motion is not in one direction, hence, the displacement is not equal to the distance traveled. Accordingly, the average speed is not equal to the magnitude of the average velocity.

Problem (9): You begin at a pillar and run towards the east (the positive $x$ direction) for $250\,\rm m$ at an average speed of $5\,\rm m/s$. After that, you run towards the west for $300\,\rm m$ at an average speed of $4\,\rm m/s$ until you reach a post. Calculate (a) your average speed from pillar to post, and (b) your average velocity from pillar to post. 

Solution : First, you traveled a distance of $L_1=250\,\rm m$ toward east (or $+x$ direction) at $5\,\rm m/s$. Time of travel in this route is obtained as follows \begin{align*} t_1&=\frac{L_1}{v_1}\\\\ &=\frac{250}{5}\\\\&=50\,\rm s\end{align*} Likewise, traveling a distance of $L_2=300\,\rm m$ at $v_2=4\,\rm m/s$ takes \[t_2=\frac{300}{4}=75\,\rm s\]  (a) Average speed is defined as the distance traveled (or path length) divided by the total time of travel \begin{align*} v&=\frac{\text{path length}}{\text{time of travel}} \\\\ &=\frac{L_1+L_2}{t_1+t_2}\\\\&=\frac{250+300}{50+75} \\\\&=4.4\,\rm m/s\end{align*} Therefore, you travel between these two pillars in $125\,\rm s$ and with an average speed of $4.4\,\rm m/s$. 

(b) Average velocity requires finding the displacement between those two points. In the first case, you move $250\,\rm m$ toward $+x$ direction, i.e., $L_1=+250\,\rm m$. Similarly, on the way back, you move $300\,\rm m$ toward the west ($-x$ direction) or $L_2=-300\,\rm m$. Adding these two gives us the total displacement between the initial point and the final point, \begin{align*} L&=L_1+L_2 \\\\&=(+250)+(-300) \\\\ &=-50\,\rm m\end{align*} The minus sign indicates that you are generally displaced toward the west. 

Finally, the average velocity is obtained as follows: \begin{align*} \text{average velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{-50}{125} \\\\&=-0.4\,\rm m/s\end{align*} A negative average velocity indicating motion to the left along the $x$-axis. 

This speed problem better makes it clear to us the difference between average speed and average speed. Unlike average speed, which is always a positive number, the average velocity in a straight line can be either positive or negative. 

Problem (10): What is the average speed for the round trip of a car moving uphill at 40 km/h and then back downhill at 60 km/h? 

Solution : Assuming the length of the hill to be $L$, the total distance traveled during this round trip is $2L$ since $L_{up}=L_{down}=L$. However, the time taken for going uphill and downhill was not provided. We can write them in terms of the hill's length $L$ as $t=\frac L v$. 

Applying the definition of average speed gives us \begin{align*} v&=\frac{\text{distance traveled}}{\text{total time}} \\\\ &=\frac{L_{up}+L_{down}}{t_{up}+t_{down}} \\\\ &=\cfrac{2L}{\cfrac{L}{v_{up}}+\cfrac{L}{v_{down}}} \end{align*} By reorganizing this expression, we obtain a formula that is useful for solving similar problems in the AP Physics 1 exams. \[\text{average speed}=\frac{2v_{up} \times v_{down}}{v_{up}+v_{down}}\] Substituting the numerical values into this, yields \begin{align*} v&=\frac{2(40\times 60)}{40+60} \\\\ &=\boxed{48\,\rm m/s}\end{align*} What if we were asked for the average velocity instead? During this round trip, the car returns to its original position, and thus its displacement, which defines the average velocity, is zero. Therefore, \[\text{average velocity}=0\,\rm m/s\]

Acceleration Problems

Problem (9): A car moves from rest to a speed of $45\,\rm m/s$ in a time interval of $15\,\rm s$. At what rate does the car accelerate? 

Solution : The car is initially at rest, $v_1=0$, and finally reaches $v_2=45\,\rm m/s$ in a time interval $\Delta t=15\,\rm s$. Average acceleration is the change in velocity, $\Delta v=v_2-v_1$, divided by the elapsed time $\Delta t$, so \[\bar{a}=\frac{45-0}{15}=\boxed{3\,\rm m/s^2} \] 

Problem (10): A car moving at a velocity of $15\,{\rm m/s}$, uniformly slows down. It comes to a complete stop in $10\,{\rm s}$. What is its acceleration?

Solution:  Let the car's uniform velocity be $v_1$ and its final velocity $v_2=0$.   Average acceleration is the difference in velocities divided by the time taken, so we have: \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{0-15}{10}\\\\ &=\boxed{-1.5\,{\rm m/s^2}}\end{align*}The minus sign indicates the direction of the acceleration vector, which is toward the $-x$ direction.

Problem (11): A car moves from rest to a speed of $72\,{\rm km/h}$ in $4\,{\rm s}$. Find the acceleration of the car.

Solution: Known: $v_1=0$, $v_2=72\,{\rm km/h}$, $\Delta t=4\,{\rm s}$.  Average acceleration is defined as the difference in velocities divided by the time interval between those points \begin{align*}\bar{a}&=\frac{v_2-v_1}{t_2-t_1}\\\\&=\frac{20-0}{4}\\\\&=5\,{\rm m/s^2}\end{align*} In above, we converted $\rm km/h$ to the SI unit of velocity ($\rm m/s$) as \[1\,\frac{km}{h}=\frac {1000\,m}{3600\,s}=\frac{10}{36}\, \rm m/s\] so we get \[72\,\rm km/h=72\times \frac{10}{36}=20\,\rm m/s\] 

Problem (12): A race car accelerates from an initial velocity of $v_i=10\,{\rm m/s}$ to a final velocity of $v_f = 30\,{\rm m/s}$ in a time interval of $2\,{\rm s}$. Determine its average acceleration.

Solution:  A change in the velocity of an object $\Delta v$ over a time interval $\Delta t$ is defined as an average acceleration. Known: $v_i=10\,{\rm m/s}$, $v_f = 30\,{\rm m/s}$, $\Delta t=2\,{\rm s}$. Applying definition of average acceleration, we get \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\&=\frac{30-10}{2}\\&=10\,{\rm m/s^2}\end{align*}

Problem (13): A motorcycle starts its trip along a straight line with a velocity of $10\,{\rm m/s}$ and ends with $20\,{\rm m/s}$ in the opposite direction in a time interval of $2\,{\rm s}$. What is the average acceleration of the car?

Solution:  Known: $v_i=10\,{\rm m/s}$, $v_f=-20\,{\rm m/s}$, $\Delta t=2\,{\rm s}$, $\bar{a}=?$. Using average acceleration definition we have \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\\\&=\frac{(-20)-10}{2}\\\\ &=\boxed{-15\,{\rm m/s^2}}\end{align*}Recall that in the definition above, velocities are vector quantities. The final velocity is in the opposite direction from the initial velocity so a negative must be included.

Problem (14): A ball is thrown vertically up into the air by a boy. After $4$ seconds, it reaches the highest point of its path. How fast does the ball leave the boy's hand?

Solution : At the highest point, the ball has zero speed, $v_2=0$. It takes the ball $4\,\rm s$ to reach that point. In this problem, our unknown is the initial speed of the ball, $v_1=?$. Here, the ball accelerates at a constant rate of $g=-9.8\,\rm m/s^2$ in the presence of gravity.

When the ball is tossed upward, the only external force that acts on it is the gravity force. 

Using the average acceleration formula $\bar{a}=\frac{\Delta v}{\Delta t}$ and substituting the numerical values into this, we will have \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{0-v_1}{4} \\\\ \Rightarrow \boxed{v_1=39.2\,\rm m/s} \end{gather*} Note that $\Delta v=v_2-v_1$. 

Problem (15): A child drops crumpled paper from a window. The paper hit the ground in $3\,\rm s$. What is the velocity of the crumpled paper just before it strikes the ground? 

Solution : The crumpled paper is initially in the child's hand, so $v_1=0$. Let its speed just before striking be $v_2$. In this case, we have an object accelerating down in the presence of gravitational force at a constant rate of $g=-9.8\,\rm m/s^2$. Using the definition of average acceleration, we can find $v_2$ as below \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{v_2-0}{3} \\\\ \Rightarrow v_2=3\times (-9.8)=\boxed{-29.4\,\rm m/s} \end{gather*} The negative shows us that the velocity must be downward, as expected!

Problem (16): A car travels along the $x$-axis for $4\,{\rm s}$ at an average velocity of $10\,{\rm m/s}$ and $2\,{\rm s}$ with an average velocity of $30\,{\rm m/s}$ and finally $4\,{\rm s}$ with an average velocity $25\,{\rm m/s}$. What is its average velocity across the whole path?

Solution: There are three different parts with different average velocities. Assume each trip is done in one dimension without changing direction. Thus, displacements associated with each segment are the same as the distance traveled in that direction and is calculated as below: \begin{align*}\Delta x_1&=v_1\,\Delta t_1\\&=10\times 4=40\,{\rm m}\\ \\ \Delta x_2&=v_2\,\Delta t_2\\&=30\times 2=60\,{\rm m}\\ \\ \Delta x_3&=v_3\,\Delta t_3\\&=25\times 4=100\,{\rm m}\end{align*}Now use the definition of average velocity, $\bar{v}=\frac{\Delta x_{tot}}{\Delta t_{tot}}$, to find it over the whole path\begin{align*}\bar{v}&=\frac{\Delta x_{tot}}{\Delta t_{tot}}\\ \\&=\frac{\Delta x_1+\Delta x_2+\Delta x_3}{\Delta t_1+\Delta t_2+\Delta t_3}\\ \\&=\frac{40+60+100}{4+2+4}\\ \\ &=\boxed{20\,{\rm m/s}}\end{align*}

Problem (17): An object moving along a straight-line path. It travels with an average velocity $2\,{\rm m/s}$ for $20\,{\rm s}$ and $12\,{\rm m/s}$ for $t$ seconds. If the total average velocity across the whole path is $10\,{\rm m/s}$, then find the unknown time $t$.

Solution: In this velocity problem, the whole path $\Delta x$ is divided into two parts $\Delta x_1$ and $\Delta x_2$ with different average velocities and times elapsed, so the total average velocity across the whole path is obtained as \begin{align*}\bar{v}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{\Delta x_1+\Delta x_2}{\Delta t_1+\Delta t_2}\\\\&=\frac{\bar{v}_1\,t_1+\bar{v}_2\,t_2}{t_1+t_2}\\\\10&=\frac{2\times 20+12\times t}{20+t}\\\Rightarrow t&=80\,{\rm s}\end{align*}

Note : whenever a moving object, covers distances $x_1,x_2,x_3,\cdots$ in $t_1,t_2,t_3,\cdots$ with constant or average velocities $v_1,v_2,v_3,\cdots$ along a straight-line without changing its direction, then its total average velocity across the whole path is obtained by one of the following formulas

• Distances and times are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{t_1+t_2+t_3+\cdots}\]
• Velocities and times are known: \[\bar{v}=\frac{v_1\,t_1+v_2\,t_2+v_3\,t_3+\cdots}{t_1+t_2+t_3+\cdots}\]
• Distances and velocities are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{\frac{x_1}{v_1}+\frac{x_2}{v_2}+\frac{x_3}{v_3}+\cdots}\]

Problem (18): A car travels one-fourth of its path with a constant velocity of $10\,{\rm m/s}$, and the remaining with a constant velocity of $v_2$. If the total average velocity across the whole path is $16\,{\rm m/s}$, then find the $v_2$?

Solution: This is the third case of the preceding note. Let the length of the path be $L$ so \begin{align*}\bar{v}&=\frac{x_1+x_2}{\frac{x_1}{v_1}+\frac{x_2}{v_2}}\\\\16&=\frac{\frac 14\,L+\frac 34\,L}{\frac{\frac 14\,L}{10}+\frac{\frac 34\,L}{v_2}}\\\\\Rightarrow v_2&=20\,{\rm m/s}\end{align*}

Problem (19): An object moves along a straight-line path. It travels for $t_1$ seconds with an average velocity $50\,{\rm m/s}$ and $t_2$ seconds with a constant velocity of $25\,{\rm m/s}$. If the total average velocity across the whole path is $30\,{\rm m/s}$, then find the ratio $\frac{t_2}{t_1}$?

Solution: the velocities and times are known, so we have \begin{align*}\bar{v}&=\frac{v_1\,t_1+v_2\,t_2}{t_1+t_2}\\\\30&=\frac{50\,t_1+25\,t_2}{t_1+t_2}\\\\ \Rightarrow \frac{t_2}{t_1}&=4\end{align*} 

Read more related articles:  

Kinematics Equations: Problems and Solutions

Position vs. Time Graphs

Velocity vs. Time Graphs

In the following section, some sample AP Physics 1 problems on acceleration are provided.

Problem (20): An object moves with constant acceleration along a straight line. If its velocity at instant of $t_1 = 3\,{\rm s}$ is $10\,{\rm m/s}$ and at the moment of $t_2 = 8\,{\rm s}$ is $20\,{\rm m/s}$, then what is its initial speed?

Solution: Let the initial speed at time $t=0$ be $v_0$. Now apply average acceleration definition in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ and equate them.\begin{align*}\text{average acceleration}\ \bar{a}&=\frac{\Delta v}{\Delta t}\\\\\frac{v_1 - v_0}{t_1-t_0}&=\frac{v_2-v_0}{t_2-t_0}\\\\ \frac{10-v_0}{3-0}&=\frac{20-v_0}{8-0}\\\\ \Rightarrow v_0 &=4\,{\rm m/s}\end{align*} In the above, $v_1$ and $v_2$ are the velocities at moments $t_1$ and $t_2$, respectively. 

Problem (21): For $10\,{\rm s}$, the velocity of a car that travels with a constant acceleration, changes from $10\,{\rm m/s}$ to $30\,{\rm m/s}$. How far does the car travel?

Solution: Known: $\Delta t=10\,{\rm s}$, $v_1=10\,{\rm m/s}$ and $v_2=30\,{\rm m/s}$. 

Method (I) Without computing the acceleration: Recall that in the case of constant acceleration, we have the following kinematic equations for average velocity and displacement:\begin{align*}\text{average velocity}:\,\bar{v}&=\frac{v_1+v_2}{2}\\\text{displacement}:\,\Delta x&=\frac{v_1+v_2}{2}\times \Delta t\\\end{align*}where $v_1$ and $v_2$ are the velocities in a given time interval. Now we have \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\\&=\frac{10+30}{2}\times 10\\&=200\,{\rm m}\end{align*}

Method (II) with computing acceleration: Using the definition of average acceleration, first determine it as below \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{30-10}{10}\\\\&=2\,{\rm m/s^2}\end{align*} Since the velocities at the initial and final points of the problem are given so use the below time-independent kinematic equation to find the required displacement \begin{align*} v_2^{2}-v_1^{2}&=2\,a\Delta x\\\\ (30)^{2}-(10)^{2}&=2(2)\,\Delta x\\\\ \Rightarrow \Delta x&=\boxed{200\,{\rm m}}\end{align*}

Problem (22): A car travels along a straight line with uniform acceleration. If its velocity at the instant of $t_1=2\,{\rm s}$ is $36\,{\rm km/s}$ and at the moment $t_2=6\,{\rm s}$ is $72\,{\rm km/h}$, then find its initial velocity (at $t_0=0$)?

Solution: Use the equality of definition of average acceleration $a=\frac{v_f-v_i}{t_f-t_i}$ in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ to find the initial velocity as below \begin{align*}\frac{v_2-v_0}{t_2-t_0}&=\frac{v_1-v_0}{t_1-t_0}\\\\ \frac{20-v_0}{6-0}&=\frac{10-v_0}{2-0}\\\\ \Rightarrow v_0&=\boxed{5\,{\rm m/s}}\end{align*}

All these kinematic problems on speed, velocity, and acceleration are easily solved by choosing an appropriate kinematic equation. Keep in mind that these motion problems in one dimension are of the uniform or constant acceleration type. Projectiles are also another type of motion in two dimensions with constant acceleration.

Author:   Dr. Ali Nemati

Date Published: 9/6/2020

Updated: Jun 28,  2023

© 2015 All rights reserved. by Physexams.com

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Velocity is the rate at which an object changes position with time. An object is displaced when it changes its position. The amount of displacement over the time in which the displacement occurred gives the velocity. It is a vector quantity that has both magnitude and direction.

The importance of velocity is that it can give you an estimated time to go from one point to another. Suppose you are traveling from place A to place B. Velocity tells you how long it will take to arrive at your destination.

How to Calculate Velocity

Velocity can be calculated by measuring the object’s displacement over the time taken to displace it.

In vectorial notation, velocity is given by

The SI unit of velocity is Newtons per second of N/s. The cgs unit is ergs per second or ergs/s. Other units include miler per hour or mph, kilometers per hour or kph, and feet per second or fps.

• A car traveling on the highway at 60 mph toward the West
• A biker riding on the road at 35 mph toward the southwest
• A bicyclist heading northwards at 14 mph
• An airplane flying eastwards at 300 mph
• A rocket launched into the sky in the south-east direction
• A train traveling in the south direction at 80 kph
• A runner runs 420 m in 70 seconds

Average and Instantaneous Velocity

An object can move from one position to another in several steps. The average velocity is simply the change in position over the time interval within which change takes place.

v avg is the average velocity

x f is the final velocity

x i is the initial velocity

t f is the final time

t i is the initial time

If the starting time is zero, then t i = 0.

The instantaneous (v ins. ) velocity is the velocity at a particular instant of time. It can be obtained by taking the limit of the above expression in the limit Δt → 0.

Speed vs. Velocity

Both speed and velocity measure rate. Both of them depict how fast an object is moving. However, there is an essential difference. While speed tells us the magnitude of the rate at which the object moves, it does not say anything about the direction. On the other hand, velocity considers the direction of motion.

Let us take an example to illustrate this. When we say a car is moving on the road at 35 mph, we specify the magnitude but not the direction. However, when we say a car is moving on the road eastwards at 35 mph, we specify both the magnitude and direction. The first sentence refers to the speed, and the second refers to the velocity. Therefore, speed is a scalar quantity, and velocity is a vector quantity.

The formula for speed is the distance traveled over the time taken to travel that distance.

Velocity vs. Time Graph

The following graph shows how velocity changes with time. At some parts of the graph, the velocity increases linearly with time. The quotient of velocity and time is called acceleration . When the velocity decreases with time, the phenomenon is known as deceleration. Regions of constant velocity are also indicated on the graph.

Solved Problems

Problem 1. Suppose a runner moves along the x-direction over some time.During a 5.00 s interval, the runner’s position changes from x 1 = 75 m to x 2 = 35 m. What was the runner’s average velocity?

Given x 1 = 75 m, x 2 = 35 m, Δt = 5 s

The average velocity is given by

v avg = (x 2 – x 1 )/Δt = (35 m – 75 m)/5s = – 8 m/s

Therefore, the runner is running in the negative x-direction.

Problem 2 . A commuter train travels from New York to Philadelphia in 1 hour and 25 minutes and from Philadelphia to New York in 1 hour and 35 minutes. The distance between the two stations is approximately 96 miles. What is (a) the average velocity of the train and (b) the average speed of the train in m/s? (1 mile = 1.6 km)

a) The average velocity of the train is zero since the displacement is zero. (The train returns to New York).

d = 96 miles = 96 x 1.6 = 153.6 km = 153.6 x 10 3 m

t 1 = 1 hr 25 mins = 60 mins + 25 mins = 85 mins = 85 x 60 = 5100 s

Therefore, v 1 = 153.6 x 10 3 m / 5100 s = 30.2 m/s

t 2 = 1 hr 35 mins = 60 mins + 35 mins = 95 mins = 95 x 60 = 5700 s

Therefore, v 2 = 153.6 x 10 3 m / 5700 s = 26.9 m/s

The average speed is (30.2 m/s + 26.9 m/s)/2 = 28.55 m/s

• Speed vs. Velocity – Physicsclassroom.com
• What is Velocity? – Khanacademy.org
• What is Velocity in Physics? – Thoughtco.com
• Speed and Velocity – Physics.info
• Instantaneous Velocity and Speed – Pressbooks.online.ucf.edu
• Time, Velocity, and Speed – Phys.libretexts.org
• Velocity – Hyperphysics.phy-astr.gsu.edu

Article was last reviewed on Friday, July 28, 2023

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6.3: Solving Problems with Newton's Laws (Part 2)

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Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

Example 6.6: What Force Must a Soccer Player Exert to Reach Top Speed?

A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. The solutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we do not need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns; apply Newton’s second law; and check to see whether the answer is reasonable.

• We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is \(\Delta\)v = 8.00 m/s . We are given the elapsed time, so \(\Delta\)t = 2.50 s. The unknown is acceleration, which can be found from its definition: $$a = \frac{\Delta v}{\Delta t} \ldotp$$Substituting the known values yields $$a = \frac{8.00\; m/s}{2.50\; s} = 3.20\; m/s^{2} \ldotp$$
• Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, $$F_{net} = ma \ldotp$$Substituting the known values of m and a gives $$F_{net} = (70.0\; kg)(3.20\; m/s^{2}) = 224\; N \ldotp$$

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Significance

This example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in the problem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.

Exercise 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7: What Force Acts on a Model Helicopter?

A 1.50-kg model helicopter has a velocity of 5.00 \(\hat{j}\) m/s at t = 0. It is accelerated at a constant rate for two seconds (2.00 s) after which it has a velocity of (6.00 \(\hat{i}\) + 12.00 \(\hat{j}\)) m/s. What is the magnitude of the resultant force acting on the helicopter during this time interval?

We can easily set up a coordinate system in which the x-axis (\(\hat{i}\) direction) is horizontal, and the y-axis (\(\hat{j}\) direction) is vertical. We know that \(\Delta\)t = 2.00s and \(\Delta\)v = (6.00 \(\hat{i}\) + 12.00 \(\hat{j}\) m/s) − (5.00 \(\hat{j}\) m/s). From this, we can calculate the acceleration by the definition; we can then apply Newton’s second law.

\[a = \frac{\Delta v}{\Delta t} = \frac{(6.00 \hat{i} + 12.00 \hat{j}\; m/s) - (5.00 \hat{j}\; m/s)}{2.00\; s} = 3.00 \hat{i} + 3.50 \hat{j}\; m/s^{2}$$ $$\sum \vec{F} = m \vec{a} = (1.50\; kg)(3.00 \hat{i} + 3.50 \hat{j}\; m/s^{2}) = 4.50 \hat{i} + 5.25 \hat{j}\; N \ldotp\]

The magnitude of the force is now easily found:

\[F = \sqrt{(4.50\; N)^{2} + (5.25\; N)^{2}} = 6.91\; N \ldotp\]

The original problem was stated in terms of \(\hat{i}\) − \(\hat{j}\) vector components, so we used vector methods. Compare this example with the previous example.

Exercise 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8: Baggage Tractor

Figure \(\PageIndex{7}\)(a) shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg, while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force acting for a brief period of time accelerates the system from rest and acts for 3.00 s. (a) If this driving force is given by F = (820.0t) N, find the speed after 3.00 seconds. (b) What is the horizontal force acting on the connecting cable between the tractor and cart A at this instant?

A free-body diagram shows the driving force of the tractor, which gives the system its acceleration. We only need to consider motion in the horizontal direction. The vertical forces balance each other and it is not necessary to consider them. For part b, we make use of a free-body diagram of the tractor alone to determine the force between it and cart A. This exposes the coupling force \(\vec{T}\), which is our objective.

• $$\sum F_{x} = m_{system} a_{x}\; and\; \sum F_{x} = 820.0t,$$so $$820.0t = (650.0 + 250.0 + 150.0)a$$ $$a = 0.7809t \ldotp$$Since acceleration is a function of time, we can determine the velocity of the tractor by using a = \(\frac{dv}{dt}\) with the initial condition that v 0 = 0 at t = 0. We integrate from t = 0 to t = 3: $$\begin{split}dv & = adt \\ \int_{0}^{3} dv & = \int_{0}^{3.00} adt = \int_{0}^{3.00} 0.7809tdt \\ v & = 0.3905t^{2} \big]_{0}^{3.00} = 3.51\; m/s \ldotp \end{split}$$
• Refer to the free-body diagram in Figure \(\PageIndex{7}\)(b) $$\begin{split} \sum F_{x} & = m_{tractor} a_{x} \\ 820.0t - T & = m_{tractor} (0.7805)t \\ (820.0)(3.00) - T & = (650.0)(0.7805)(3.00) \\ T & = 938\; N \ldotp \end{split}$$

Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving Figure \(\PageIndex{7}\)(a), whereas only the mass of the truck (since it supplied the force) was of use in Figure \(\PageIndex{7}\)(b).

Recall that v = \(\frac{ds}{dt}\) and a = \(\frac{dv}{dt}\). If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have dt = \(\frac{ds}{v}\) and dt = \(\frac{dv}{a}\). Now, equating these expressions, we have \(\frac{ds}{v}\) = \(\frac{dv}{a}\). We can rearrange this to obtain a ds = v dv.

Example 6.9: Motion of a Projectile Fired Vertically

A 10.0-kg mortar shell is fired vertically upward from the ground, with an initial velocity of 50.0 m/s (see Figure \(\PageIndex{8}\)). Determine the maximum height it will travel if atmospheric resistance is measured as F D = (0.0100 v 2 ) N, where v is the speed at any instant.

The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

Initially, y 0 = 0 and v 0 = 50.0 m/s. At the maximum height y = h, v = 0. The free-body diagram shows F D to act downward, because it slows the upward motion of the mortar shell. Thus, we can write

\[\begin{split} \sum F_{y} & = ma_{y} \\ -F_{D} - w & = ma_{y} \\ -0.0100 v^{2} - 98.0 & = 10.0 a \\ a & = -0.00100 v^{2} - 9.80 \ldotp \end{split}\]

The acceleration depends on v and is therefore variable. Since a = f(v), we can relate a to v using the rearrangement described above,

\[a ds = v dv \ldotp\]

We replace ds with dy because we are dealing with the vertical direction,

\[\begin{split} ady & = vdv \\ (−0.00100v^{2} − 9.80)dy & = vdv \ldotp \end{split}\]

We now separate the variables (v’s and dv’s on one side; dy on the other):

\[\begin{split} \int_{0}^{h} dy & = \int_{50.0}^{0} \frac{vdv}{(-0.00100 v^{2} - 9.80)} \\ & = - \int_{50.0}^{0} \frac{vdv}{(-0.00100 v^{2} + 9.80)} \\ & = (-5 \times 10^{3}) \ln(0.00100v^{2} + 9.80) \Big|_{50.0}^{0} \ldotp \end{split}\]

Thus, h = 114 m.

Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

Exercise 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

Speed and Velocity

Practice problem 1.

Average speed

This problem is deceptively easy. Averaging is taught in elementary school, which makes this an elementary problem. Right?

The add-and-divide method of averaging only works when averaging items of equal weight. The average age of the students in a classroom is the sum of their ages divided by the number of students only because each student is considered to have the same significance (a student, is a student, is a student, … ). In this problem, however, the two segments of the walk are significantly different. The second "half" was actually the majority of the walk. It carries more weight than the shorter first "half". Thus, the add-and-divide method won't work.

Let's return to our definition. Since speed is the rate of change of distance with time, we'll need both the distance traveled and the time it took to complete the walk. After we determine both of these numbers, the rest is easy.

Look closely at the calculations on the right side. Notice that the formula contains ∆ (delta) symbols and yet I added the distances in the numerator and the times in the denominator. That's because ∆ doesn't mean difference, it means change. During the walk my position didn't change from 6.0 km to 10 km, it changed first by 6.0 km and then by 10 km for a total change of 16 km.

Average velocity

Velocity is the rate of change of displacement with respect to time. Velocity is a vector, which means the problem should be solved graphically. Draw an arrow pointing toward the top of the page (north). Label it 6 km. Draw another arrow to the left (west) starting from the previous one (arranged head to tail). Make it slightly longer and label it 10 km. Draw a third arrow starting on the tail of the first and ending on the head of the second. Since north and west are at right angles to one another, the resultant displacement is the hypotenuse of a right triangle. Use Pythagorean theorem to find its magnitude and tangent to find its direction.

Divide displacement by time to get velocity.

practice problem 2

Notice that no numbers are stated in this problem. When a numerical value is needed to solve a problem and that number is not given, it could mean one of several things.

• Look it up! It may appear somewhere in the textbook you are using — on the inside covers, in an appendix , or in the text of the chapter you are currently working on. It may be found in the reference table that some teachers distribute. Standardized exams usually also have their own reference table .
• Know it! Some numbers are numbers that you should just know. In this problem, there is one relevant number that nearly everyone knows. You may also be expected to memorize certain numbers by an instructor or professor.
• Calculate it! Maybe there's a way to find the number you need to know using other numbers given in the problem.
• Forget about it! Maybe you don't really need the number you think you do. Maybe you are on the wrong track. Especially under test conditions, it is highly unlikely that you could be asked a question that requires a numerical value that you can not find, do not know, or can not calculate. Perhaps there is another method to solve this problem.

In order to calculate speed, you will need distance and time. What distance does a point on the equator move in a convenient period of time? Well, I hope you know that the Earth rotates once on its axis every day. You should also know how to calculate the length of a day in seconds. (A day is the period of the Earth's rotation, for which an upper case T is the symbol.) During a day, a point on the Earth's equator would have traveled a distance equal to the circumference of the Earth. The radius of the Earth is a number commonly found in textbooks and on reference tables. The problem can now be solved.

That's about 75% faster than the speed of sound in air at room temperature (343 m/s). An interesting problem to be dealt with later is that if the Earth is spinning so rapidly, how come things on the equator don't fly off into space?

practice problem 3

Astronomical distances are so large that using meters is cumbersome. For really large distances the light year is the best unit. A light year is the distance that light would travel in one year in a vacuum. Since the speed of light is fast, and a year is long, the light year is a pretty good unit for astronomy. One light year is about ten petameters (ten quadrillion meters) as the following calculation shows.

Start with the definition of speed and solve it for distance. The traditional symbol for the speed of light is c from the Latin word for swiftness — celeritas .

Numbers in, answer out.

Since both the speed of light and the year have exactly defined values in the International System of Units, the light year can be stated with an unnecessarily large number of significant digits.

Some distances in light years are provided below.

• The distance to Proxima Centauri (the star nearest the Sun) is 4.3 light years .
• The diameter of the Milky Way (the collection of stars that includes the Sun and all the stars visible to the naked eye) is about 100,000 light years .
• The distance to Andromeda (the nearest galaxy outside the Milky Way) is about 2 million light  years.
• The distance to the edge of the universe (the observable part of it) is 13.77 billion light years .

practice problem 4

Everyone should know (or at least realize after a bit of thought) that there are…

∆ t  =  60 × 60 = 3,600 s

in an hour. Many Americans who are fans of track and field know that four laps around a 400 m outdoor track is almost one mile.

∆ s  =  1 mile ≈ 4 × 400 m = 1,600 m = 1.6 km

More precisely… actually, most precisely… actually, exactly by definition…

∆ s  =  1 mile = 1,609.344 m = 1.609344 km

The first answer…

For comparison, the speed limit on many Canadian highways is 100 km/h.

The second answer…

You should note that the number with International units is a little bit less than half the value of the number with British-American units. I've gotten used to mph, but I have to be conversant in m/s for my job. A good rule of thumb for comparing speeds is to…

divide by 2 and subtract a little when converting from mph to m/s

multiply by 2 and add a little when converting from m/s to mph.

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Speed, Velocity, and Acceleration Problems

• by Abnurlion
• December 26, 2022 November 2, 2023

What is Speed?

Definition of Speed : Speed is the rate of change of distance with time . Speed is different from velocity because it’s not in a specified direction. In this article, you will learn how to solve speed, velocity, and acceleration problems.

Additionally, you need to know that speed is a scalar quantity and we can write its symbol as S. The formula for calculating the speed of an object is:

Speed, S = Distance (d) / Time (t)

Thus, s = d/t

Note: In most cases, we also use S as a symbol for distance.

The S.I unit for speed is meter per second (m/s) or ms -1

Non-Uniform or Average Speed : This is a non-steady distance covered by an object at a particular period of time. We can also define non-uniform speed as the type of distance that an object covered at an equal interval of time.

The formula for calculating non-uniform speed is

Average speed = Total distance covered by the object / Total time taken

Actual speed: This is also known as the instantaneous speed of an object which is the distance covered by an object over a short interval of time.

You may also like to read:

How to Find Displacement in Physics

also How to Find a Position in Physics

and How to Calculate Bearing in Physics

What is Velocity?

Definition of Velocity: Velocity is the rate of displacement with time. Velocity is the speed of an object in a specified direction. The unit of velocity is the same as that of speed which is meter per second (ms -1 ). We use V as a symbol for velocity.

Note: We often use U to indicate initial speed, and V to indicate final speed.

The formula for calculating Velocity (V) = displacement (S) / time (t)

The difference between velocity and speed is the presence of displacement and distance respectively. Because displacement is a measure of separation in a specified direction, while distance is not in a specified direction. Velocity is a vector quantity.

Uniform Velocity

Definition of uniform velocity: The rate of change of displacement is constant no matter how small the time interval may be. Also, uniform velocity is the distance covered by an object in a specified direction in an equal time interval.

The formula for uniform velocity = Total displacement / Total time taken

What is Acceleration?

Definition of Acceleration: Acceleration is the rate of change of velocity with time. Acceleration is measured in meters per second square (ms -2 ). The symbol for acceleration is a. Acceleration is also a vector quantity.

The formula for acceleration , a = change in velocity (v)/time taken (t)

Thus, a = v/t

We can also write acceleration as

a = change in velocity/time = ΔV/Δt = (v – u)/t

[where v = final velocity, u = initial velocity, and t = time taken]

Uniform Acceleration

In the case of uniform acceleration , the rate of change of velocity with time is constant.

Average velocity of the object = (Initial velocity + final velocity)/2

Average velocity = (v + u)/2

What is Retardation?

Retardation is the decreasing rate of change in velocity moved, covered, or traveled by an object.

The formula for calculating retardation is

Retardation = Change in a decrease in velocity/time taken

Equations of Motion

You can also apply the following equations of motion to calculate the speed, velocity, and acceleration of the body:

• s = [ (v + u)/2 ]t
• v 2 = u 2 + 2as

s = ut + (1/2)at 2

Where v = final velocity, u = initial velocity, t = time, a = acceleration, and s = distance

How to Calculate Maximum height

What are Prefixes in Physics

How to Calculate Dimension in Physics

Solved Problems of Speed, Velocity, and Acceleration

Here are solved problems to help you understand how to calculate speed, velocity, and acceleration:

A train moves at a speed of 54km/h for a one-quarter minute. Find the distance travelled by train.

From the question above

Speed = 54 km/h = (54 x 1000)/(60 x 60) = 54,000/3,600 = 15 m/s

Time = one quarter minute = 1/4 minute = (1/4) x 60 = 15 seconds

Since we have

speed = distance/time

After cross-multiplication, we will now have

Distance = speed x time

We can now insert our data into the above expression

Distance = 15 m/s x 15 s = 225 m

Therefore, the distance travelled by train is 225 meters.

A car travelled a distance of 5km in 50 seconds. Find the speed in meters per second.

Distance = 5km = 5 x 1000m = 5,000m

Time = 50 seconds

and the formula for speed

speed = distance/time = 5000/50 = 100m/s

A motorcycle starting from rest moves with a uniform acceleration until it attains a speed of 108 kilometres per hour after 15 seconds. Find its acceleration.

Initial velocity, u = 0 (because the motorcycle starts from rest)

Final velocity, v = speed = 108 km/h = (108 x 1000m) / (60 x 60s) = 108,000/3,600 = 30m/s

Time taken, t = 15 seconds

Therefore, we can now apply the formula that says

Acceleration = change in velocity/ time = (v-u)/t = (30-0)/15 =30/15 = 2ms -2

Therefore, the acceleration of the motorcycle is 2ms -2

A bus covers 50 kilometres in 1 hour. What is it is the average speed?

Total distance covered = 50 km = 50 x 1000m = 50,000m

Time taken = 1 hour = 1 x 60 x 60s =3,600s

Therefore, we can now calculate the average speed of the bus by substituting our data into the above formula

Average speed = 50,000/3,600 = 13.9 m/s

Therefore, the average speed of the bus is 13.9m/s or approximately 14 meters per second.

A car travels 80 km in 1 hour and then another 20 km in the next hour. Find the average velocity of the car.

Initial displacement of the car = 80km

Final displacement of the car = 20km

The total displacement of the car = initial displacement of the car + final displacement of the car

The total displacement of the car = 80km + 20km = 100km

The time for 80km is 1hr

And the time for 20km is 1hr

Total time taken = The time for 80km (1hr) + The time for 20km (1hr)

Total time taken = 1hr + 1hr = 2hrs

Now, to calculate the average velocity of the car, we apply the formula that says

Average velocity = total displacement/total time taken = 100km/2hrs = 50km/h

We can further convert the above answer into meters per second

Average velocity = 50km/h = (50 x 1000m)/(1 x 60 x 60s) = 50,000/3,600 = 13.9m/s or 14ms -1

Therefore, the average velocity of the car is 14ms -1

How to Conduct Physics Practical

and How to Calculate Velocity Ratio of an Inclined Plane

A body falls from the top of a tower 100 meters high and hits the ground in 5 seconds. Find its acceleration.

Displacement = 100m

Time = 5 seconds

and we can apply the formula for acceleration that says

acceleration, a = Displacement/time 2 = 100/5 2 = 100/25 = 4ms -2

Therefore the acceleration due to the gravity of the body is 4ms -2

Note: The acceleration due to the gravity of a body on the surface of the earth is constant (9.8ms -2 ), even though there may be a slight difference due to the mass and altitude of the body.

An object is thrown vertically upward at an initial velocity of 10ms -1 and reaches a maximum height of 50 meters. Find its initial upward acceleration.

Initial velocity, u = 10ms -1

Final velocity, v = 0

maximum height = displacement = 50m

Initial upward acceleration, a =?

When we apply the formula that says a = (v 2 – u 2 )/2d we will have

a = (0 – 10 2 )/(2 x 50) = -100/100 = -1 ms -2

Hence, since our acceleration is negative, we can now say that we are dealing with retardation or deceleration.

Therefore, the retardation is -1ms -2

Note: Retardation is the negative of acceleration, thus it is written in negative form.

A car is traveling at a velocity of 8ms -1 and experiences an acceleration of 5ms -2 . How far does it travel in 4 seconds?

Initial velocity, u = 8ms -1

acceleration, a = 5ms -2

Distance, s =?

time, t = 4s

We can apply the formula that says

s = 8 x 4 + (1/2) x 5 x 4 2

s = 32 + (1/2) x 80 = 32 + 40 = 72m

Therefore, the distance covered by the car in 4s is 72 meters.

A body is traveling at a velocity of 10m/s and experiences a deceleration of 5ms -2 . How long does it take the body to come to a complete stop?

Initial velocity, u = 10m/s

acceleration , a = retardation = -5ms -2

time, t = ?

We already know that acceleration, a = change in velocity/time

This implies that

Time, t = change in velocity/acceleration

t = (v – u)/t = (0-10)/-5 = -10/-5 = 2s

Therefore, the time it takes the car to stop is 2 seconds .

A body is traveling at a velocity of 20 m/s and has a mass of 10 kg. How much force is required to change its velocity by 10 m/s in 5 seconds?

Change in velocity, v =10 m/s

mass of the object, m = 10kg

time, t = 5s

We can apply newton’s second law of motion which says f = ma

and since a = change in velocity/time

we will have an acceleration equal to

a = 10/5 = 2ms -2

Therefore, to find the force, we can now say

f = ma = 10 x 2 = 20N

Therefore, the force that can help us to change the velocity by 10 m/s in 5 seconds is 20-Newton.

Drop a comment if there is anything you don’t understand about speed, velocity, and acceleration Problems.

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Classical mechanics.

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The equations of 1D Kinematics are very useful in many situations. While they may seem minimal and straightforward at first glance, a surprising amount of subtlety belies these equations. And the number of physical scenarios to which they can be applied is vast. These problems may not be groundbreaking advances in modern physics, but they do represent very tangible everyday experiences: cars on roads, balls thrown in the air, hockey pucks on ice, and countless more examples can be modeled with these three relatively simple equations.

Equation Review

1d kinematics problems: easy, 1d kinematics problems: medium.

The three fundamental equations of kinematics in one dimension are:

\[v = v_0 + at,\]

\[x = x_0 + v_0 t + \frac12 at^2,\]

\[v^2 = v_0^2 + 2a(x-x_0).\]

The first gives the change in velocity under a constant acceleration given a change in time, the second gives the change in position under a constant acceleration given a change in time, and the third gives the change in velocity under a constant acceleration given a change in distance.

Here, the subscript "0" always refers to "initial". So, \(v_0\) is the initial velocity, and \(x_0\) is the initial position. Letters with no subscript indicate the quantity value after some time, \(t\). So, in the first equation, \(v\) is the velocity of an object that began at velocity \(v_0\) and has moved with constant acceleration \(a\) for an amount of time \(t\).

Very often, rather than using the initial and final positions, we simply want to know the total change in position, the distance traveled. This change in position is always merely the initial position subtracted from the final position: \(x-x_0\), often called \(d\) for distance. In many problems, this simplifies things and makes it simpler to see what is being asked. With this change, the second and third equations are sometimes rewritten:

\[d = v_0 t + \frac12 at^2,\]

\[v^2 = v_0^2 + 2ad.\]

A ball is dropped from rest off a cliff of height \(100 \text{ m}\). Assuming gravity accelerates masses uniformly on Earth's surface at \(g = 9.8 \text{ m}/\text{s}^2\), how fast is the ball going when it hits the ground? How long does it take to hit the ground? Solution: The third kinematics equation gives the final speed as: \[v_f^2 = 2( 9.8 \text{ m}/\text{s}^2)(100 \text{ m}) \implies v_f \approx 44.3 \text{ m}/\text{s}.\] The first kinematical equation gives the time to accelerate up to this speed: \[t = \frac{v}{a} = \frac{44.3 \text{ m}/\text{s}}{9.8 \text{ m}/\text{s}^2} \approx 4.5 \text{ s}.\]

A soccer ball is kicked from rest at the penalty spot into the net \(11 \text{ m}\) away. It takes \(0.4 \text{ s}\) for the ball to hit the net. If the soccer ball does not accelerate after being kicked, how fast was it traveling immediately after being kicked? Solution: This is a straightforward application of the second equation of motion with \(a = 0\), i.e \(d = vt\): \[v = \frac{d}{t} = \frac{11 \text{ m}}{0.4 \text{ s}} = 27.5 \text{ m}/\text{s}.\]

A continuously accelerating car starts from rest as it zooms over a span of \(100 \text{ m}\). If the final velocity of the car is \(30 \text{ m}/\text{s}\), what is the acceleration of the car? Solution: Applying the third kinematical equation with \(v_0 = 0\), \[v^2 = 2ad \implies a = \frac{v^2}{2d} = \frac{900}{200} \text{ m}/\text{s}^2 = 4.5 \text{ m}/\text{s}^2.\]

A basketball is dropped from a height of \(10 \text{ m}\) above the surface of the moon, accelerating downwards at \(1.6 \text { m}/\text{s}^2\). How long does it take to hit the surface, in seconds to the nearest tenth?

A train traveling at \(40 \text{ m}/\text{s}\) is heading towards a station \(400 \text{ m}\) away. If the train must slow down with constant deceleration \(a\) into the station, how long does it take to come to a complete stop, in seconds? Answer to the nearest integer.

Sometimes kinematics problems require multiple steps of computation, which can make them more difficult. Below, some more challenging problems are explored.

A projectile is launched with speed \(v_0\) at an angle \(\theta\) to the horizontal and follows a trajectory under the influence of gravity. Find the range of the projectile. Solution: The projectile begins with velocity in the vertical direction of \(v_0 \sin \theta\). To reach the apex of its trajectory, where the projectile is at rest, thus requires a time: \[t = \frac{v_0 \sin \theta}{g}.\] The time that it takes to fall back to the ground is therefore double this time, \[t = \frac{2v_0 \sin \theta}{g}.\] The range is the total distance in the horizontal direction traveled during this time. This is just the velocity in the x-direction times the time: \[R = v_x t = v_0 \cos \theta t = \frac{2v_0^2 \sin \theta \cos \theta}{g} = \frac{v_0^2 \sin 2 \theta}{g}.\]

A package is dropped from a cargo plane which is traveling at an altitude of \(10000 \text{ m}\) with a horizontal velocity of \(250 \text{ m}/\text{s}\) and no vertical component of the velocity. The package is initially at rest with respect to the plane. On the ground, a man is speeding along parallel to the plane in a \(5 \text{ m}\) wide car traveling \(40 \text{ m}/\text{s}\) trying to catch the package. The car starts a distance \(X \text{ m}\) ahead of the plane. What does \(X\) need to be for the man to succeed in catching the package? Solution: First, compute how long it takes for the package to hit the ground: \[d = \frac12gt^2 \implies t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{20000 \text{ m}}{9.8 \text{ m}/\text{s}^2}} = 45.2 \text{ s}.\] How far does the package travel horizontally during that time? \[d_{\text{package}} = v_x t = (250 \text{ m}/\text{s})(45.2 \text{ s}) = 11300 \text{ m}.\] How far does the car travel during that time? \[d_{\text{car}} = v_x t = (40 \text{ m}/\text{s})(45.2 \text{ s}) = 1808 \text{ m}.\] If the package is caught, then \(d_{\text{car}} + X = d_{\text{package}}\). This requires: \[X = (11300 - 1808) \text{ m} = 9492 \text{ m},\] or nearly \(10\) kilometers! To be exact, the above quantity for \(X\) can be shifted by up to \(2.5 \text{ m}\) and still make contact with the car, because of the nonzero width of the car, but this is a negligible correction; \(X\) is very large in comparison.

A pitcher throws a baseball towards home plate, a distance of \(18 \text{ m}\) away, at \(v = 40 \text{ m}/\text{s}\). Suppose the batter takes \(.2 \text {s}\) to react before swinging. In swinging, the batter accelerates the end of the bat from rest through \(2 \text{ m}\) at some constant acceleration \(a\). Assuming that the end of the bat hits the ball if it crosses the plate within \(. 05 \text{ s}\) of the ball crossing the plate, what is the minimum required \(a\) in \(\text{m}/\text{s}^2\) to the nearest tenth for the batter to hit the ball?

SpaceX is trying to land their next reusable rocket back on a drone ship. The drone ship is traveling due west in an ocean current at a constant speed of \(5 \text{ m}/\text{s}\). The rocket is \(2000 \text{ m}\) east of the drone ship and \(5000 \text{ m}\) vertically above it, traveling vertically downwards at \(100 \text{ m}/\text{s}\). If the rocket can apply vertical and horizontal thrusts to change the acceleration of the rocket in the vertical and horizontal directions, and must accelerate constantly in both directions, find the magnitude of the net acceleration (vector) required to land on the ship with no vertical velocity. Answer in \(\text{ m}/\text{s}^2\) to the nearest tenth.

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6.1 Solving Problems with Newton’s Laws

Learning objectives.

By the end of this section, you will be able to:

• Apply problem-solving techniques to solve for quantities in more complex systems of forces
• Use concepts from kinematics to solve problems using Newton’s laws of motion
• Solve more complex equilibrium problems
• Solve more complex acceleration problems
• Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy

Applying newton’s laws of motion.

• Identify the physical principles involved by listing the givens and the quantities to be calculated.
• Sketch the situation, using arrows to represent all forces.
• Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
• Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
• Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 6.2 (a). Then, as in Figure 6.2 (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 6.2 (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure 6.2 (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure 6.2 (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set a y = 0 . a y = 0 . ) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Example 6.1

Different tensions at different angles.

Thus, as you might expect,

This gives us the following relationship:

Note that T 1 T 1 and T 2 T 2 are not equal in this case because the angles on either side are not equal. It is reasonable that T 2 T 2 ends up being greater than T 1 T 1 because it is exerted more vertically than T 1 . T 1 .

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for T 2 T 2 in terms of T 1 T 1 reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of T 1 T 1 to be

Finally, we find the magnitude of T 2 T 2 by using the relationship between them, T 2 = 1.225 T 1 T 2 = 1.225 T 1 , found above. Thus we obtain

Significance

Particle acceleration.

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Example 6.2

Drag force on a barge.

The drag of the water F → D F → D is in the direction opposite to the direction of motion of the boat; this force thus works against F → app , F → app , as shown in the free-body diagram in Figure 6.4 (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x - and y -axes are in the same direction as F → 1 F → 1 and F → 2 . F → 2 . The problem quickly becomes a one-dimensional problem along the direction of F → app F → app , since friction is in the direction opposite to F → app . F → app . Our strategy is to find the magnitude and direction of the net applied force F → app F → app and then apply Newton’s second law to solve for the drag force F → D . F → D .

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that F → D F → D is in the opposite direction of F → app , F → app , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F → app , F → app , but its magnitude is slightly less than F → app . F → app . The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water F D F D in terms of known quantities:

Substituting known values gives

The direction of F → D F → D has already been determined to be in the direction opposite to F → app , F → app , or at an angle of 53 ° 53 ° south of west.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

Example 6.3

What does the bathroom scale read in an elevator.

From the free-body diagram, we see that F → net = F → s − w → , F → net = F → s − w → , so we have

Solving for F s F s gives us an equation with only one unknown:

or, because w = m g , w = m g , simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes F s − w = − m a . F s − w = − m a . )

• We have a = 1.20 m/s 2 , a = 1.20 m/s 2 , so that F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) yielding F s = 825 N . F s = 825 N .
• Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because a = Δ v Δ t a = Δ v Δ t and Δ v = 0 . Δ v = 0 . Thus, F s = m a + m g = 0 + m g F s = m a + m g = 0 + m g or F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , which gives F s = 735 N . F s = 735 N .

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure 6.5 (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding 6.1

Now calculate the scale reading when the elevator accelerates downward at a rate of 1.20 m/s 2 . 1.20 m/s 2 .

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Example 6.4

Two attached blocks.

For block 1: T → + w → 1 + N → = m 1 a → 1 T → + w → 1 + N → = m 1 a → 1

For block 2: T → + w → 2 = m 2 a → 2 . T → + w → 2 = m 2 a → 2 .

Notice that T → T → is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

When block 1 moves to the right, block 2 travels an equal distance downward; thus, a 1 x = − a 2 y . a 1 x = − a 2 y . Writing the common acceleration of the blocks as a = a 1 x = − a 2 y , a = a 1 x = − a 2 y , we now have

From these two equations, we can express a and T in terms of the masses m 1 and m 2 , and g : m 1 and m 2 , and g :

Check Your Understanding 6.2

Calculate the acceleration of the system, and the tension in the string, when the masses are m 1 = 5.00 kg m 1 = 5.00 kg and m 2 = 3.00 kg . m 2 = 3.00 kg .

Example 6.5

Atwood machine.

• We have For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . (The negative sign in front of m 2 a m 2 a indicates that m 2 m 2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . Solving for a : a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 . a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 .
• Observing the first block, we see that T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N . T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N .

Check Your Understanding 6.3

Determine a general formula in terms of m 1 , m 2 m 1 , m 2 and g for calculating the tension in the string for the Atwood machine shown above.

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

Example 6.6

What force must a soccer player exert to reach top speed.

• We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δ v = 8.00 m/s Δ v = 8.00 m/s . We are given the elapsed time, so Δ t = 2.50 s . Δ t = 2.50 s . The unknown is acceleration, which can be found from its definition: a = Δ v Δ t . a = Δ v Δ t . Substituting the known values yields a = 8.00 m/s 2.50 s = 3.20 m/s 2 . a = 8.00 m/s 2.50 s = 3.20 m/s 2 .
• Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, F net = m a . F net = m a . Substituting the known values of m and a gives F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N . F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N .

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Check Your Understanding 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7

What force acts on a model helicopter.

The magnitude of the force is now easily found:

Check Your Understanding 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8

Baggage tractor.

• ∑ F x = m system a x ∑ F x = m system a x and ∑ F x = 820.0 t , ∑ F x = 820.0 t , so 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . Since acceleration is a function of time, we can determine the velocity of the tractor by using a = d v d t a = d v d t with the initial condition that v 0 = 0 v 0 = 0 at t = 0 . t = 0 . We integrate from t = 0 t = 0 to t = 3 : t = 3 : d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s . d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s .
• Refer to the free-body diagram in Figure 6.8 (b). ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N . ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N .

Recall that v = d s d t v = d s d t and a = d v d t a = d v d t . If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have d t = d s v d t = d s v and d t = d v a . d t = d v a . Now, equating these expressions, we have d s v = d v a . d s v = d v a . We can rearrange this to obtain a d s = v d v . a d s = v d v .

Example 6.9

Motion of a projectile fired vertically.

The acceleration depends on v and is therefore variable. Since a = f ( v ) , a = f ( v ) , we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, h = 114 m . h = 114 m .

Check Your Understanding 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Interactive

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

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3.1: Velocity and Acceleration

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• Joel Feldman, Andrew Rechnitzer and Elyse Yeager
• University of British Columbia

If you are moving along the \(x\)–axis and your position at time \(t\) is \(x(t)\text{,}\) then your velocity at time \(t\) is \(v(t)=x'(t)\) and your acceleration at time \(t\) is \(a(t)=v'(t) = x''(t)\text{.}\)

Example 3.1.1 Velocity as derivative of position.

Suppose that you are moving along the \(x\)–axis and that at time \(t\) your position is given by

\begin{align*} x(t)&=t^3-3t+2. \end{align*}

We're going to try and get a good picture of what your motion is like. We can learn quite a bit just by looking at the sign of the velocity \(v(t)=x'(t)\) at each time \(t\text{.}\)

• If \(x'(t) \gt 0\text{,}\) then at that instant \(x\) is increasing, i.e. you are moving to the right.
• If \(x'(t)=0\text{,}\) then at that instant you are not moving at all.
• If \(x'(t) \lt 0\text{,}\) then at that instant \(x\) is decreasing, i.e. you are moving to the left.

From the given formula for \(x(t)\) it is straight forward to work out the velocity

\begin{align*} v(t) = x'(t) &=3t^2-3=3(t^2-1)=3(t+1)(t-1) \end{align*}

This is zero only when \(t=-1\) and when \(t=+1\text{;}\) at no other value 1  of \(t\) can this polynomial be equal zero. Consequently in any time interval that does not include either \(t=-1\) or \(t=+1\text{,}\) \(v(t)\) takes only a single sign 2 . So

• For all \(t \lt -1\text{,}\) both \((t+1)\) and \((t-1)\) are negative (sub in, for example, \(t=-10\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \gt 0\text{.}\)
• For all \(-1 \lt t \lt 1\text{,}\) the factor \((t+1) \gt 0\) and the factor \((t-1) \lt 0\) (sub in, for example, \(t=0\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \lt 0\text{.}\)
• For all \(t \gt 1\text{,}\) both \((t+1)\) and \((t-1)\) are positive (sub in, for example, \(t=+10\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \gt 0\text{.}\)

The figure below gives a summary of the sign information we have about \(t-1\text{,}\) \(t+1\) and \(x'(t)\text{.}\)

It is now easy to put together a mental image of your trajectory.

• For \(t\) large and negative (i.e. far in the past), \(x(t)\) is large and negative and \(v(t)\) is large and positive. For example 3 , when \(t=-10^6\text{,}\) \(x(t)\approx t^3=- 10^{18}\) and \(v(t)\approx 3t^2 = 3\cdot 10^{12}\text{.}\) So you are moving quickly to the right.
• For \(t \lt -1\text{,}\) \(v(t)=x'(t) \gt 0\) so that \(x(t)\) is increasing and you are moving to the right.
• At \(t=-1\text{,}\) \(v(-1)=0\) and you have come to a halt at position \(x(-1)=(-1)^3-3(-1)+2=4\text{.}\)
• For \(-1 \lt t \lt 1\text{,}\) \(v(t)=x'(t) \lt 0\) so that \(x(t)\) is decreasing and you are moving to the left.
• At \(t=+1\text{,}\) \(v(1)=0\) and you have again come to a halt, but now at position \(x(1)=1^3-3+2=0\text{.}\)
• For \(t \gt 1\text{,}\) \(v(t)=x'(t) \gt 0\) so that \(x(t)\) is increasing and you are again moving to the right.
• For \(t\) large and positive (i.e. in the far future), \(x(t)\) is large and positive and \(v(t)\) is large and positive. For example 4 , when \(t=10^6\text{,}\) \(x(t)\approx t^3= 10^{18}\) and \(v(t)\approx 3t^2 = 3\cdot 10^{12}\text{.}\) So you are moving quickly to the right.

Here is a sketch of the graphs of \(x(t)\) and \(v(t)\text{.}\) The heavy lines in the graphs indicate when you are moving to the right — that is where \(v(t)=x'(t)\) is positive.

And here is a schematic picture of the whole trajectory.

Example 3.1.2 Position and velocity from acceleration.

In this example we are going to figure out how far a body falling from rest will fall in a given time period.

• time in seconds by \(t\text{,}\)
• mass in kilograms by \(m\text{,}\)
• distance fallen (in metres) at time \(t\) by \(s(t)\text{,}\) velocity (in m/sec) by \(v(t)=s'(t)\) and acceleration (in m/sec\(^2\)) by \(a(t)=v'(t)=s''(t)\text{.}\)

It makes sense to choose a coordinate system so that the body starts to fall at \(t=0\text{.}\)

\begin{gather*} \text{the force applied to the body at time } t = m \cdot a(t). \end{gather*}

\begin{align*} \text{the force due to gravity acting on a body of mass } m &= m \cdot g. \end{align*}

\begin{align*} v(0) &= 0. \end{align*}

\begin{align*} m\cdot a(t) &= \text{force due to gravity}\\ m \cdot v'(t) &= m \cdot g & \text{ cancel the } m\\ v'(t) &=g \end{align*}

\begin{gather*} v(t) = gt + c \end{gather*}

for any constant \(c\text{.}\) One can then verify 6  that \(v'(t)=g\text{.}\) Using the fact that \(v(0)=0\) we must then have \(c=0\) and so

\begin{align*} v(t) &= gt. \end{align*}

\begin{align*} s'(t) &= v(t) = g \cdot t. \end{align*}

\begin{align*} s(t) &= \frac{g}{2} t^2 + c \end{align*}

\begin{align*} s(t) &= \frac{g}{2} t^2 & \text{but $g=9.8$, so}\\ &= 4.9 t^2, \end{align*}

Let's now do a similar but more complicated example.

Example 3.1.3 Stoping distance of a braking car.

A car's brakes can decelerate the car at 64000\(\textrm{km/hr}^2\text{.}\) How fast can the car be driven if it must be able to stop within a distance of 50m?

Before getting started, notice that there is a small “trick” in this problem — several quantities are stated but their units are different. The acceleration is stated in kilometres per hour\(^2\text{,}\) but the distance is stated in metres. Whenever we come across a “real world” problem 8  we should be careful of the units used.

• time (in hours) by \(t\text{,}\)
• the position of the car (in kilometres) at time \(t\) by \(x(t)\text{,}\) and
• the velocity (in kilometres per hour) by is \(v(t)\text{.}\)

We can also choose a coordinate system such that \(x(0)=0\) and the car starts braking at time \(t=0\text{.}\)

• We are told that, at maximum braking, the acceleration \(v'(t)=x''(t)\) of the car is \(-64000\text{.}\)
• We need to determine the maximum initial velocity \(v(0)\) so that the stopping distance is at most \(50m = 0.05km\) (being careful with our units). Let us call the stopping distance \(x_{stop}\) which is really \(x(t_{stop})\) where \(t_{stop}\) is the stopping time.
• In order to determine \(x_{stop}\) we first need to determine \(t_{stop}\text{,}\) which we will do by assuming maximum braking from a, yet to be determined, initial velocity of \(v(0)=q\) m/sec.

\begin{align*} v'(t) &= -64000 \end{align*}

This equation is very similar to the ones we had to solve in Example 3.1.2 just above.

As we did there  9  Now is a good time to go back and have a read of that example. , we are going to just guess \(v(t)\text{.}\) First, we just guess one function whose derivative is \(-64000\text{,}\) namely \(-64000 t\text{.}\) Next we observe that, since the derivative of a constant is zero, any function of the form

\begin{gather*} v(t) = -64000\,t + c \end{gather*}

with constant \(c\text{,}\) has the correct derivative. Finally, the requirement that the initial velocity \(v(0)=q\)" forces \(c=q\text{,}\) so

\begin{gather*} v(t) = q - 64000\,t \end{gather*}

\begin{align*} 0 = v(t_{stop}) &= q-64000 \cdot t_{stop} & \text{ and so}\\ t_{stop} &= \frac{q}{64000}. \end{align*}

\begin{align*} x'(t) &= v(t) = q - 64000t. \end{align*}

\begin{align*} x(t) &= qt - 32000t^2 + \text{constant} \end{align*}

\begin{align*} x(t) &= qt - 32000 t^2. \end{align*}

\begin{align*} x_{stop} &= x(t_{stop}) = q t_{stop} - 32000 t_{stop}^2\\ &= \frac{q^2}{64000} - \frac{32000 q^2}{64000^2}\\ &= \frac{q^2}{64000} \left(1 - \frac{1}{2} \right)\\ &= \frac{q^2}{2 \times 64000} \end{align*}

\begin{align*} x_{stop} = \frac{q^2}{2 \times 64000} &\leq \frac{5}{100}\\ q^2 &\leq \frac{2 \times 64000 \times 5}{100} = \frac{64000 \times 10}{100} = 6400 \end{align*}

Exercise \(\PageIndex{1}\)

Suppose you throw a ball straight up in the air, and its height from \(t=0\) to \(t=4\) is given by \(h(t)=-4.9t^2+19.6t\text{.}\) True or false: at time \(t=2\text{,}\) the acceleration of the ball is 0.

Exercise \(\PageIndex{2}\)

Suppose an object is moving with a constant acceleration. It takes ten seconds to accelerate from \(1\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(2\;\frac{\mathrm{m}}{\mathrm{s}}\text{.}\) How long does it take to accelerate from \(2\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(3\;\frac{\mathrm{m}}{\mathrm{s}}\text{?}\) How long does it take to accelerate from \(3\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(13\;\frac{\mathrm{m}}{\mathrm{s}}\text{?}\)

Exercise \(\PageIndex{3}\)

Let \(s(t)\) be the position of a particle at time \(t\text{.}\) True or false: if \(s''(a) \gt 0\) for some \(a\text{,}\) then the particle's speed is increasing when \(t=a\text{.}\)

Exercise \(\PageIndex{4}\)

Let \(s(t)\) be the position of a particle at time \(t\text{.}\) True or false: if \(s'(a) \gt 0\) and \(s''(a) \gt 0\) for some \(a\text{,}\) then the particle's speed is increasing when \(t=a\text{.}\)

For this section, we will ask you a number of questions that have to do with objects falling on Earth. Unless otherwise stated, you should assume that an object falling through the air has an acceleration due to gravity of 9.8 meters per second per second.

Exercise \(\PageIndex{5}\)

A flower pot rolls out of a window 10m above the ground. How fast is it falling just as it smacks into the ground?

Exercise \(\PageIndex{6}\)

You want to know how deep a well is, so you drop a stone down and count the seconds until you hear it hit bottom.

• If the stone took \(x\) seconds to hit bottom, how deep is the well?
• Suppose you think you dropped the stone down the well, but actually you tossed it down, so instead of an initial velocity of 0 metres per second, you accidentally imparted an initial speed of \(1\) metres per second. What is the actual depth of the well, if the stone fell for \(x\) seconds?

Exercise \(\PageIndex{7}\)

You toss a key to your friend, standing two metres away. The keys initially move towards your friend at 2 metres per second, but slow at a rate of 0.25 metres per second per second. How much time does your friend have to react to catch the keys? That is--how long are the keys flying before they reach your friend?

Exercise \(\PageIndex{8}\)

A car is driving at 100 kph, and it brakes with a deceleration of \(50000 \frac{\mathrm{km}}{\mathrm{hr}^2}\text{.}\) How long does the car take to come to a complete stop?

Exercise \(\PageIndex{9}\)

You are driving at 120 kph, and need to stop in 100 metres. How much deceleration do your brakes need to provide? You may assume the brakes cause a constant deceleration.

Exercise \(\PageIndex{10}\)

You are driving at 100 kph, and apply the brakes steadily, so that your car decelerates at a constant rate and comes to a stop in exactly 7 seconds. What was your speed one second before you stopped?

Exercise \(\PageIndex{11}\)

About 8.5 minutes after liftoff, the US space shuttle has reached orbital velocity, 17 500 miles per hour. Assuming its acceleration was constant, how far did it travel in those 8.5 minutes?

Source: http://www.nasa.gov/mission_pages/shuttle/shuttlemissions/sts121/launch/qa-leinbach.html

Exercise \(\PageIndex{12}\)

A pitching machine has a dial to adjust the speed of the pitch. You rotate it so that it pitches the ball straight up in the air. How fast should the ball exit the machine, in order to stay in the air exactly 10 seconds?

You may assume that the ball exits from ground level, and is acted on only by gravity, which causes a constant deceleration of 9.8 metres per second.

Exercise \(\PageIndex{13}\)

A peregrine falcon can dive at a speed of 325 kph. If you were to drop a stone, how high up would you have to be so that the stone reached the same speed in its fall?

Exercise \(\PageIndex{14}\)

You shoot a cannon ball into the air with initial velocity \(v_0\text{,}\) and then gravity brings it back down (neglecting all other forces). If the cannon ball made it to a height of 100m, what was \(v_0\text{?}\)

Exercise \(\PageIndex{15}\)

Suppose you are driving at 120 kph, and you start to brake at a deceleration of \(50 000\) kph per hour. For three seconds you steadily increase your deceleration to \(60 000\) kph per hour. (That is, for three seconds, the rate of change of your deceleration is constant.) How fast are you driving at the end of those three seconds?

Exercise \(\PageIndex{16}\)

You jump up from the side of a trampoline with an initial upward velocity of \(1\) metre per second. While you are in the air, your deceleration is a constant \(9.8\) metres per second per second due to gravity. Once you hit the trampoline, as you fall your speed decreases by \(4.9\) metres per second per second. How many seconds pass between the peak of your jump and the lowest part of your fall on the trampoline?

Exercise \(\PageIndex{17}\)

Suppose an object is moving so that its velocity doubles every second. Give an expression for the acceleration of the object.

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Relative Velocity and Riverboat Problems

• Vectors and Direction
• What is a Resultant?
• Vector Components
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• Relative Velocity and River Boat Problems
• Independence of Perpendicular Components of Motion

Tailwinds, Headwinds, and Side Winds

To illustrate this principle, consider a plane flying amidst a tailwind . A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity. If the plane is traveling at a velocity of 100 km/hr with respect to the air, and if the wind velocity is 25 km/hr, then what is the velocity of the plane relative to an observer on the ground below? The resultant velocity of the plane (that is, the result of the wind velocity contributing to the velocity due to the plane's motor) is the vector sum of the velocity of the plane and the velocity of the wind. This resultant velocity is quite easily determined if the wind approaches the plane directly from behind. As shown in the diagram below, the plane travels with a resulting velocity of 125 km/hr relative to the ground.

If the plane encounters a headwind, the resulting velocity will be less than 100 km/hr. Since a headwind is a wind that approaches the plane from the front, such a wind would decrease the plane's resulting velocity. Suppose a plane traveling with a velocity of 100 km/hr with respect to the air meets a headwind with a velocity of 25 km/hr. In this case, the resultant velocity would be 75 km/hr; this is the velocity of the plane relative to an observer on the ground. This is depicted in the diagram below.

Now consider a plane traveling with a velocity of 100 km/hr, South that encounters a side wind of 25 km/hr, West. Now what would the resulting velocity of the plane be? This question can be answered in the same manner as the previous questions. The resulting velocity of the plane is the vector sum of the two individual velocities. To determine the resultant velocity, the plane velocity (relative to the air) must be added to the wind velocity. This is the same procedure that was used above for the headwind and the tailwind situations; only now, the resultant is not as easily computed. Since the two vectors to be added - the southward plane velocity and the westward wind velocity - are at right angles to each other, the Pythagorean theorem can be used. This is illustrated in the diagram below.

In this situation of a side wind, the southward vector can be added to the westward vector using the usual methods of vector addition . The magnitude of the resultant velocity is determined using Pythagorean theorem. The algebraic steps are as follows:

10 000 km 2 /hr 2 + 625 km 2 /hr 2 = R 2

10 625 km 2 /hr 2 = R 2

SQRT(10 625 km 2 /hr 2 ) = R

103.1 km/hr = R

tan (theta) = (25/100)

theta = invtan (25/100)

theta = 14.0 degrees

If the resultant velocity of the plane makes a 14.0 degree angle with the southward direction (theta in the above diagram), then the direction of the resultant is 256 degrees. Like any vector, the resultant's direction is measured as a counterclockwise angle of rotation from due East.

Analysis of a Riverboat's Motion

The effect of the wind upon the plane is similar to the effect of the river current upon the motorboat. If a motorboat were to head straight across a river (that is, if the boat were to point its bow straight towards the other side), it would not reach the shore directly across from its starting point. The river current influences the motion of the boat and carries it downstream. The motorboat may be moving with a velocity of 4 m/s directly across the river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the downstream direction. While the speedometer of the boat may read 4 m/s, its speed with respect to an observer on the shore will be greater than 4 m/s.

The resultant velocity of the motorboat can be determined in the same manner as was done for the plane. The resultant velocity of the boat is the vector sum of the boat velocity and the river velocity. Since the boat heads straight across the river and since the current is always directed straight downstream, the two vectors are at right angles to each other. Thus, the Pythagorean theorem can be used to determine the resultant velocity. Suppose that the river was moving with a velocity of 3 m/s, North and the motorboat was moving with a velocity of 4 m/s, East. What would be the resultant velocity of the motorboat (i.e., the velocity relative to an observer on the shore)? The magnitude of the resultant can be found as follows:

16 m 2 /s 2 + 9 m 2 /s 2 = R 2

25 m 2 /s 2 = R 2

SQRT (25 m 2 /s 2 ) = R

5.0 m/s = R

tan (theta) = (3/4)

theta = invtan (3/4)

theta = 36.9 degrees

Given a boat velocity of 4 m/s, East and a river velocity of 3 m/s, North, the resultant velocity of the boat will be 5 m/s at 36.9 degrees.

Motorboat problems such as these are typically accompanied by three separate questions:

• What is the resultant velocity (both magnitude and direction) of the boat?
• If the width of the river is X meters wide, then how much time does it take the boat to travel shore to shore?
• What distance downstream does the boat reach the opposite shore?

The first of these three questions was answered above; the resultant velocity of the boat can be determined using the Pythagorean theorem (magnitude) and a trigonometric function (direction). The second and third of these questions can be answered using the average speed equation (and a lot of logic).

Consider the following example.

The solution to the first question has already been shown in the above discussion . The resultant velocity of the boat is 5 m/s at 36.9 degrees. We will start in on the second question.

The river is 80-meters wide. That is, the distance from shore to shore as measured straight across the river is 80 meters. The time to cross this 80-meter wide river can be determined by rearranging and substituting into the average speed equation.

The distance of 80 m can be substituted into the numerator. But what about the denominator? What value should be used for average speed? Should 3 m/s (the current velocity), 4 m/s (the boat velocity), or 5 m/s (the resultant velocity) be used as the average speed value for covering the 80 meters? With what average speed is the boat traversing the 80 meter wide river? Most students want to use the resultant velocity in the equation since that is the actual velocity of the boat with respect to the shore. Yet the value of 5 m/s is the speed at which the boat covers the diagonal dimension of the river. And the diagonal distance across the river is not known in this case. If one knew the distance C in the diagram below, then the average speed C could be used to calculate the time to reach the opposite shore. Similarly, if one knew the distance B in the diagram below, then the average speed B could be used to calculate the time to reach the opposite shore. And finally, if one knew the distance A in the diagram below, then the average speed A could be used to calculate the time to reach the opposite shore.

In our problem, the 80 m corresponds to the distance A, and so the average speed of 4 m/s (average speed in the direction straight across the river) should be substituted into the equation to determine the time.

It requires 20 s for the boat to travel across the river. During this 20 s of crossing the river, the boat also drifts downstream. Part c of the problem asks "What distance downstream does the boat reach the opposite shore?" The same equation must be used to calculate this downstream distance . And once more, the question arises, which one of the three average speed values must be used in the equation to calculate the distance downstream? The distance downstream corresponds to Distance B on the above diagram. The speed at which the boat covers this distance corresponds to Average Speed B on the diagram above (i.e., the speed at which the current moves - 3 m/s). And so the average speed of 3 m/s (average speed in the downstream direction) should be substituted into the equation to determine the distance.

distance = 60 m

The boat is carried 60 meters downstream during the 20 seconds it takes to cross the river.

The mathematics of the above problem is no more difficult than dividing or multiplying two numerical quantities by each other. The mathematics is easy! The difficulty of the problem is conceptual in nature; the difficulty lies in deciding which numbers to use in the equations. That decision emerges from one's conceptual understanding (or unfortunately, one's misunderstanding) of the complex motion that is occurring. The motion of the riverboat can be divided into two simultaneous parts - a motion in the direction straight across the river and a motion in the downstream direction. These two parts (or components) of the motion occur simultaneously for the same time duration (which was 20 seconds in the above problem). The decision as to which velocity value or distance value to use in the equation must be consistent with the diagram above . The boat's motor is what carries the boat across the river the Distance A ; and so any calculation involving the Distance A must involve the speed value labeled as Speed A (the boat speed relative to the water). Similarly, it is the current of the river that carries the boat downstream for the Distance B ; and so any calculation involving the Distance B must involve the speed value labeled as Speed B (the river speed). Together, these two parts (or components) add up to give the resulting motion of the boat. That is, the across-the-river component of displacement adds to the downstream displacement to equal the resulting displacement. And likewise, the boat velocity (across the river) adds to the river velocity (down the river) to equal the resulting velocity. And so any calculation of the Distance C or the Average Speed C ("Resultant Velocity") can be performed using the Pythagorean theorem.

Now to illustra te an important point, let's try a second example problem that is similar to the first example problem . Make an attempt to answer the three questions and then click the button to c heck your answer.

a. The resultant velocity can be found using the Pythagorean theorem. The resultant is the hypotenuse of a right triangle with sides of 4 m/s and 7 m/s. It is

Its direction can be determined using a trigonometric function.

b. The time to cross the river is t = d / v = (80 m) / (4 m/s) = 20 s

c. The distance traveled downstream is d = v • t = (7 m/s) • (20 s) = 140 m

An import ant concept emerges from the analysis of the two example problems above. In Example 1, the time to cross the 80-meter wide river (when moving 4 m/s) was 20 seconds. This was in the presence of a 3 m/s current velocity. In Example 2, the current velocity was much greater - 7 m/s - yet the time to cross the river remained unchanged. In fact, the current velocity itself has no effect upon the time required for a boat to cross the river. The river moves downstream parallel to the banks of the river. As such, there is no way that the current is capable of assisting a boat in crossing a river. While the increased current may affect the resultant velocity - making the boat travel with a greater speed with respect to an observer on the ground - it does not increase the speed in the direction across the river. The component of the resultant velocity that is increased is the component that is in a direction pointing down the river. It is often said that "perpendicular components of motion are independent of each other." As applied to riverboat problems, this would mean that an across-the-river variable would be independent of (i.e., not be affected by) a downstream variable. The time to cross the river is dependent upon the velocity at which the boat crosses the river. It is only the component of motion directed across the river (i.e., the boat velocity) that affects the time to travel the distance directly across the river (80 m in this case). The component of motion perpendicular to this direction - the current velocity - only affects the distance that the boat travels down the river. This concept of perpendicular components of motion will be investigated in more detail in the next part of Lesson 1 .

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Check your understanding.

1. A plane can travel with a speed of 80 mi/hr with respect to the air. Determine the resultant velocity of the plane (magnitude only) if it encounters a

a. 10 mi/hr headwind. b. 10 mi/hr tailwind. c. 10 mi/hr crosswind. d. 60 mi/hr crosswind.

a. A headwind would decrease the resultant velocity of the plane to 70 mi/hr .

b. A tailwind would increase the resultant velocity of the plane to 90 mi/hr .

c. A 10 mi/hr crosswind would increase the resultant velocity of the plane to 80.6 mi/hr .

This can be determined using the Pythagorean theorem: SQRT[ (80 mi/hr) 2 + (10 mi/hr) 2 ] )

d. A 60 mi/hr crosswind would increase the resultant velocity of the plane to 100 mi/hr .

This can be determined using the Pythagorean theorem: SQRT[ (80 mi/hr) 2 + (60 mi/hr) 2 ] )

2. A motorboat traveling 5 m/s, East encounters a current traveling 2.5 m/s, North.

a. What is the resultant velocity of the motor boat? b. If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore? c. What distance downstream does the boat reach the opposite shore?

a. The resultant velocity can be found using the Pythagorean theorem. The resultant is the hypotenuse of a right triangle with sides of 5 m/s and 2.5 m/s. It is

b. The time to cross the river is t = d / v = (80 m) / (5 m/s) = 16.0 s

c. The distance traveled downstream is d = v • t = (2.5 m/s)*(16.0 s) = 40 m

3. A motorboat traveling 5 m/s, East encounters a current traveling 2.5 m/s, South.

NOTE: the direction of the resultant velocity (like any vector) is expressed as the counterclockwise angle of rotation from due East.

c. The distance traveled downstream is d = v • t = (2.5 m/s) • (16.0 s) = 40 m

4. A motorboat traveling 6 m/s, East encounters a current traveling 3.8 m/s, South.

a. What is the resultant velocity of the motor boat? b. If the width of the river is 120 meters wide, then how much time does it take the boat to travel shore to shore? c. What distance downstream does the boat reach the opposite shore?

a. The resultant velocity can be found using the Pythagorean theorem. The resultant is the hypotenuse of a right triangle with sides of 6 m/s and 3.8 m/s. It is

NOTE: the direction of the resultant velocity (like any vector) is expressed as the counterclockwise direction of rotation from due East.

b. The time to cross the river is t = d / v = (120 m) / (6 m/s) = 20.0 s

c. The distance traveled downstream is d = v • t = (3.8 m/s) • (20.0 s) = 76 m

5. If the current velocity in question #4 were increased to 5 m/s, then

a. how much time would be required to cross the same 120-m wide river? b. what distance downstream would the boat travel during this time?

a. It would require the same amount of time as before ( 20 s ). Changing the current velocity does not affect the time required to cross the river since perpendicular components of motion are independent of each other.

b. The distance traveled downstream is

Note that an alteration in the current velocity would only affect the distance traveled downstream (and the resultant velocity).

• What is a Projectile?

Physics for You – Dare to know!

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Position, velocity, acceleration problems and solutions

When solving a Physics problem in general and one of Kinematics in particular, it is important that you follow an order . Get used to being organized when you solve problems, and you will see how it gives good results. It is worth spending a little time on the previous analysis of a problem before you tackle it.

When solving a Kinematics problem, follow the following steps:

• Read carefully the Problem Statement.
• Draw a picture of the physical situation depicted in the problem statement, clearly indicating which reference frame you will use to solve it. You must indicate where the origin of the reference system is located and the positive direction of the axes you are going to use.
• Write in your notebook the givens in the problem statement.
• Identify which are the initial conditions of the problem: what is the initial position of the particle, its initial velocity and its acceleration.
• If the acceleration is a given, you will have to integrate it to obtain the velocity and integrate this velocity to obtain the position vector. In both integrals, you must take into account the initial conditions.
• The components of each vector must have the corresponding sign depending on your choice of the Cartesian axes.
• Do not forget to include the units in your results.
• Review the problem and check that the results you have obtained make sense.

On the following pages you will find some Kinematics problems with solutions. Try to do them yourself before looking at the resolution.

• Position Velocity Acceleration vectors - Integrals
• Position Velocity Acceleration vectors - Initial conditions
• Position Velocity Acceleration vectors - Derivatives
• Position Velocity Acceleration vectors - Parametric equations
• Position Velocity Acceleration vectors - Two-dimensional motion
• Position Velocity Acceleration vectors - Parabolic motion and free fall
• Position Velocity Acceleration vectors - Parabolic and uniform motions
• Position Velocity Acceleration vectors - Changing reference frame
• Position Velocity Acceleration vectors - Parabolic motion

JEE-IIT-NCERT Physics & Math

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Do You Understand the Problem You’re Trying to Solve?

To solve tough problems at work, first ask these questions.

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Problem solving skills are invaluable in any job. But all too often, we jump to find solutions to a problem without taking time to really understand the dilemma we face, according to Thomas Wedell-Wedellsborg , an expert in innovation and the author of the book, What’s Your Problem?: To Solve Your Toughest Problems, Change the Problems You Solve .

In this episode, you’ll learn how to reframe tough problems by asking questions that reveal all the factors and assumptions that contribute to the situation. You’ll also learn why searching for just one root cause can be misleading.

Key episode topics include: leadership, decision making and problem solving, power and influence, business management.

HBR On Leadership curates the best case studies and conversations with the world’s top business and management experts, to help you unlock the best in those around you. New episodes every week.

• Listen to the original HBR IdeaCast episode: The Secret to Better Problem Solving (2016)
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HANNAH BATES: Welcome to HBR on Leadership , case studies and conversations with the world’s top business and management experts, hand-selected to help you unlock the best in those around you.

Problem solving skills are invaluable in any job. But even the most experienced among us can fall into the trap of solving the wrong problem.

Thomas Wedell-Wedellsborg says that all too often, we jump to find solutions to a problem – without taking time to really understand what we’re facing.

He’s an expert in innovation, and he’s the author of the book, What’s Your Problem?: To Solve Your Toughest Problems, Change the Problems You Solve .

  In this episode, you’ll learn how to reframe tough problems, by asking questions that reveal all the factors and assumptions that contribute to the situation. You’ll also learn why searching for one root cause can be misleading. And you’ll learn how to use experimentation and rapid prototyping as problem-solving tools.

This episode originally aired on HBR IdeaCast in December 2016. Here it is.

SARAH GREEN CARMICHAEL: Welcome to the HBR IdeaCast from Harvard Business Review. I’m Sarah Green Carmichael.

Problem solving is popular. People put it on their resumes. Managers believe they excel at it. Companies count it as a key proficiency. We solve customers’ problems.

The problem is we often solve the wrong problems. Albert Einstein and Peter Drucker alike have discussed the difficulty of effective diagnosis. There are great frameworks for getting teams to attack true problems, but they’re often hard to do daily and on the fly. That’s where our guest comes in.

Thomas Wedell-Wedellsborg is a consultant who helps companies and managers reframe their problems so they can come up with an effective solution faster. He asks the question “Are You Solving The Right Problems?” in the January-February 2017 issue of Harvard Business Review. Thomas, thank you so much for coming on the HBR IdeaCast .

THOMAS WEDELL-WEDELLSBORG: Thanks for inviting me.

SARAH GREEN CARMICHAEL: So, I thought maybe we could start by talking about the problem of talking about problem reframing. What is that exactly?

THOMAS WEDELL-WEDELLSBORG: Basically, when people face a problem, they tend to jump into solution mode to rapidly, and very often that means that they don’t really understand, necessarily, the problem they’re trying to solve. And so, reframing is really a– at heart, it’s a method that helps you avoid that by taking a second to go in and ask two questions, basically saying, first of all, wait. What is the problem we’re trying to solve? And then crucially asking, is there a different way to think about what the problem actually is?

SARAH GREEN CARMICHAEL: So, I feel like so often when this comes up in meetings, you know, someone says that, and maybe they throw out the Einstein quote about you spend an hour of problem solving, you spend 55 minutes to find the problem. And then everyone else in the room kind of gets irritated. So, maybe just give us an example of maybe how this would work in practice in a way that would not, sort of, set people’s teeth on edge, like oh, here Sarah goes again, reframing the whole problem instead of just solving it.

THOMAS WEDELL-WEDELLSBORG: I mean, you’re bringing up something that’s, I think is crucial, which is to create legitimacy for the method. So, one of the reasons why I put out the article is to give people a tool to say actually, this thing is still important, and we need to do it. But I think the really critical thing in order to make this work in a meeting is actually to learn how to do it fast, because if you have the idea that you need to spend 30 minutes in a meeting delving deeply into the problem, I mean, that’s going to be uphill for most problems. So, the critical thing here is really to try to make it a practice you can implement very, very rapidly.

There’s an example that I would suggest memorizing. This is the example that I use to explain very rapidly what it is. And it’s basically, I call it the slow elevator problem. You imagine that you are the owner of an office building, and that your tenants are complaining that the elevator’s slow.

Now, if you take that problem framing for granted, you’re going to start thinking creatively around how do we make the elevator faster. Do we install a new motor? Do we have to buy a new lift somewhere?

The thing is, though, if you ask people who actually work with facilities management, well, they’re going to have a different solution for you, which is put up a mirror next to the elevator. That’s what happens is, of course, that people go oh, I’m busy. I’m busy. I’m– oh, a mirror. Oh, that’s beautiful.

And then they forget time. What’s interesting about that example is that the idea with a mirror is actually a solution to a different problem than the one you first proposed. And so, the whole idea here is once you get good at using reframing, you can quickly identify other aspects of the problem that might be much better to try to solve than the original one you found. It’s not necessarily that the first one is wrong. It’s just that there might be better problems out there to attack that we can, means we can do things much faster, cheaper, or better.

SARAH GREEN CARMICHAEL: So, in that example, I can understand how A, it’s probably expensive to make the elevator faster, so it’s much cheaper just to put up a mirror. And B, maybe the real problem people are actually feeling, even though they’re not articulating it right, is like, I hate waiting for the elevator. But if you let them sort of fix their hair or check their teeth, they’re suddenly distracted and don’t notice.

But if you have, this is sort of a pedestrian example, but say you have a roommate or a spouse who doesn’t clean up the kitchen. Facing that problem and not having your elegant solution already there to highlight the contrast between the perceived problem and the real problem, how would you take a problem like that and attack it using this method so that you can see what some of the other options might be?

THOMAS WEDELL-WEDELLSBORG: Right. So, I mean, let’s say it’s you who have that problem. I would go in and say, first of all, what would you say the problem is? Like, if you were to describe your view of the problem, what would that be?

SARAH GREEN CARMICHAEL: I hate cleaning the kitchen, and I want someone else to clean it up.

THOMAS WEDELL-WEDELLSBORG: OK. So, my first observation, you know, that somebody else might not necessarily be your spouse. So, already there, there’s an inbuilt assumption in your question around oh, it has to be my husband who does the cleaning. So, it might actually be worth, already there to say, is that really the only problem you have? That you hate cleaning the kitchen, and you want to avoid it? Or might there be something around, as well, getting a better relationship in terms of how you solve problems in general or establishing a better way to handle small problems when dealing with your spouse?

SARAH GREEN CARMICHAEL: Or maybe, now that I’m thinking that, maybe the problem is that you just can’t find the stuff in the kitchen when you need to find it.

THOMAS WEDELL-WEDELLSBORG: Right, and so that’s an example of a reframing, that actually why is it a problem that the kitchen is not clean? Is it only because you hate the act of cleaning, or does it actually mean that it just takes you a lot longer and gets a lot messier to actually use the kitchen, which is a different problem. The way you describe this problem now, is there anything that’s missing from that description?

SARAH GREEN CARMICHAEL: That is a really good question.

THOMAS WEDELL-WEDELLSBORG: Other, basically asking other factors that we are not talking about right now, and I say those because people tend to, when given a problem, they tend to delve deeper into the detail. What often is missing is actually an element outside of the initial description of the problem that might be really relevant to what’s going on. Like, why does the kitchen get messy in the first place? Is it something about the way you use it or your cooking habits? Is it because the neighbor’s kids, kind of, use it all the time?

There might, very often, there might be issues that you’re not really thinking about when you first describe the problem that actually has a big effect on it.

SARAH GREEN CARMICHAEL: I think at this point it would be helpful to maybe get another business example, and I’m wondering if you could tell us the story of the dog adoption problem.

THOMAS WEDELL-WEDELLSBORG: Yeah. This is a big problem in the US. If you work in the shelter industry, basically because dogs are so popular, more than 3 million dogs every year enter a shelter, and currently only about half of those actually find a new home and get adopted. And so, this is a problem that has persisted. It’s been, like, a structural problem for decades in this space. In the last three years, where people found new ways to address it.

So a woman called Lori Weise who runs a rescue organization in South LA, and she actually went in and challenged the very idea of what we were trying to do. She said, no, no. The problem we’re trying to solve is not about how to get more people to adopt dogs. It is about keeping the dogs with their first family so they never enter the shelter system in the first place.

In 2013, she started what’s called a Shelter Intervention Program that basically works like this. If a family comes and wants to hand over their dog, these are called owner surrenders. It’s about 30% of all dogs that come into a shelter. All they would do is go up and ask, if you could, would you like to keep your animal? And if they said yes, they would try to fix whatever helped them fix the problem, but that made them turn over this.

And sometimes that might be that they moved into a new building. The landlord required a deposit, and they simply didn’t have the money to put down a deposit. Or the dog might need a $10 rabies shot, but they didn’t know how to get access to a vet.

And so, by instigating that program, just in the first year, she took her, basically the amount of dollars they spent per animal they helped went from something like $85 down to around $60. Just an immediate impact, and her program now is being rolled out, is being supported by the ASPCA, which is one of the big animal welfare stations, and it’s being rolled out to various other places.

And I think what really struck me with that example was this was not dependent on having the internet. This was not, oh, we needed to have everybody mobile before we could come up with this. This, conceivably, we could have done 20 years ago. Only, it only happened when somebody, like in this case Lori, went in and actually rethought what the problem they were trying to solve was in the first place.

SARAH GREEN CARMICHAEL: So, what I also think is so interesting about that example is that when you talk about it, it doesn’t sound like the kind of thing that would have been thought of through other kinds of problem solving methods. There wasn’t necessarily an After Action Review or a 5 Whys exercise or a Six Sigma type intervention. I don’t want to throw those other methods under the bus, but how can you get such powerful results with such a very simple way of thinking about something?

THOMAS WEDELL-WEDELLSBORG: That was something that struck me as well. This, in a way, reframing and the idea of the problem diagnosis is important is something we’ve known for a long, long time. And we’ve actually have built some tools to help out. If you worked with us professionally, you are familiar with, like, Six Sigma, TRIZ, and so on. You mentioned 5 Whys. A root cause analysis is another one that a lot of people are familiar with.

Those are our good tools, and they’re definitely better than nothing. But what I notice when I work with the companies applying those was those tools tend to make you dig deeper into the first understanding of the problem we have. If it’s the elevator example, people start asking, well, is that the cable strength, or is the capacity of the elevator? That they kind of get caught by the details.

That, in a way, is a bad way to work on problems because it really assumes that there’s like a, you can almost hear it, a root cause. That you have to dig down and find the one true problem, and everything else was just symptoms. That’s a bad way to think about problems because problems tend to be multicausal.

There tend to be lots of causes or levers you can potentially press to address a problem. And if you think there’s only one, if that’s the right problem, that’s actually a dangerous way. And so I think that’s why, that this is a method I’ve worked with over the last five years, trying to basically refine how to make people better at this, and the key tends to be this thing about shifting out and saying, is there a totally different way of thinking about the problem versus getting too caught up in the mechanistic details of what happens.

SARAH GREEN CARMICHAEL: What about experimentation? Because that’s another method that’s become really popular with the rise of Lean Startup and lots of other innovation methodologies. Why wouldn’t it have worked to, say, experiment with many different types of fixing the dog adoption problem, and then just pick the one that works the best?

THOMAS WEDELL-WEDELLSBORG: You could say in the dog space, that’s what’s been going on. I mean, there is, in this industry and a lot of, it’s largely volunteer driven. People have experimented, and they found different ways of trying to cope. And that has definitely made the problem better. So, I wouldn’t say that experimentation is bad, quite the contrary. Rapid prototyping, quickly putting something out into the world and learning from it, that’s a fantastic way to learn more and to move forward.

My point is, though, that I feel we’ve come to rely too much on that. There’s like, if you look at the start up space, the wisdom is now just to put something quickly into the market, and then if it doesn’t work, pivot and just do more stuff. What reframing really is, I think of it as the cognitive counterpoint to prototyping. So, this is really a way of seeing very quickly, like not just working on the solution, but also working on our understanding of the problem and trying to see is there a different way to think about that.

If you only stick with experimentation, again, you tend to sometimes stay too much in the same space trying minute variations of something instead of taking a step back and saying, wait a minute. What is this telling us about what the real issue is?

SARAH GREEN CARMICHAEL: So, to go back to something that we touched on earlier, when we were talking about the completely hypothetical example of a spouse who does not clean the kitchen–

THOMAS WEDELL-WEDELLSBORG: Completely, completely hypothetical.

SARAH GREEN CARMICHAEL: Yes. For the record, my husband is a great kitchen cleaner.

You started asking me some questions that I could see immediately were helping me rethink that problem. Is that kind of the key, just having a checklist of questions to ask yourself? How do you really start to put this into practice?

THOMAS WEDELL-WEDELLSBORG: I think there are two steps in that. The first one is just to make yourself better at the method. Yes, you should kind of work with a checklist. In the article, I kind of outlined seven practices that you can use to do this.

But importantly, I would say you have to consider that as, basically, a set of training wheels. I think there’s a big, big danger in getting caught in a checklist. This is something I work with.

My co-author Paddy Miller, it’s one of his insights. That if you start giving people a checklist for things like this, they start following it. And that’s actually a problem, because what you really want them to do is start challenging their thinking.

So the way to handle this is to get some practice using it. Do use the checklist initially, but then try to step away from it and try to see if you can organically make– it’s almost a habit of mind. When you run into a colleague in the hallway and she has a problem and you have five minutes, like, delving in and just starting asking some of those questions and using your intuition to say, wait, how is she talking about this problem? And is there a question or two I can ask her about the problem that can help her rethink it?

SARAH GREEN CARMICHAEL: Well, that is also just a very different approach, because I think in that situation, most of us can’t go 30 seconds without jumping in and offering solutions.

THOMAS WEDELL-WEDELLSBORG: Very true. The drive toward solutions is very strong. And to be clear, I mean, there’s nothing wrong with that if the solutions work. So, many problems are just solved by oh, you know, oh, here’s the way to do that. Great.

But this is really a powerful method for those problems where either it’s something we’ve been banging our heads against tons of times without making progress, or when you need to come up with a really creative solution. When you’re facing a competitor with a much bigger budget, and you know, if you solve the same problem later, you’re not going to win. So, that basic idea of taking that approach to problems can often help you move forward in a different way than just like, oh, I have a solution.

I would say there’s also, there’s some interesting psychological stuff going on, right? Where you may have tried this, but if somebody tries to serve up a solution to a problem I have, I’m often resistant towards them. Kind if like, no, no, no, no, no, no. That solution is not going to work in my world. Whereas if you get them to discuss and analyze what the problem really is, you might actually dig something up.

Let’s go back to the kitchen example. One powerful question is just to say, what’s your own part in creating this problem? It’s very often, like, people, they describe problems as if it’s something that’s inflicted upon them from the external world, and they are innocent bystanders in that.

SARAH GREEN CARMICHAEL: Right, or crazy customers with unreasonable demands.

THOMAS WEDELL-WEDELLSBORG: Exactly, right. I don’t think I’ve ever met an agency or consultancy that didn’t, like, gossip about their customers. Oh, my god, they’re horrible. That, you know, classic thing, why don’t they want to take more risk? Well, risk is bad.

It’s their business that’s on the line, not the consultancy’s, right? So, absolutely, that’s one of the things when you step into a different mindset and kind of, wait. Oh yeah, maybe I actually am part of creating this problem in a sense, as well. That tends to open some new doors for you to move forward, in a way, with stuff that you may have been struggling with for years.

SARAH GREEN CARMICHAEL: So, we’ve surfaced a couple of questions that are useful. I’m curious to know, what are some of the other questions that you find yourself asking in these situations, given that you have made this sort of mental habit that you do? What are the questions that people seem to find really useful?

THOMAS WEDELL-WEDELLSBORG: One easy one is just to ask if there are any positive exceptions to the problem. So, was there day where your kitchen was actually spotlessly clean? And then asking, what was different about that day? Like, what happened there that didn’t happen the other days? That can very often point people towards a factor that they hadn’t considered previously.

SARAH GREEN CARMICHAEL: We got take-out.

THOMAS WEDELL-WEDELLSBORG: S,o that is your solution. Take-out from [INAUDIBLE]. That might have other problems.

Another good question, and this is a little bit more high level. It’s actually more making an observation about labeling how that person thinks about the problem. And what I mean with that is, we have problem categories in our head. So, if I say, let’s say that you describe a problem to me and say, well, we have a really great product and are, it’s much better than our previous product, but people aren’t buying it. I think we need to put more marketing dollars into this.

Now you can go in and say, that’s interesting. This sounds like you’re thinking of this as a communications problem. Is there a different way of thinking about that? Because you can almost tell how, when the second you say communications, there are some ideas about how do you solve a communications problem. Typically with more communication.

And what you might do is go in and suggest, well, have you considered that it might be, say, an incentive problem? Are there incentives on behalf of the purchasing manager at your clients that are obstructing you? Might there be incentive issues with your own sales force that makes them want to sell the old product instead of the new one?

So literally, just identifying what type of problem does this person think about, and is there different potential way of thinking about it? Might it be an emotional problem, a timing problem, an expectations management problem? Thinking about what label of what type of problem that person is kind of thinking as it of.

SARAH GREEN CARMICHAEL: That’s really interesting, too, because I think so many of us get requests for advice that we’re really not qualified to give. So, maybe the next time that happens, instead of muddying my way through, I will just ask some of those questions that we talked about instead.

THOMAS WEDELL-WEDELLSBORG: That sounds like a good idea.

SARAH GREEN CARMICHAEL: So, Thomas, this has really helped me reframe the way I think about a couple of problems in my own life, and I’m just wondering. I know you do this professionally, but is there a problem in your life that thinking this way has helped you solve?

THOMAS WEDELL-WEDELLSBORG: I’ve, of course, I’ve been swallowing my own medicine on this, too, and I think I have, well, maybe two different examples, and in one case somebody else did the reframing for me. But in one case, when I was younger, I often kind of struggled a little bit. I mean, this is my teenage years, kind of hanging out with my parents. I thought they were pretty annoying people. That’s not really fair, because they’re quite wonderful, but that’s what life is when you’re a teenager.

And one of the things that struck me, suddenly, and this was kind of the positive exception was, there was actually an evening where we really had a good time, and there wasn’t a conflict. And the core thing was, I wasn’t just seeing them in their old house where I grew up. It was, actually, we were at a restaurant. And it suddenly struck me that so much of the sometimes, kind of, a little bit, you love them but they’re annoying kind of dynamic, is tied to the place, is tied to the setting you are in.

And of course, if– you know, I live abroad now, if I visit my parents and I stay in my old bedroom, you know, my mother comes in and wants to wake me up in the morning. Stuff like that, right? And it just struck me so, so clearly that it’s– when I change this setting, if I go out and have dinner with them at a different place, that the dynamic, just that dynamic disappears.

SARAH GREEN CARMICHAEL: Well, Thomas, this has been really, really helpful. Thank you for talking with me today.

THOMAS WEDELL-WEDELLSBORG: Thank you, Sarah.  

HANNAH BATES: That was Thomas Wedell-Wedellsborg in conversation with Sarah Green Carmichael on the HBR IdeaCast. He’s an expert in problem solving and innovation, and he’s the author of the book, What’s Your Problem?: To Solve Your Toughest Problems, Change the Problems You Solve .

We’ll be back next Wednesday with another hand-picked conversation about leadership from the Harvard Business Review. If you found this episode helpful, share it with your friends and colleagues, and follow our show on Apple Podcasts, Spotify, or wherever you get your podcasts. While you’re there, be sure to leave us a review.

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This episode was produced by Anne Saini, and me, Hannah Bates. Ian Fox is our editor. Music by Coma Media. Special thanks to Maureen Hoch, Adi Ignatius, Karen Player, Ramsey Khabbaz, Nicole Smith, Anne Bartholomew, and you – our listener.

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