kinematics problem solving with answers

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Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep) 

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

• (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use:  v f  = v i  + a*t

0 m/s = v i  + (-9.8  m/s 2 )*(3.13 s)

0 m/s = v i  - 30.7 m/s

v i  = 30.7 m/s  (30.674 m/s)

Now use:  v f 2  = v i 2  + 2*a*d

(0 m/s) 2  = (30.7 m/s) 2  + 2*(-9.8  m/s 2 )*(d)

0 m 2 /s 2  = (940 m 2 /s 2 ) + (-19.6  m/s 2 )*d

-940  m 2 /s 2  = (-19.6  m/s 2 )*d

(-940  m 2 /s 2 )/(-19.6  m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

Kinematic Equations: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: April 29, 2022

Now that you’ve learned about displacement, velocity, and acceleration, you’re well on your way to being able to describe just about any motion you could observe around you with physics. All that’s left is to learn how these values really play into each other. We know a few ways to move between them, but they’re all pretty limited. What happens if you need to find displacement, but only know acceleration and time? We don’t have a way to combine all of those values yet. Enter the four kinematic equations. 

What We Review

The Kinematic Equations

The following four kinematic equations come up throughout physics from the earliest high school class to the highest level college course:

Don’t let all of these numbers and symbols intimidate you. We’ll talk through each one – what they mean and when we use them. By the end of this post, you’ll be a master of understanding and implementing each of these physics equations. Let’s start with defining what all of those symbols mean. 

The First Kinematic Equation

This physics equation would be read as “the final velocity is equal to the initial velocity plus acceleration times time”. All it means is that if you have constant acceleration for some amount of time, you can find the final velocity. You’ll use this one whenever you’re looking at changing velocities with a constant acceleration.

The Second Kinematic Equation

This one is read as “displacement equals final velocity plus initial velocity divided by two times time”. You’ll use this one whenever you don’t have an acceleration to work with but you need to relate a changing velocity to a displacement.

The Third Kinematic Equation

This one may look a bit scarier as it is longer than the others, but it is read as “displacement equals initial velocity times time plus one half acceleration times time squared”. All it means is that our displacement can be related to our initial velocity and a constant acceleration without having to find the final velocity. You’ll use this one when final velocity is the only value you don’t know yet.

It is worth noting that this kinematic equation has another popular form: x=x_{0}+v_{0}t+\frac{1}{2}at^{2} . While that may seem even more intimidating, it’s actually exactly the same. The only difference here is that we have split up \Delta x into x-x_{0} and then solved to get x on its own. This version can be particularly helpful if you’re looking specifically for a final or initial position rather than just an overall displacement.

The Fourth Kinematic Equation

Our last kinematic equation is read as “final velocity squared equals initial velocity squared plus two times acceleration times displacement”. It’s worth noting that this is the only kinematic equation without time in it. Many starting physicists have been stumped by reaching a problem without a value for time. While staring at an equation sheet riddled with letters and numbers can be overwhelming, remembering you have this one equation without time will come up again and again throughout your physics career.

It may be worth noting that all of these are kinematic equations for constant acceleration. While this may seem like a limitation, we learned before that high school physics courses generally utilize constant acceleration so we don’t need to worry about it changing yet. If you do find yourself in a more advanced course, new physics equations will be introduced at the appropriate times.

How to Approach a Kinematics Problem

So now that we have all of these different kinematic equations, how do we know when to use them? How can we look at a physics word problem and know which of these equations to apply? You must use problem-solving steps. Follow these few steps when trying to solve any complex problems, and you won’t have a problem.

Step 1: Identify What You Know

This one probably seems obvious, but skipping it can be disastrous to any problem-solving endeavor. In physics problems, this just means pulling out values and directions. If you can add the symbol to go with the value (writing t=5\text{ s} instead of just 5\text{ s} , for example), even better. It’ll save time and make future steps even easier.

Step 2: Identify the Goal

In physics, this means figuring out what question you’re actually being asked. Does the question want you to find the displacement? The acceleration? How long did the movement take? Figure out what you’re being asked to do and then write down the symbol of the value you’re solving for with a question mark next to it ( t=\text{?} , for example). Again, this feels obvious, but it’s also a vital step.

Step 3: Gather Your Tools

Generally, this means a calculator and an equation. You’ll want to look at all of the symbols you wrote down and pick the physics equation for all of them, including the unknown value. Writing everything down beforehand will make it easier to pull a relevant equation than having to remember what values you need while searching for the right equation. You can use the latter method, but you’re far more likely to make a mistake and feel frustrated that way.

Step 4: Put it all Together

Plug your values into your equation and solve for the unknown value. This will usually be your last step, though you may find yourself having to repeat it a few times for exceptionally complex problems. That probably won’t come up for quite a while, though. After you’ve found your answer, it’s generally a good idea to circle it to make it obvious. That way, whoever is grading you can find it easily and you can easily keep track of which problems you’ve already completed while flipping through your work.

Kinematic Equation 1: Review and Examples

To learn how to solve problems with these new, longer equations, we’ll start with v=v_{0}+at . This kinematic equation shows a relationship between final velocity, initial velocity, constant acceleration, and time. We will explore this equation as it relates to physics word problems. This equation is set up to solve for velocity, but it can be rearranged to solve for any of the values it contains. For this physics equation and the ones following, we will look at one example finding the variable that has already been isolated and one where a new variable needs to be isolated using the steps we just outlined. So, let’s jump into applying this kinematic equation to a real-world problem.

A car sits at rest waiting to merge onto a highway. When they have a chance, they accelerate at 4\text{ m/s}^2 for 7\text{ s} . What is the car’s final velocity?

We have a clearly stated acceleration and time, but there’s no clearly defined initial velocity here. Instead, we have to take this from context. We know that the car “sits at rest” before it starts moving. This means that our initial velocity in this situation is zero. Other context clues for an object starting at rest is if it is “dropped” or if it “falls”. Our other known values will be even easier to pull as we were actually given numerical values. Now it’s time to put everything into a list.

• v_{0}=0\text{ m/s}
• a=4\text{ m/s}^2
• t=7\text{ s}

Our goal here was clearly stated: find the final velocity. We’ll still want to list that out so we can see exactly what symbols we have to work with on this problem.

We already know which of the kinematic equations we’re using, but if we didn’t, this would be where we search our equation sheet for the right one. Regardless, we’ll want to write that down too.

Step 4: Put it All Together

At this point, we’ll plug all of our values into our kinematic equation. If you’re working on paper, there’s no need to repeat anything we’ve put above. That being said, for the purposes of digital organization and so you can see the full problem in one spot, we will be rewriting things here.

Now let’s get a bit trickier with a problem that will require us to rearrange our kinematic equation.

A ball rolls toward a hill at 3\text{ m/s} . It rolls down the hill for 5\text{ s} and has a final velocity of 18\text{ m/s} . What was the ball’s acceleration as it rolled down the hill?

Just like before, we’ll make a list of our known values:

• v_{0}=3\text{ m/s}
• t=5\text{ s}
• v=18\text{ m/s}

Again, our goal was clearly stated, so let’s add it to our list:

We already know which equation we’re using, but let’s pretend we didn’t. We know that we need to solve for acceleration, but if you look at our original list of kinematic equations, there isn’t one that’s set up to solve for acceleration:

This begs the question, how to find acceleration (or any value) that hasn’t already been solved for? The answer is to rearrange an equation. First, though, we need to pick the right one. We start by getting rid of the second equation in this list as it doesn’t contain acceleration at all. Our options are now:

• \Delta x=v_{0}t+\dfrac{1}{2}at^{2}
• v^{2}=v_{0}^{2}+2a\Delta x

Now we’ll need to look at the first list we made of what we know. We know the initial velocity, time, and final velocity. There’s only one equation that has all the values we’re looking for and all of the values we know with none that we don’t. This is the first kinematic equation:

In this case, we knew the kinematic equation coming in so this process of elimination wasn’t necessary, but that won’t often be the case in the future. You’ll likely have to find the correct equation far more often than you’ll have it handed to you. It’s best to practice finding it now while we only have a few equations to work with.

Like before, we’ll be rewriting all of our relevant information below, but you won’t need to if you’re working on paper.

Although you can plug in values before rearranging the equation, in physics, you’ll usually see the equation be rearranged before values are added. This is mainly done to help keep units where they’re supposed to be and to avoid any mistakes that could come from moving numbers and units rather than just a variable. We’ll be taking the latter approach here. Follow the standard PEMDAS rules for rearranging the equation and then write it with the variable we’ve isolated on the left. While that last part isn’t necessary, it is a helpful organizational practice:

For a review of solving literal equations, visit this post ! Now we can plug in those known values and solve:

Kinematic Equation 2: Review and Examples

Next up in our four kinematics equations is \Delta x=\dfrac{v+v_{0}}{2} t . This one relates an object’s displacement to its average velocity and time. The right-hand side shows the final velocity plus the initial velocity divided by two – the sum of some values divided by the number of values, or the average. Although this equation doesn’t directly show a constant acceleration, it still assumes it. Applying this equation when acceleration isn’t constant can result in some error so best not to apply it if a changing acceleration is mentioned.

A car starts out moving at 10\text{ m/s} and accelerates to a velocity of 24\text{ m/s} . What displacement does the car cover during this velocity change if it occurs over 10\text{ s} ?

• v_{0}=10\text{ m/s}
• v=24\text{ m/s}
• t=10\text{ s}
• \Delta x=\text{?}
• \Delta x=\dfrac{v+v_{0}}{2} t

This time around we won’t repeat everything here. Instead, We’ll jump straight into plugging in our values and solving our problem:

A ball slows down from 15\text{ m/s} to 3\text{ m/s} over a distance of 36\text{ m} . How long did this take?

• v_{0}=15\text{ m/s}
• v=3\text{ m/s}
• \Delta x=36\text{ m}

We don’t have a kinematic equation for time specifically, but we learned before that we can rearrange certain equations to solve for different variables. So, we’ll pull the equation that has all of the values we need and isolate the variable we want later:

Again, we won’t be rewriting anything, but we will begin by rearranging our equation to solve for time:

Now we can plug in our known values and solve for time.

Kinematic Equation 3: Review and Examples

Our next kinematic equation is \Delta x=v_{0}t+\frac{1}{2}at^{2} . This time we are relating our displacement to our initial velocity, time, and acceleration. The only odd thing you may notice is that it doesn’t include our final velocity, only the initial. This equation will come in handy when you don’t have a final velocity that was stated either directly as a number or by a phrase indicating the object came to rest. Just like before, we’ll use this equation first to find a displacement, and then we’ll rearrange it to find a different value.

A rocket is cruising through space with a velocity of 50\text{ m/s} and burns some fuel to create a constant acceleration of 10\text{ m/s}^2 . How far will it have traveled after 5\text{ s} ?

• v_{0}=50\text{ m/s}
• a=10\text{ m/s}^2
• \Delta x=v_{0}t+\frac{1}{2}at^{2}

At this point, it appears that these problems seem to be quite long and take several steps. While that is an inherent part of physics in many ways, it will start to seem simpler as time goes on. This problem presents the perfect example. While it may have been easy to combine lines 4 and 5 mathematically, they were shown separately here to make sure the process was as clear as possible. While you should always show all of the major steps of your problem-solving process, you may find that you are able to combine some of the smaller steps after some time of working with these kinematic equations.

Later in its journey, the rocket is moving along at 20\text{ m/s} when it has to fire its thrusters again. This time it covers a distance of 500\text{ m} in 10\text{ s} . What was the rocket’s acceleration during this thruster burn?

• v_{0}=20\text{ m/s}
• \Delta x=500\text{ m}

As usual, we’ll begin by rearranging the equation, this time to solve for acceleration.

Now we can plug in our known values to find the value of our acceleration.

Kinematic Equation 4: Review and Examples

The last of the kinematic equations that we will look at is v^{2}=v_{0}^{2}+2a\Delta x . This one is generally the most complicated looking, but it’s also incredibly important as it is our only kinematic equation that does not involve time. It relates final velocity, initial velocity, acceleration, and displacement without needing a time over which a given motion occurred. For this equation, as with the others, let’s solve it as is and then rearrange it to solve for a different variable.

A car exiting the highway begins with a speed of 25\text{ m/s} and travels down a 100\text{ m} long exit ramp with a deceleration (negative acceleration) of 3\text{ m/s}^2 . What is the car’s velocity at the end of the exit ramp?

• v_{0}=25\text{ m/s}
• \Delta x=100\text{ m}
• a=-3\text{ m/s}^2

Note that our acceleration here is a negative value. That is because our problem statement gave us a deceleration instead of an acceleration. Whenever you have a deceleration, you’ll make the value negative to use it as an acceleration in your problem-solving. This also tells us that our final velocity should be less than our initial velocity so we can add that to the list of what we know as well.

• Final velocity will be less than initial.

Being able to know something to help check your answer at the end is what makes this subject a bit easier than mathematics for some students.

While we generally try to not have any operations going on for the isolated variable, sometimes it’s actually easier that way. Having your isolated variable raised to a power is generally a time to solve before simplifying. This may seem like an arbitrary rule, and in some ways it is, but as you continue through your physics journey you’ll come up with your own practices for making problem-solving easier.

Now that we have both sides simplified, we’ll take the square root to eliminate the exponent on the left-hand side:

If we remember back at the beginning, we said that our final velocity would have to be less than our initial velocity because the problem statement told us that we were decelerating. Our initial velocity was 25\text{ m/s} which is, indeed, greater than 5\text{ m/s} so our answer checks out.

A ghost is sliding a wrench across a table to terrify the mortal onlooker. The wrench starts with a velocity of 2\text{ m/s} and accelerates to a velocity of 5\text{ m/s} over a distance of 7\text{ m} . What acceleration did the ghost move the wrench with?

• v_{0}=2\text{ m/s}
• v=5\text{ m/s}
• \Delta x=7\text{ m}

We can also make an inference about our acceleration here – that it will be positive. Not every problem will tell you clearly the direction of the acceleration, but if your final velocity is greater than your initial velocity, you can be sure that your acceleration will be positive.

• Positive acceleration

You’ll get better at picking up on subtle hints like this as you continue your physics journey and your brain starts naturally picking up on some patterns. You’ll likely find this skill more and more helpful as it develops and as problems get more difficult.

We’ll start by rearranging our equation to solve for acceleration.

As usual, now that we’ve rearranged our equation, we can plug in our values.

Again, we can go back to the beginning when we said our acceleration would be a positive number and confirm that it is. 

Problem-Solving Strategies

At this point, you’re likely getting the sense that physics will be a lot of complex problem-solving. If so, your senses are correct. In many ways, physics is the science of explaining nature with mathematical equations. There’s a lot that goes into developing and applying these equations, but at this point in your physics career, you’ll find that the majority of your time will likely be spent on applying equations to word problems. If you feel that your problem-solving skills could still use some honing, check out more examples and strategies from this post by the Physics Classroom or through this video-guided tutorial from Khan Academy.

That was a lot of equations and examples to take in. Eventually, whether you’re figuring out how to find a constant acceleration or how to solve velocity when you don’t have a value for time, you’ll know exactly which of the four kinematic equations to apply and how. Just keep the problem-solving steps we’ve used here in mind, and you’ll be able to get through your physics course without any unsolvable problems.

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2.6 Problem-Solving Basics for One-Dimensional Kinematics

Learning objectives.

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

• Apply problem-solving steps and strategies to solve problems of one-dimensional kinematics.
• Apply strategies to determine whether or not the result of a problem is reasonable, and if not, determine the cause.

Problem-solving skills are obviously essential to success in a quantitative course in physics. More importantly, the ability to apply broad physical principles, usually represented by equations, to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations, whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both for solving problems in this text and for applying physics in everyday and professional life.

Problem-Solving Steps

While there is no simple step-by-step method that works for every problem, the following general procedures facilitate problem solving and make it more meaningful. A certain amount of creativity and insight is required as well.

Examine the situation to determine which physical principles are involved . It often helps to draw a simple sketch at the outset. You will also need to decide which direction is positive and note that on your sketch. Once you have identified the physical principles, it is much easier to find and apply the equations representing those principles. Although finding the correct equation is essential, keep in mind that equations represent physical principles, laws of nature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numerical solution is meaningless.

Make a list of what is given or can be inferred from the problem as stated (identify the knowns) . Many problems are stated very succinctly and require some inspection to determine what is known. A sketch can also be very useful at this point. Formally identifying the knowns is of particular importance in applying physics to real-world situations. Remember, “stopped” means velocity is zero, and we often can take initial time and position as zero.

Identify exactly what needs to be determined in the problem (identify the unknowns) . In complex problems, especially, it is not always obvious what needs to be found or in what sequence. Making a list can help.

Find an equation or set of equations that can help you solve the problem . Your list of knowns and unknowns can help here. It is easiest if you can find equations that contain only one unknown—that is, all of the other variables are known, so you can easily solve for the unknown. If the equation contains more than one unknown, then an additional equation is needed to solve the problem. In some problems, several unknowns must be determined to get at the one needed most. In such problems it is especially important to keep physical principles in mind to avoid going astray in a sea of equations. You may have to use two (or more) different equations to get the final answer.

Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units . This step produces the numerical answer; it also provides a check on units that can help you find errors. If the units of the answer are incorrect, then an error has been made. However, be warned that correct units do not guarantee that the numerical part of the answer is also correct.

Check the answer to see if it is reasonable: Does it make sense? This final step is extremely important—the goal of physics is to accurately describe nature. To see if the answer is reasonable, check both its magnitude and its sign, in addition to its units. Your judgment will improve as you solve more and more physics problems, and it will become possible for you to make finer and finer judgments regarding whether nature is adequately described by the answer to a problem. This step brings the problem back to its conceptual meaning. If you can judge whether the answer is reasonable, you have a deeper understanding of physics than just being able to mechanically solve a problem.

When solving problems, we often perform these steps in different order, and we also tend to do several steps simultaneously. There is no rigid procedure that will work every time. Creativity and insight grow with experience, and the basics of problem solving become almost automatic. One way to get practice is to work out the text’s examples for yourself as you read. Another is to work as many end-of-section problems as possible, starting with the easiest to build confidence and progressing to the more difficult. Once you become involved in physics, you will see it all around you, and you can begin to apply it to situations you encounter outside the classroom, just as is done in many of the applications in this text.

Unreasonable Results

Physics must describe nature accurately. Some problems have results that are unreasonable because one premise is unreasonable or because certain premises are inconsistent with one another. The physical principle applied correctly then produces an unreasonable result. For example, if a person starting a foot race accelerates at 0 . 40 m/s 2 0 . 40 m/s 2 for 100 s, his final speed will be 40 m/s (about 150 km/h)—clearly unreasonable because the time of 100 s is an unreasonable premise. The physics is correct in a sense, but there is more to describing nature than just manipulating equations correctly. Checking the result of a problem to see if it is reasonable does more than help uncover errors in problem solving—it also builds intuition in judging whether nature is being accurately described.

Use the following strategies to determine whether an answer is reasonable and, if it is not, to determine what is the cause.

Solve the problem using strategies as outlined and in the format followed in the worked examples in the text . In the example given in the preceding paragraph, you would identify the givens as the acceleration and time and use the equation below to find the unknown final velocity. That is,

Check to see if the answer is reasonable . Is it too large or too small, or does it have the wrong sign, improper units, …? In this case, you may need to convert meters per second into a more familiar unit, such as miles per hour.

This velocity is about four times greater than a person can run—so it is too large.

If the answer is unreasonable, look for what specifically could cause the identified difficulty . In the example of the runner, there are only two assumptions that are suspect. The acceleration could be too great or the time too long. First look at the acceleration and think about what the number means. If someone accelerates at 0 . 40 m/s 2 0 . 40 m/s 2 , their velocity is increasing by 0.4 m/s each second. Does this seem reasonable? If so, the time must be too long. It is not possible for someone to accelerate at a constant rate of 0 . 40 m/s 2 0 . 40 m/s 2 for 100 s (almost two minutes).

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Kinematics in Two Dimensions

Practice problem 1.

• the total distance of the entire trip
• the total displacement of the entire trip
• the average speed of the entire trip
• the average velocity of the entire trip
• the average acceleration of the entire trip

Distance? No problem. First I walked 6.0 km and then I walked 10 km for a total of 16 km. Distance is a scalar quantity, so the individual distances add just like regular numbers.

Displacement is a bit more challenging. Displacement is a vector and vectors have direction, so it's best to diagram this problem (a procedure that's remarkably useful in general). The resultant displacement is the vector sum of the two displacements experienced during the trip. Since they're perpendicular to one another, the resultant is the hypotenuse of a right triangle. Its magnitude can be found using pythagorean theorem and its direction can be found using the tangent function.

r  =  11.7 km at 59° west of north

The speed was 6.0 km/h for the first 6.0 km and 5 km/h for the last 10 km. The naive solution is to average the speeds using the add-and-divide method taught in junior high school. This method is wrong, not because the method itself is wrong, but because it doesn't apply to this situation.

The weights of the two segments are not equal. The second segment lasted twice as long as the first (as you will soon see).

Go back to the definition to solve this problem. Average speed is the total distance (which we've already found) divided by the total time (which we need to find). Since time is a scalar, add the times for each leg of the journey to get the total time.

The average speed is then…

The velocity was 6.0 km/h  north over the first 6.0 km and 5 km/h  west over the last 10 km. Average velocity is the total displacement divided by the total time. Both of these quantities have already been determined.

Acceleration in this context is relatively meaningless. It would be better to illustrate acceleration in two dimensions with a different problem (like the one below).

practice problem 2

• the speed of the current
• the magnitude of the swimmer's resultant velocity
• the direction of the swimmer's resultant velocity
• the time it takes the swimmer to cross the river

Since distance and velocity are directly proportional, this begins as a similar triangles problem.

Since speed and distance are directly proportional, the ratio of the downstream distance to the width of the river is the same as the ratio of the current speed to the swimmer's speed.

Determining the resultant velocity is a simple application of Pythagorean theorem.

Direction angles are often best determined using the tangent function. This problem is no exception. The only thing open to discussion is our choice of angle. I suggest using the angle between the resultant velocity and the displacement vector that points directly across the river, but this is just my preference. Be sure to indicate that the resultant lies on a particular side of this vector for clarity.

This is where it gets interesting. By now you should understood that time is the ratio of displacement to velocity. This is a vector problem, so direction matters. This is why we should probably use the words displacement and velocity instead of distance and speed. The only question is which distance and which speed should we use? The simple answer is pick the pair you like the best, just be sure they point in the same direction . It works along either of the component directions…

It also works along the resultant direction…

There's an interesting sideline to this question that astute readers might have noticed when looking at the first ratio in the chain of three shown above. The time it takes to cross a river by a swimmer swimming straight across is independent of the speed of the river. The only factors that matter are the speed of the swimmer and the width of the river. This swimmer will always cross the river in 50 s regardless of the speed of the river. 1 m/s, 10 m/s, 100 m/s, it doesn't matter. This example is a perfect illustration of an idea to be presented in the next section of this book. Motion in two dimensions can be thoroughly described with two independent one-dimensional equations. This idea is central to the field of analytical geometry.

practice problem 3

Finding the change in velocity is complicated in this problem by the change in direction. A diagram is indispensable. Let's assume that the initial direction of the car is 0° (to the right in standard position) and that the final velocity will be 90° (toward the top of the page in standard position). The difference of two vectors drawn this way would then connect the the head of the initial vector to the head of the final vector. Use Pythagorean Theorem for magnitude and tangent for direction as usual. Only after we have done all of this can we then plug numbers into the definition.

a  =  20 m/s 2 at 143°

practice problem 4

Start with a diagram.

Strip it down to its essence.

Two sides of this triangle are given ( v asteroid and v impact ). None of the angles are known. The third side ( v earth ) can be determined from basic knowledge. The average speed of the Earth is the distance covered in one orbit (the circumference) divided by the time it takes to complete that orbit (one year). We could do this on a hand held calculator…

or use an online calculator that knows the Earth-Sun distance…

We now have three sides of a triangle and can find the desired angle using the law of cosines.

a 2  =  b 2  + c 2  − 2 bc  cos  A

Solve algebraically, substitute numerical values, and compute the answer.

Physics Problems with Solutions

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Wolfram Demonstrations Project

1d kinematics problem solver.

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Contributed by: Daniel M. Smith, Jr.   (April 2011) South Carolina State University, [email protected] Open content licensed under CC BY-NC-SA

Example: A ball is thrown straight up with speed of 13.0 m/s from the top of a 40.0 m tall building. (a) How high above the building's rooftop does the ball rise? (b) If the ball misses the building on the way down, how much time does it take for the ball to strike the ground after it is thrown? (c) What is the ball's velocity as it strikes the ground?

Snapshot 1: (a) 8.6 m

This Demonstration lets you solve problems without using equations.

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Daniel M. Smith, Jr. "1D Kinematics Problem Solver" http://demonstrations.wolfram.com/1DKinematicsProblemSolver/ Wolfram Demonstrations Project Published: April 11 2011

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2D Kinematics Problem-solving

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Kinematics is used in astrophysics to describe the motion of celestial bodies and collections of such bodies. In mechanical engineering, robotics, and biomechanics, kinematics is used to describe the motion of systems composed of joined parts (multi-link systems) such as an engine, a robotic arm, or the human skeleton.

The most important thing to remember in 2D kinematics problems is that the two dimensions are entirely independent of each other. So, that means you are never actually doing a 2D kinematics problem, but you are always doing two 1D kinematics problems at the same time. Literally, it is the same in the two problems; that's all that connects them.

An airplane is taking off on the runway. At the moment the wheels leave the ground, the plane is traveling at \(60\text{ m/s}\) horizontally. The wings generate a lift, which causes a vertical acceleration of \(1\text{ m/s}^2,\) and the jets produce a horizontal acceleration of \(2\text{ m/s}^2.\) When the airplane is flying at an altitude of \(3200\text{ m},\) what is its horizontal distance from the airport? Suppose the plane starts from rest and gains altitude (ignore the horizontal traveling of the plane for now). Then using the formula \[ v^2-u^2 = 2as,\] we can get the vertical velocity. As the plane starts from rest, the initial velocity \(u\) will be zero for the plane. Therefore, \[v^2 =2as.\] Since \(a \text{ (vertical)} =1 \text{ m/s}^2\) and \(s = 3200\text{ m},\) it follows that \(v= 80\text{ m/s}.\) Use this \(v\) to obtain the time of flight: \[\begin{align} v&=u +at \\ &=0+at\\\\ \Rightarrow t&= 80 \text{ sec}. \end{align}\] Now, by the formula \[s = u \text{ (horizontal)}\, t + \frac{1}{2} at^2,\] the horizontal distance is \[s = 60\times 80 + \frac{1}{2}\times 2\times 80^2 = 11200\text{ m}.\ _\square\]

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9.8: Problem Solving

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learning objectives

• Develop and apply a strong problem-solving strategy for rotational kinematics

Problem-Solving Strategy For Rotational Kinematics

When solving problems on rotational kinematics:

• Examine the situation to determine that rotational kinematics (rotational motion) is involved. Rotation must be involved, but without the need to consider forces or masses that affect the motion.
• Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful.
• Make a list of what is given or can be inferred from the problem as stated (identify the knowns).
• Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog because by now you are familiar with such motion.
• Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units. Be sure to use units of radians for angles.
• Check your answer to see if it is reasonable: Does your answer make sense?

Example \(\PageIndex{1}\):

Suppose a large freight train accelerates from rest, giving its 0.350 m radius wheels an angular acceleration of 0.250 rad/s 2 . After the wheels have made 200 revolutions (assume no slippage): (a) How far has the train moved down the track? (b) What are the final angular velocity of the wheels and the linear velocity of the train?

In part (a), we are asked to find x, and in (b) we are asked to find ω and v. We are given the number of revolutions θ, the radius of the wheels r, and the angular acceleration α.

The distance x is very easily found from the relationship between distance and rotation angle: \(\mathrm{θ=\frac{x}{r}}\).

Solving this equation for x yields \(\mathrm{x=rθ}\).

Before using this equation, we must convert the number of revolutions into radians, because we are dealing with a relationship between linear and rotational quantities:

\[\mathrm{θ=(200 rev)(\dfrac{2π \; rad}{1 \; rev})=1257 \; rad.}\]

Substitute the known values into \(\mathrm{x=rθ}\) to find the distance the train moved down the track:

\[\mathrm{x=rθ=(0.350 \;m)(1257 \; rad)=440 \; m.}\]

We cannot use any equation that incorporates t to find ω, because the equation would have at least two unknown values. The equation \(\mathrm{ω^2=ω_0^2+2 \alpha \theta}\) will work, because we know the values for all variables except ω. Taking the square root of this equation and entering the known values gives

\[\begin{align} \mathrm{ω} & \mathrm{=\sqrt{0+2(0.250 \; rad/s^2)(1257 \; rad)}} \\ & \mathrm{=25.1 \; rad/s} \end{align}\]

One may find the linear velocity of the train, v, through its relationship to ω:

\[\mathrm{v=rω=(0.350 \; m)(25.1 \; rad/s)=8.77 \; m/s}\]

Rotational motion : Part of a series of videos on physics problem-solving. The problems are taken from “The Joy of Physics. ” This one deals with angular motion. The viewer is urged to pause the video at the problem statement and work the problem before watching the rest of the video.

Equation list : Rotational and translational kinematic equations.

• Examine the situation to determine that rotational kinematics (rotational motion ) is involved, and identify exactly what needs to be determined.
• Make a list of what is given or can be inferred from the problem as stated and solve the appropriate equations.
• Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units.
• kinematics : The branch of mechanics concerned with objects in motion, but not with the forces involved.

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CC LICENSED CONTENT, SPECIFIC ATTRIBUTION

• OpenStax College, Kinematics of Rotational Motion. September 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m42178/latest/ . License : CC BY: Attribution
• kinematics. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/kinematics . License : CC BY-SA: Attribution-ShareAlike
• Rotational motion. Located at : http://www.youtube.com/watch?v=FZ8U_3qZQqs . License : Public Domain: No Known Copyright . License Terms : Standard YouTube license
• OpenStax College, Kinematics of Rotational Motion. February 8, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m42178/latest/ . License : CC BY: Attribution