acceleration problem solving grade 7

Physics Problems with Solutions

• Acceleration: Tutorials with Examples

Examples with explanations on the concepts of acceleration of moving object are presented. More problems and their solutions can also be found in this website.

Average Acceleration

An object with initial velocity v 0 at time t 0 and final velocity v at time t has an average acceleration between t 0 and t given by

Examples with soltutions

What is the acceleration of an object that moves with uniform velocity? Solution: If the velocity is uniform, let us say V, then the initial and final velocities are both equal to V and the definition of the acceleration gives

A car accelerates from rest to a speed of 36 km/h in 20 seconds. What is the acceleration of the car in m/s 2 ? Solution: The initial velocity is 0 (from rest) and the final velocity is 36 km/h. Hence

More References and links

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

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• Physics Formulas

Acceleration Formula

One may have perceived that pushing a terminally ill bus can give it a sudden start. That’s because lift provides an upward push when it starts. Here velocity changes and this is acceleration! Henceforth, the frame accelerates. Acceleration is described as the rate of change of velocity of an object. A body’s acceleration is the final result of all the forces being applied to the body, as defined by Newton’s second law. Acceleration is a vector quantity that is described as the frequency at which a body’s velocity changes.

Formula of Acceleration

Acceleration is the rate of change in velocity to the change in time. It is denoted by symbol a and is articulated as-

The  S.I  unit for acceleration is meter per second square or m/s 2 .

• Final Velocity is v
• Initial velocity is u
• Acceleration is a
• Time taken is t
• Distance traveled is s

Acceleration Solved Examples

Underneath we have provided some sample numerical based on acceleration which might aid you to get an idea of how the formula is made use of:

Problem 1:  A toy car accelerates from 3 m/s to 5 m/s in 5 s. What is its acceleration? Answer:

Given: Initial Velocity u = 3  m/s, Final Velocity v = 5m/s, Time taken t = 5s.

Problem 2:  A stone is released into the river from a bridge. It takes 4s for the stone to touch the river’s water surface. Compute the height of the bridge from the water level.

(Initial Velocity) u = 0 (because the stone was at rest), t = 4s (t is Time taken) a = g = 9.8 m/s 2 , (a is Acceleration due to gravity) distance traveled by stone = Height of bridge  = s The distance covered is articulated by

s = 0 + 1/2 × 9.8 × 4 = 19.6 m/s 2

Therefore, s = 19.6 m/s 2

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Grade 7- Quarter 3- Acceleration

Quiz   by Maria Alma D. Feliña

Feel free to use or edit a copy

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Measures 1 skill from Grade 7 Science Philippines Curriculum: Grades K-10 (MELC)

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• Q 1 / 3 Score 0 What is a  change in velocity over time? 29 Users re-arrange answers into correct order

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• Q 1 What is a  change in velocity over time? Users re-arrange answers into correct order Jumble 30 s S7FE-IIIa-1
• Q 2 If an object has a constant velocity but it changes its direction, it still accelerates. true false True or False 30 s S7FE-IIIa-1
• Q 3 An car accelerates from 20 m/s ( Initial Velocity) to 40 m/s ( Final velocity) in 4 s. Calculate its acceleration. 8 m/s 2 4 m/s 2 10 m/s 2 5 m/s 2 120 s S7FE-IIIa-1

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Grade 7 Acceleration

Displaying top 8 worksheets found for - Grade 7 Acceleration .

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Found worksheet you are looking for? To download/print, click on pop-out icon or print icon to worksheet to print or download. Worksheet will open in a new window. You can & download or print using the browser document reader options.

1. Acceleration worksheet pdf

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Acceleration

Practice problem 1.

• A car is said to go "zero to sixty in six point six seconds". What is its acceleration in m/s 2 ?
• The driver can't release his foot from the gas pedal (a.k.a. the accelerator). How many additional seconds would it take for the driver to reach 80 mph assuming the aceleration remains constant?
• A car moving at 80 mph has a speed of 35.8 m/s. What acceleration would it have if it took 5.0 s to come to a complete stop?

Well first of all, we shouldn't be dealing with English units. They're difficult to work with, so let's convert them straight away and then do the old "plug and chug".

Since the question asked for acceleration and acceleration is a vector quantity this answer is not complete. A proper answer must include a direction as well. This is quite easy to do. Since the car is starting from rest and moving forward, its acceleration must also be forward. The ultimate, complete answer to this problem is the car is accelerating at…

a  =  4.06 m/s 2  forward

We should convert the final speed to SI units.

Use the fact that change equals rate times time, and then add that change to our velocity at the end of the previous problem. Algebra will do the rest for us.

Alternate solution. We don't need no stinkin' conversions with this method. The ratio of eighty to sixty is a simple one, namely 4 3 . From our definition of acceleration, it should be apparent that time is directly proportional to change in velocity when acceleration is constant. Thus…

This is not the answer. It is the time elapsed from the moment when the car began to move. The question was about the additional time needed, so we should subtract the time required to go from zero to sixty. Thus…

∆ t  =  8.8 s − 6.6 s  =  2.2 s

The two methods give essentially the same answer.

Quite simple. Let's do it.

Nothing surprising there except the negative sign. When a vector quantity is negative what does it mean? There are several interpretations of this, but I think mine is the best. When a vector has a negative value, it means that it points in a direction opposite that of the positive vectors. In this problem, since the positive vectors are assumed to point forward (What other direction would a normal car drive?) the acceleration must be backward. Thus the complete answer to this problem is that the car's acceleration is…

a  =  7.16 m/s 2 backward

Although it is common to assign deceleration a negative value, negative acceleration does not automatically imply deceleration. When dealing with vector quantities, any direction can be assumed positive…

up, down, right, left, forward, backward, north, south, east, west

and the corresponding opposite direction assumed negative…

down, up, left, right, backward, forward, south, north, west, east.

It won't matter which you chose as long as you are consistent throughout a problem. Don't learn any rules for assigning signs to particular directions and don't let anyone tell you that a certain direction must be positive or must be negative.

practice problem 2

Acceleration is the rate of change of velocity with time. Since velocity is a vector, this definition means acceleration is also a vector. When it comes to vectors, direction matters as much as size. In a simple one-dimensional problem like this one, directions are indicated by algebraic sign. Every quantity that points away from the batter will be positive. Every quantity that points toward him will be negative. Thus, the ball comes in at −40 m/s and goes out at +50 m/s. If we didn't pay attention to this detail, we wouldn't get the right answer.

practice problem 3

Practice problem 4.

Acceleration

Acceleration is how fast velocity changes:

• Speeding up
• Slowing down (also called deceleration )
• Changing direction

It is usually shown as:

m/s 2   "meters per second squared"

What is this " per second squared " thing? An example will help:

A runner accelerates from 5 m/s (5 meters per second) to 6 m/s in just one second

So they accelerate by 1 meter per second per second

See how "per second" is used twice?

It can be thought of as (m/s)/s but is usually written m/s 2

So their acceleration is 1 m/s 2

The formula is:

Acceleration = Change in Velocity (m/s) Time (s)

Example: A bike race!

You are cruising along in a bike race, going a steady 10 meters per second (10 m/s).

Acceleration : Now you start cycling faster! You increase to 14 m/s over the next 2 seconds (still heading in the same direction):

Your speed  increases by 4 m/s , over a time period of 2 seconds , so:

= 4 m/s 2 s = 2 m/s 2

Your speed changes by 2 meters per second per second .

Or more simply "2 meters per second squared".

Example: You are running at 7 m/s, and skid to a halt in 2 seconds.

You went from 7 m/s to 0, so that is a decrease in speed:

= −7 m/s 2 s = −3.5 m/s 2

We don't always say it, but acceleration has direction (making it a vector ):

A car is heading West at 16 m/s . The driver flicks the wheel, and within 4 seconds has the car headed East at 16 m/s .   What is the acceleration?

The numbers are the same, but the direction is different!

Acceleration = From 16 m/s West to 16 m/s East 4 s

From 16 m/s West to 16 m/s East is a total change of 32 m/s towards the East.

Acceleration = 32 m/s East 4 s = 8 m/s 2 East

For more complicated direction changes read vectors .

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Speed Velocity Acceleration Grade 7

Speed Velocity Acceleration Grade 7 - Displaying top 8 worksheets found for this concept.

Some of the worksheets for this concept are Educator guide motion speed velocity and acceleration, Science topic, Acceleration and speed problems answer, Acceleration work, Speed velocity and acceleration calculations work, Council rock school district overview, Motion speed velocity acceleration, Motion distance and displacement.

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1. Educator Guide Motion Speed Velocity And Acceleration

2. science topic, 3. acceleration and speed problems answer sheet, 4. acceleration worksheet, 5. speed, velocity and acceleration calculations worksheet, 6. council rock school district / overview, 7. motion; speed; velocity; acceleration, 8. motion distance and displacement.

Acceleration Examples

Acceleration is a vector quantity that measures a change in speed or direction. It is defined as a change in velocity per unit of time. Given below is the acceleration examples problems with solution for your reference to calculate acceleration in m / s 2 .

Acceleration Example Problems

Let us consider the acceleration practice problem: A car accelerates uniformly from 22.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car.

We can calculate the Acceleration using the given formula.

Substituting the values in the formula,

Acceleration = (46.1 – 22.5 ) / 2.47 = 23.6 / 2.47 = 9.554 m/s 2

Therefore, the value of Acceleration is 9.554 m/s 2 .

Refer the below example of acceleration with solution. When an airplane accelerates down a runway at 3.20 m/s 2 to 5.41 m/s 2 for 28 s until is finally lifts off the ground, calculate its acceleration before its take off.

Substituting the values in the above given formula,

Acceleration = (5.41 – 3.20 ) / 28 = 2.21 / 28 = 0.078 m/s 2

Therefore, Acceleration is 0.078 m/s 2 .

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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

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Grade 7 Math Acceleration

• Intended Audience: Administrators, Parents, Teachers
• Subject: Math

Description

In this course, students will build understanding of the following modules: Transforming Geometric Objects, Developing Function Foundations, Modeling Linear Equations, Applying Powers, and Analyzing Populations and Probabilities.

Each module is broken up into topics where you will find teacher materials to guide the instruction and the student materials both used in the classroom for learning together and learning individually.

The agency developed these learning resources as a contingency option for school districts during COVID. All resources are optional. Prior to publication, materials go through a rigorous third-party review. Review criteria include TEKS alignment, support for all learners, progress monitoring, implementation supports, and more. Products also are subject to a focus group of Texas educators. 

Book Outline

• Program Materials
• Module 1 : Transforming Geometric Objects
• Module 2: Developing Function Foundations
• Module 3: Modeling Linear Equations
• Module 4: Applying Powers
• Module 5: Analyzing Populations, Probabilities, and Potential
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