sample problems and solutions222222

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Kinematics /Physics-

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The Physics Classroom(/) Physics Tutorial(/class) 1-D Kinematics(/Physics-Tutorial/1-D-Kinematics) Sample

Problems and Solutions

1-D Kinematics - Lesson 6 - Describing Motion with Equations

Sample Problems and SolutionsKinematic Equations(/class/1DKin/Lesson-6/Kinematic-Equations)

Kinematic Equations and Problem-Solving(/class/1DKin/Lesson-6/Kinematic-Equations-and-

Problem-Solving)

Kinematic Equations and Free Fall(/class/1DKin/Lesson-6/Kinematic-Equations-and-Free-Fall)

Sample Problems and Solutions

Kinematic Equations and Graphs(/class/1DKin/Lesson-6/Kinematic-Equations-and-Graphs)

Earlier in Lesson 6, four kinematic

equations(http://www.physicsclassroom.com/Class/1DKin/U1L6a.cfm) were introduced

and discussed. A useful problem-solving

strategy(http://www.physicsclassroom.com/Class/1DKin/U1L6b.cfm) was presented for

use with these equations and two examples were given that illustrated the use of the

strategy. Then, the application of the kinematic equations and the problem-solving

strategy to free-fall motion(http://www.physicsclassroom.com/Class/1DKin/U1L6c.cfm)

Student Extras Teacher's Guides
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was scusse an us ra e . n s par o esson , severa samp e pro ems w epresented. These problems allow any student of physics to test their understanding of

the use of the four kinematic equations to solve problems involving the one-dimensional

motion of objects. You are encouraged to read each problem and practice the use of the

strategy in the solution of the problem. Then click the button to check the answer or use

the link to view the solution.

Check Your Understanding

1. An airplane accelerates down a runway at 3.20 m/s for 32.8 s until is finally lifts offthe ground. Determine the distance traveled before takeoff.

See nswer

See solution below.

(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol1)

2. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a

distance of 110 m. Determine the acceleration of the car.

See nswer See solution below.


3. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60

seconds, what will be his final velocity and how far will he fall?

See nswer

See solution below.


4. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds.

Determine the acceleration of the car and the distance traveled.

2
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Share The News(/Share)

ee nswer

See solution below.


5. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of

gravity on the moon is 1.67 m/s . Determine the time for the feather to fall to the

surface of the moon.See nswer

See solution below.

6. Rocket-powered sleds are used to test the human response to acceleration. If a

rocket-powered sled is accelerated to a speed of 444 m/s in 1.83 seconds, then what

is the acceleration and what is the distance that the sled travels?

See nswer

See solution below.


7. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4

m. Determine the acceleration of the bike.

See nswer

See solution below.


8. An engineer is designing the runway for an airport. Of the planes that will use the

airport the lowest acceleration rate is likely to be 3 m/s The takeoff speed for this

2

2
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plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum

allowed length for the runway?

See nswer

See solution below.


9. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance

of the car (assume uniform acceleration).

See nswer

See solution below.


10. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speedof the kangaroo.

See nswer

See solution below.


11. If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and hishang time (total time to move upwards to the peak and then return to the ground)?

See nswer

See solution below.


12. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through

h b l f h ifl h b ll di f 0 840 D i h
http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol11http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol10http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol9http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol8


5/16

the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the

acceleration of the bullet (assume a uniform acceleration).

See nswer

See solution below.


13. A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determinethe height to which the ball rises before it reaches its peak. (Hint: the time to rise to

the peak is one-half the total hang-time.)

See nswer

See solution below.


14. The observation deck of tall skyscraper 370 m above the street. Determine the time

required for a penny to free fall from the deck to the street below.

See nswer

See solution below.

15. A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. Thebullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet

while moving into the clay. (Assume a uniform acceleration.)

See nswer

See solution below.

http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol15http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol13http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol12


6/16

16. A stone is dropped into a deep well and is heard to hit the water 3.41 s after being

dropped. Determine the depth of the well.

See nswer

See solution below.


17. It was once recorded that a Jaguar left skid marks that were 290 m in length.

Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90

m/s , determine the speed of the Jaguar before it began to skid.

See nswer

See solution below.


18. A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed.

Determine the acceleration of the plane and the time required to reach this speed.

See nswer

See solution below.


19. A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the

acceleration (assume uniform) of the dragster.

See nswer

See solution below.


2
http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol19http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol18http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol17http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol16


7/16

20. With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a

height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.

See nswer

See solution below.


Solutions to bove Problems

Given:

a = +3.2 m/s t = 32.8 s v = 0 m/s

Find:

d = ??

d = v *t + 0.5*a*t

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

d = 1720 m

Return to Problem 1

Given:

d = 110 m t = 5.21 s v = 0 m/s

Find:

a = ??

d = v *t + 0.5*a*t

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

110 m = (13.57 s )*a

a = (110 m)/(13.57 s )

a = 8.10 m/ s

Return to Problem 2

2i

i2

2 2

i

i2

2

2

2

2
http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#sol20


8/16

Given:

a = -9.8 m t = 2.6 s v = 0 m/s

Find:

d = ??

v = ??

d = v *t + 0.5*a*t

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

d = -33.1 m (- indicates direction)

v = v + a*t

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

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

Return to Problem 3

Given:

v = 18.5 m/s v = 46.1 m/s t = 2.47 s

Find:

d = ??

a = ??

a = (Delta v)/t

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

a = 11.2 m/s

d = v *t + 0.5*a*t

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

d = 45.7 m + 34.1 m

d = 79.8 m

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

Return to Problem

4(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q4)

Given:

0 / d 1 40 1 67 /

Find:

if

i2

2 2

f i

f2

f

i f

2

i2

2 2

f2

i2

2
http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q4


9/16

d = v *t + 0.5*a*t

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

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

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

1.68 s = t

t = 1.29 s

Return to Problem5(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q5)

Given:

v = 0 m/s v = 444 m/s t = 1.83 s

Find:

a = ??

d = ??

a = (Delta v)/t

a = (444 m/s - 0 m/s)/(1.83 s)a = 243 m/s

d = v *t + 0.5*a*t

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

d = 0 m + 406 m

d = 406 m

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

Return to Problem


Given:

v = 0 m/s v = 7.10 m/s d = 35.4 m

Find:

a = ??

v = v + 2*a*d

i2

2 2

2 2

2 2

2 2

i f

2

i2

2 2

f2

i2

i f

f2

i2

m s = m s + a m
http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q6http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q5


10/16

. m s = m s + a . m

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

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

a = 0.712 m/s

Return to Problem


Given:

v = 0 m/s v = 65 m/s a = 3 m/s

Find:

d = ??

v = v + 2*a*d

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

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

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

d = 704 mReturn to Problem


Given:

v = 22.4 m/s v = 0 m/s t = 2.55 s

Find:

d = ??

d = (v + v )/2 *t

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

d = (11.2 m/s)*2.55 s

d = 28.6 m

Return to Problem


2 2 2

2 2

2

i f2

f2

i2

2 2 2

2 2 2 2

2 2 2

i f

i f

n :
http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q9http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q8http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q7


11/16

a = -9.8 m/s v = 0 m/s d = 2.62 m

n :

v = ??

v = v + 2*a*d

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

0 m /s = v - 51.35 m /s

51.35 m /s = v

v = 7.17 m/s

Return to Problem10(http://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm#q10)

Given:

a = -9.8 m/s v = 0 m/s d = 1.29 m

Find:

v = ??

t = ??

v = v + 2*a*d

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

0 m /s = v - 25.28 m /s

25.28 m /s = v

v = 5.03 m/s

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

v = v + a*t

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

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

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

t = 0.513 s

hang time = 1.03 s

Return to Problem


2f i

f2

i2

2i2 2

2 2i2 2 2

2 2i2

i

2f

i

f2

i2

2 i2 2

2 2i2 2 2

2 2i2

i

f i

2up

2up

2up

up


12/16

Given:

v = 0 m/s v = 521 m/s d = 0.840 m

Find:

a = ??

v = v + 2*a*d

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

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

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

a = 1.62*10 m /s

Return to Problem


Given:

a = -9.8 m/s v = 0 m/s t = 3.13 s

Find:

d = ??

a. (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 = v + a*t

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

0 m/s = v - 30.7 m/s

v = 30.7 m/s (30.674 m/s)

Now use: v = v + 2*a*d

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

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

-940 m /s = (-19.6 m/s )*d

(-940 m /s )/(-19.6 m/s ) = d

d = 48.0 m

Return to Problem

13(htt // h i l /Cl /1DKi /U1L6d f # 13)

i f

f2

i2

2 2

2 2 2

2 2

5 2

2f

f i

i2

i

i

f

2

i

2

2 2 2

2 2 2 2 2

2 2 2

2 2 2


13/16

Given:

v = 0 m/s d = -370 m a = -9.8 m/s

Find:

t = ??

d = v *t + 0.5*a*t

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

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

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

75.5 s = t

t = 8.69 s

Return to Problem


Given:

v = 367 m/s v = 0 m/s d = 0.0621 m

Find:

a = ??

v = v + 2*a*d

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

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

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

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

a = -1.08*10 m /s

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

Return to Problem


Given:

9 8 / t 3 41 0 /

Find:

i2

i2

2 2

2 2

2 2

2 2

i f

f2

i2

2 2

2 2 2 2

2 2

2 2

6 2

2


14/16

d = v *t + 0.5*a*t

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

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

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem


Given:

a = -3.90 m/s v = 0 m/s d = 290 m

Find:

v = ??

v = v + 2*a*d

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

0 m /s = v - 2262 m /s

2262 m /s = v

v = 47.6 m /s

Return to Problem


Given:

v = 0 m/s v = 88.3 m/s d = 1365 m

Find:

a = ??t = ??

v = v + 2*a*d

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

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

7797 m /s = (2730 m)*a

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

a = 2.86 m/s

i2

2 2

2 2

2f i

f2

i2

2i2 2

2 2i2 2 2

2 2 i2

i

i f

f2

i2

2 2

2 2 2 2

2 2

2 2

2

*t


15/16

v = v + a*t

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

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

t = 30. 8 s

Return to Problem


Given:

v = 0 m/s v = 112 m/s d = 398 m

Find:

a = ??

v = v + 2*a*d

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

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

12544 m /s = (796 m)*a

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

a = 15.8 m/s

Return to Problem


Given:

a = -9.8 m/s v = 0 m/s d = 91.5 m

Find:

v = ??t = ??

First, find speed in units of m/s:

v = v + 2*a*d

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

0 m /s = v - 1793 m /s

1793 m /s = v

v = 42.3 m/s

f i

2

2

i f

f2

i2

2 2

2 2 2 2

2 2

2 2

2

2f

i

f2

i2

2i2 2

2 2i2 2 2

2 2i2

i

f / i/h


16/16

Now convert from m/s to mi/hr:

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

v = 94.4 mi/hr

Return to Problem


Next Section:

Kinematic Equations and Graphs(/class/1DKin/Lesson-6/Kinematic-Equations-and-Graphs)

i

i

home(/) about(/about) 1996-2016 The Physics Classroom, All rights reserved.
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