examview - kinematics practice test 2013teacher.cgs.k12.va.us/kbywaters/rhs ap physics 1/lesson...

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Kinematics Practice Test 2013

Multiple ChoiceIdentify the choice that best completes the statement or answers the question.

1. Given a velocity time graph for a student skipping, literally skipping to the office to leave early, how would you figure her displacement in the first 30 seconds?a. can’t tell c. areab. y intercept d. slope

2. How far will a car travel in 43.3 seconds if it has a velocity of 18.1 m/s?a. 784 m c. –424 mb. 0.42 m d. 2.4 m

3. A honey bee takes off from rest and accelerates at 10.0 m/s2 for 0.8 seconds. How far does she fly in this time?a. 6.4 m c. 3.2 mb. 8.0 m d. 4.0 m

Short Answer

4. A car starts from rest and reaches 12 m/s after 3 seconds.

What does the word “rest” imply?

5. What base units of mass, length and time are used in the metric system to obtain results in standard derived metric units?

_____

6. A bus driver picks up his bus at the county schools transportation lot, drives his route, bringing 44 excited high school students anticipating a wonderful day of knowledge gathering. He drops them off at school then, completes his elementary school route. He does this again taking them all home in the afternoon and returns the bus to the same parking spot in the county lot. He was gone for 7.3 hours and the odometer reading in the bus increased by 107 miles. His top speed on the trip was 45.0 mph. What was his average velocity?

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7. A shark accelerates from 3.1 m/s to 20.1 m/s in 1.6 sec. How far did the shark swim during this time?

_______

8. An airplane decelerates from 245 m/s to 194 m/s after traveling 1400 meters.

a. What was its average acceleration over that interval?

b. How much time did this take?

_____

9. How much time is required to travel 78 meters with a velocity of 16.9 m/s?

_______

10. What velocity is required to travel 250 miles in 4.0 hours?



Name: ________________________ ID: D

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11. The graph above shows velocity as a function of time for a toy car.

a. How far did the car travel in the 14 seconds depicted?

b. What is the acceleration during the first 5 seconds?

______



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12. a. What is the velocity of the object whose position time graph is given above?

b. If the time to change velocity is 0.11 seconds, what is teh acceleration of the object?

________

13. A launch angle of 28 deg strikes a target, what other angle could be used to hit the same target?

_________

14. What are the conditions necessary to apply the range equation?

R v 0

2 sin2g

15. Our kinematics equations apply only to constant acceleration. How do we use them to analyze displacement of objects subject to changes in acceleration?

_______



Name: ________________________ ID: D

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Problem

16.

In a modern version of jousting, two fighter pilots face off with radar guided missiles. They are heading toward each other at the same altitude, 40 km apart when the “good” guy fires his missile. The “bad” guy is having a bad day, he doesn’t even know he’s in a fight and just keeps on flying toward impending doom at 210 m/s. The good guy’s airspeed was 200 m/s when he fired.

The missile’s performance is summarized below. acceleration 75.0 m/s2 top speed 800 m/s engine run time at top speed 25.0 sec Acceleration after motor burn out = –30.0 m/s2

a. The missile will only be able to fly level and maneuver for 54.7 sec. What is the maximum effective range of the missile?

1) How far will it fly while accelerating, and how long will it take to get to top speed?

2) How far will it fly at its max speed, and what time will the motor burn out?

3) How much further will the missile fly before it can’t maneuver anymore?



Name: ________________________ ID: D

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b. Now use the same data above to determine where the enemy plane is when it gets hit.



Name: ________________________ ID: D

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17. Not liking his cereal a whole lot, Mikey shoves it off the table. It leaves the edge of the table doing 3.7 m/s horizontally. The height of the table is 1.15 meters.

a. How long does it take for the bowl to hit the floor?

________

b. How far from the edge of the table does the bowl hit the floor?

________



ID: D

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Kinematics Practice Test 2013Answer Section

MULTIPLE CHOICE

1. ANS: Carea

PTS: 5 2. ANS: A

d=vt=18.1 x 43.3 = 784 m

PTS: 5 3. ANS: C

Have time so use:

d d 0 v 0 t 12 at2

d 0 0t 12 10.0 0.82 3.2 m

PTS: 5

SHORT ANSWER

4. ANS: initial velocity equals zero

PTS: 7 5. ANS:

kg, m, sUsing kilograms, meters and seconds will result in proper metric units with special names such as...Newton, joule, watt, Tesla, etc.

PTS: 7 6. ANS:

zeroSpeed = path length/time= 400/7.3=14.7 m/s.

PTS: 7



ID: D

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7. ANS: 19 mDd=v0t+1/2at^2

a=(vf-vo)/t

...or

d=(v0+vf)/2*t=(3.1+20.1)/2*1.6 =19 meters

PTS: 7 8. ANS:

–8.0 m/s2

a v f

2 v 02

2x 1942 2452

2 1400 8.0m / s2

t va

194 245 8.0 6.4sec

PTS: 7 9. ANS:

4.6 secd=vtt=d/v=78 / 16.9 =4.6 sec

PTS: 7 10. ANS:

63 mphd=vt

v=d/t = 250 / 4.0 = 63 mph

PTS: 7 11. ANS:

34.0 metersa. calculate the area

b. a=deltaV/time=slope=0.80 m/s2

PTS: 7



ID: D

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12. ANS: 2.0 m/sa. slope is velocity

b. a=Dv/dt=2.0/0.11=18.18 m/s2

PTS: 7 13. ANS:

90-28=72 deg

PTS: 1 14. ANS:

no air resistance, constant g, lands at same height as launched.

PTS: 1 15. ANS:

divide the motion into time intervals of constant acceleration. analyze them, add em up.

PTS: 1



ID: D

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PROBLEM

16. ANS:

Graph it to help visualize what’s happening.

Givens:

vbp=210 m/s, vgp=200 m/svgm=800 m/sagm=75.0 m/s2

R0=40 km=40000 m



ID: D

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max range=boost range+sustain range+motor out to stall speed.

=

max range=vmax

2 -v plane2

2a +vmax(t max-vmax-v plane

am)+ 1

2 a neg(tmax-tsust-vmax-v plane

am)=34,292 m

Equations:enemy plane



ID: D

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x bp x 0bp v bp t

missile

boost

xm v gp t 12 a gm t2 x 0bp v bp t

int erceptinboostphase

t (v gp v bp) (v gp v bp)2 2am x 0bp )

am 21.861sec

tboost vgm v gp

am

sustainphase

xm v gm

2 v gp2

2am vm (t tboost) x 0bp v bp t

int erceptinsustainphase

t x 0bp vm tboost

v gm2 vgp

2

2am

(vm v bp ) 41.980sec

motorburnedout

t2 tboost tsust

xm v max

2 v gp2

2am v max(t tboost )

12 a neg(t t2)2

x 0bp v bp t v max

2 v gp2

2am v max(t tboost)

12 a neg (t t2)2

12 a neg(t t2)2 v max(t tboost )

v max2 v gp

2

2am x 0bp v bp t 0

12 a neg t2 (a neg t2 v max v bp )t 1

2 a neg t22 v maxtboost v max

2 v gp2

2a m x 0bp 0

a 12 a neg 15

b a neg t2 v max v bp 2000

2 2



ID: D

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PTS: 1 17. ANS:

v 0x vo cos(0) 3.7

v 0y v0sin(0) 0

h h 0 v oyt 4.9t2 R v oxt

0 1.15 0t 4.9t2

t h4.9 0.5sec R=3.7 0.5 1.79m

PTS: 1



examview - kinematics practice test 2013teacher.cgs.k12.va.us/kbywaters/rhs ap physics 1/lesson...

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