AP Physics

Chapter 2

Review

2

€

v =ΔxΔt

=21+ 66 −54( ) − 21+ 22− 6( )[ ]m

2sec= −2 m

sec

€

v =dvdt

=d 21+ 22t − 6t 2( )

dt= 22−12t

v =v1sec + v3 sec

2=

10−14( ) msec

2= −2 m

sec

1. The position of a particle moving along the x axis is given by x = (21 + 22t - 6.0 t2) m, where t is in seconds. What is the average velocity during the time interval t = 1.0 sec to t = 3.0 sec?

3

€

v f2 = vi

2 + 2aΔx

v f2 − vi

2

2Δx= a

220 msec( )2

− 450 msec( )2

2 0.14m( )= a

−550357 msec2 = a

2. A bullet is fired through a board, 14.0 cm thick, with its line of motion perpendicular to the face of the board. If it enters with a speed of 450

€

msec and emerges with a speed of 220

€

msec , what is

the bullet's acceleration as it passes through the board?

4

€

v =dvdt

=12t − 3t 2

a =dvdt

=12 −6t

At max v, set a = 0.

€

t = 2sec

€

x = 6 2( )2− 23

x =16m

3. The position of a particle moving along the x axis is given by x = 6.0t2 - 1.0t3, where x is in meters and t in seconds. What is the position of the particle when it achieves its maximum speed in the positive x direction?

5€

v =dxdt

= 24 − 6t 2

0 = 24 − 6t 2

t = 2sec

€

x = 24 t − 2t 3( )mx = 48m −16mx = 32m

4. A particle moving along the x axis has a position given by x = (24t - 2.0t3) m, where t is measured in seconds. How far is the particle from the origin (x = 0) when the particle is not moving?

Find the area under the line from t = 1 sec to t = 6 sec and add the results to x = 2m

6

5. Vx is the velocity of a particle moving along the x axis as shown below. If x = 2.0 m at t = 1.0 sec, what is the position of the particle at t = 6.0 sec?

€

t sec( )€

v msec( )

At t = 1 sec, x = 2m.

The two red areas cancel out and the green area = -3m.

€

2m − 3m = −1m7

€

v msec( )

€

t sec( )

Area is equal to the ∆v.

€

Δv = 12 6 m

sec2( ) 5sec( ) = 15 msec

€

v5sec =15 msec +15 m

sec = 30 msec

8

€

a msec2( )6. At t = 0, a particle is

located at x = 25m and has a velocity of 15

€

msec in the

positive x direction. The acceleration of the particle varies with time as shown in the diagram above. What is the velocity of the particle at t = 5.0 sec?

€

v = ΔdΔt

=6m

2.5sec= 2.4 m

sec

v =vi + v f

22v − v f = vi

2 2.4 msec( ) −2.8 m

sec = vi

vi = 2 msec

€

a =v f − vi

t=

2.8 msec − 2 m

sec

2.5seca = 0.32 m

sec2

9

7. A particle confined to motion along the x axis moves with constant acceleration from x = 2.0 m to x = 8.0 meters during a 2.5 second time interval. The velocity of the particle at x = 8.0 m is 2.8

€

msec . What is the acceleration during this time

interval?

10

€

v f2 = vi

2 + 2aΔx

a =v f

2 − vi2

2Δx

a =80 m

sec( )2− 40 m

sec( )2

2 200m( )a =12 m

sec2

8. An automobile moving along a straight track changes its velocity from 40

€

msec to 80

€

msec in a distance of 200 m. What is

the (constant) acceleration of the vehicle during this time?

€

Δx = v f t − 12 at 2

2 Δx − v f t( )t2 = a

2 40m − 10 msec( ) 2sec( )( )

2sec( )2 = a

a = −10 msec2

11

9. In 2.0 seconds, a particle moving with constant acceleration along the x axis goes from x = 10 m to x = 50 meters. The velocity at the end of this time interval is 10

€

msec . What is the

acceleration of the particle?

12

€

Δx = 12 at 2

2Δxt 2 = a

2 0.02m( )0.005sec( )2 = a

a =1600 msec2

10. An electron, starting from rest and moving with a constant acceleration, travels 2.0 cm in 5.0 msec. What is the magnitude of this acceleration?

11. A rocket, initially at rest, is fired vertically with an upward acceleration of 10

€

msec2 . At an altitude of 0.50 km, the engine

of the rocket cuts off. What is the maximum altitude it achieves?

€

vo = 0

€

vo = 0

€

vHighest = 0 We’re going to find the velocity of the rocket when the engine runs out of fuel (quits), first. Then we will find the height the rocket achieves when moving under only the influence of gravity.

€

v f2 = vi

2 + 2aΔy

v f2 = 2aΔy

v f = 2aΔy

v f = 2 10 msec2( ) 500m( )

v f =100 msec

14

€

vo = 0

€

vHighest = 0

€

v =100 msec

€

A€

B

Now we’ll find distance B and then add distance B to distance A.

€

v f2 = vi

2 + 2gΔy

0 = vi2 + 2gΔy

−vi2

2g=ΔyB

− 100 msec( )2

2 −9.8 msec2( )

= ΔyB

ΔyB = 510m€

H = ΔyA + ΔyB

H =1010m

15

16

The ball, after two seconds:

€

v f = vi + gt = 20 msec + −10 m

sec2( ) 2sec( ) = 0

Now, the ball is starting down with an initial velocity of zero while the stone is starting up.

€

Hball = 20m

12. A ball is thrown vertically upward with an initial speed of 20

€

msec . Two seconds later, a stone is thrown vertically (from

the same initial height as the ball) with an initial speed of 24

€

msec . At what height above the release point will the ball

and stone pass each other?

€

20− Hs = 12 gt 2

€

Hs = 24t − 12 gt 2

€

×

€

20m

€

20− 12 gt 2 = 24t − 1

2 gt 2

20 = 24t t = 0.83sec

€

Hs = 24t − 12 gt 2

Hs = 24 0.83( )− 12 10 m

sec2( ) 0.83( )2

Hs =16.53m

17

€

vo

€

v 14

= 18 msec

€

vTop = 0

€

34 H

€

v f2 = v 1

4

2 + 2g 34 H( )

0 = v 14

2 + 2g 34 H( )

0 = v 14

2 + g 32 H( )

2 v 14

2( )3g

= H = 22m

18

13. An object is thrown vertically and has an upward velocity of 18

€

msec when it reaches

one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object?

€

v f2 = vo

2 + 2g H( )0 = vo

2 + 2g H( )vo = 2gH

vo = 2 9.8 msec2( ) 22m( )

vo = 20.7 msec

13.2

19

14

€

4.9t 2 − vot ± Δy = 04.9t 2 + 20t − 60 = 0

€

t =−20 + 202 − 4 4.9( ) −60( )

9.8t = 2.01sec

20

14. A stone is thrown from the top of a building with an initial velocity of 20

€

msec downward. The top of the building is 60 m

above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

€

Hs = 40t − 4.9t 2

€

×

€

60m

€

60−10t − 4.9t 2 = 40t − 4.9t 2

60 = 50tt = 1.2sec

€

Hs = 40 1.2( )− 4.9 1.2( )2

H = 40.944m

21

15. An object is thrown downward with an initial (t = 0) speed of 10

€

msec from a height of 60 m

above the ground. At the same instant (t = 0), a second object is propelled vertically upward from ground level with a speed of 40

€

msec . At what height above

the ground will the two objects pass each other?

€

− 60−Hs( ) = −10t − 4.9t 2

€

−Δy = −10t − 4.9t2

Δy =10 3sec( ) + 4.9 3sec( )2

Δy = 74.1m

1622

16. A rock is thrown downward from an unknown height above the ground with an initial speed of 10

€

msec . It strikes the

ground 3.0 seconds later. Determine the initial height of the rock above the ground.

€

Δy = vot − 4.9t 2

Δy + 4.9t 2

t= vo

14m + 4.9 msec2 3sec( )2

3sec= vo

vo = 19.36 msec

23

17. A ball thrown vertically from ground level is caught 3.0 seconds later by a person on a balcony which is 14 meters above the ground. Determine the initial speed of the ball.

€

vo

€

v 23

= 25 msec

€

vTop = 0

€

13 H

€

v f2 = v 2

3

2 + 2g 13 H( )

0 = v 23

2 + 2g 13 H( )

3 v 23

2( )2g

= H = 95.7m€

23 H

24

18. An object is thrown vertically upward such that it has a speed of 25

€

msec when it reaches

two thirds of its maximum height above the launch point. Determine this maximum height.