University of the Fraser Valley

Physics 100

PHYS 100 MidTerm Practice

Name: __________________________

Directions: Fill in the scantron form with the following information:

1. ID number (student number)

2. Name at top of form

3. Name bubbled into the columns labelled "C1C2C3C4"

4. Test Version Number (shown below); bubble in spaced labelled "version"on the form

Version # 0

Use the following information to answer the next 2 question(s).

Two objects X and Y, are moving along a straight line. The graph below shows the displacements

of both objects during a 10 sec time interval. Both objects begin moving north from the same

starting point at time t = 0.

1. At time t = 8 s, the displacement of object Y relative to object X is:

1 m south 3 m south 7 m south 4 m north 7 m northa) b) c) d) e)

Solution:

2. At time t = 8 s, the velocity of object Y relative to object X is:

zero 1 m/s north 2 m/s north 1 m/s south 2 m/s southa) b) c) d) e)

Solution:

Use the following information to answer the next 1 question(s).

The graph below was drawn from data obtained when a styrofoam ball was dropped

from rest.

KInematicsProblems Page 1 Version 0 Mr. Furse

3. Of the following graphs, which one best represents the speed of the ball as a function of

time?

a) b) c)

d) e)

Solution:

Careful examination shows the velocity (slope of tangent line) to be positive, and not increasing

with each successive time interval. Slope of tangent line becomes constant, so velocity becomes

constant.

4. Which acceleration-time graph represents the same motion as the velocity-time graph

shown below?

a) b)

c) d)

Solution:

Examine time intervals and estimate the slope of v-t graph to get acceleration.

t = 0-1 s, acceleration goes from 0 to maximum negative value.

t = 2-3 s, acceleration goes from maximum negative value to 0.

t = 3-4 s, acceleration goes from 0 to positive value.

5. Which one of the following graphs can be derived from the velocity-time graph shown in the

diagram below?

a) b) c) d)

slope of v-t graph is decreasing, so the acceleration is decreasing. Acceleration starts out with a

high value and then decreases to zero.

6. The following graph represents a displacement-time graph for an object during a 5 s

time interval.

Which one of the following graphs of velocity versus time would best represent the

velocity of the object during the same time interval ?

a) b)

c) d)

e)

Solution:

Examine the graphs for proper time intervals.

t = 0-1 s would have a constant positive velocity

t = 1-3 s would have zero velocity

t = 3-5 s would have a constant negative velocity

Use the following information to answer the next 2 question(s).

The graph below shows the motion of three cars X, Y, and Z along a straight road.

Car X is travelling at the speed limit while car Y is travelling at a speed in excess of the limit. The

two cars pass a stationary police car Z at time t=0 and continue with uniform speed. The police car

Z immediately gives chase with a constant acceleration until it reaches car Y.

7. The speed of the police car Z at the instant it overtakes car Y is:

10 m/s 20 m/s 30 m/s 40 m/s 50 m/sa) b) c) d) e)

Solution: Find the slope of the tangent line to the Z curve at the point of intersection with Y.

8. The acceleration of the police car is:

2 m/s2 4 m/s2 6 m/s2 8 m/s2 10 m/s2a) b) c) d) e)

Solution: Initially, the slope (velocity) of the police car is zero. a = ∆v/∆t = 4 m/s2

9. A car, stopped at a traffic light, accelerates constantly at 0.65 m/s2 after the light turns green.

While accelerating, the car travels one block and then crosses a bridge 120 m long. If the car

takes 6.9 s to cross the bridge, how long was the block?

61 m

180 m

230 m

470 m

a)

b)

c)

d)

Equations for the block: Equations for the bridge:

Know: vi = 0, a = 0.65 m/s2 Know: v

i = final velocity for block

d = 0.5 at2 Know d = 120 m, t = 6.9 s, a = 0.65 m/s2

vf = 0+ at d = v

it + 0.5 at2

So rearranging the displacement equation for the bridge we get:

which gives vi = 15.14 m/s

Since this initial velocity for the bridge equals the final velocity for the block, we can find the

time to travel the block

, which gives t = 23.3 s

using the displacement equation d = 0.5 at2 we get the length of the block d = 176 m

10. In an experiment to determine the acceleration due to gravity a student rolls a cart down

a 2.5 m long board that has one end raised to a height of 1.2 m. If the cart takes 1.1 s to

roll down the ramp, which of the following is the best calculated result for the

acceleration due to gravity?

2.0 m/s2 4.1 m/s2 8.6 m/s2 9.8 m/s2a) b) c) d)

Solution:

d = vit + 0.5 at2, where v

i = 0

a = 2d÷t2

a = 2 x 2.5 ÷ 1.12

a = 4.1 m/s2 This is the acceleration down the ramp.

To find the acceleration due to gravity, use:

a/g = h/d (similar triangles) so, a = g h÷d. Where h = height or ramp, d = length of ramp.

(One could also use the values to calculate the angle, then use the component values calculated

with trig functions).

g = a x d ÷ h

g = 4.1 x 2.5 ÷ 1.2

g = 8.6 m/s2

11. A girl cyclist riding at a constant speed of 5.00 m/s sees a flash of lightning. Wanting

to get home quickly, she accelerates at 1.00 m/s2 for 4.00 s and then rides at a constant

speed for 30.0 s. If she then takes a further 12.0 s to slow down uniformly to rest, the

total distance travelled after seeing the lightning flash is:

342 m. 346 m. 352 m. 406 m.a) b) c) d)

Solution:

Break distances into three segments:

1) d = vit + 0.5 at2 , where v

i = 5 m/s, a = 1 m/s2, t = 4 s

d = 5 x 4 + 0.5 x 1 x 42

d = 28 m

Now find vf at the end of the first acceleration period.

vf = v

i + at

vf = 5 + 1 x 4

vf = 9 m/s

2) d = v x t

d = 9 x 30

d = 270 m

3) d = (vf + v

i) t ÷ 2, where v

f = 0 for the third interval and vi = 9 m/s from the second

interval

d = (0 + 9) x 12 ÷ 2

d = 54 m

Now add all the distance values to get d = 352 m

12. A speeding motorist travelling at 28 m/s passes a parked police officer. The officer begins to

chase the motorist the moment the motorist passes. The officer accelerates at a constant

1.8 m/s2 and the motorist travels at a constant speed of 28 m/s, how much time will it take the

officer to catch up?

5.6 s 16 s 31 s 64 sa) b) c) d)

dm

= dp

vm

t = vip

t + 0.5 at2

28 t = 0 + 0.5 x 1.8 x t2

0.9t2 = 28 t

0.9 t = 28

t = 31.1 s

13. Construct a distance graph that represents the motion of the car described below, over

the 11 seconds time interval. Assume the car's starting position is at zero.

(5.00 marks)

• the car is stationary at a stop light for 2 seconds

• the car accelerates up to a speed of 15 m/s during the next 4 seconds

• the car continues at that constant speed of 15 m/s for the next 4 seconds

Solution:

1 mark for a horizontal straight line

2 marks for curved line

2 marks for straight line with slope of 15 m/s

14. Using the graph shown below, construct a corresponding distance-time graph. Be sure to label

an appropriate scale on the distance scale of your graph.

(5.00 marks)

(5.00 marks)

15. The rocket engines of a spaceship were fired for 7.20 seconds causing the spaceship,

which had been travelling at 7680 m/s, to accelerate at 15.0 m/s2. (6.00 marks)

a) How fast would the spaceship be going after 7.20 seconds?

b) How far would the spaceship travel in this time?

Solution:

(a) vf = v

i + at, where v

i = 7680 m/s, a = 15 m/s2

vf = 7680 + 7.2 x 15

vf = 7788 m/s, rounding to 3 sig figs, v = 7.79 x 103 m/s (3 marks)

(b) d = vit + 0.5 at2 , where v

i = 7680 m/s, t = 7.20 s, a = 15.0 m/s2

d = 7680 x 7.2 + 0.5 x 15 x 7.22

d = 55685 m, round to 3 sig figs, d = 5.57 x 104 m (3 marks)

16. Normally the "Rockey Mountain Express" passes right on through the town of Bassano

Alberta at the crusing speed of 30.0 m/s. On days when there are passengers to entrain,

or detrain, the engineer stops as quickly as possible ( –1. m/s2), waits 30.0 seconds, and

then accelerates away as quickly as possible (1.2 m/s2). How much time does the train

"lose" for such a flag-stop as compared to not stopping?

(5.00 marks)

Solution:

For stopping:

vf = v

i + at, where v

f = 0, v

i = 30, a = –1 m/s2

t = ∆v ÷ a

t = -30 ÷ –1

t = 30 sec (2 marks)

Waiting t = 30 sec (0.5 mark)

For getting back upto speed:

vf = v

i + at, where v

f = 30, v

i = 0, a = 1.2 m/s2

t = ∆v ÷ a

t = -30 ÷ 1.2

t = 25 sec (2 marks)

Total time = 85 sec (0.5 marks)

17. In a certain lab experiment, a spring gun was placed on a table and fired a steel ball horizontally

outward. A student determines that the ball starts 1.0 m above the floor and travels 2.07 m

horizontally before striking the floor.

a) Determine the time that the ball is in the air.

b) Determine the initial velocity of the ball. (6.00 marks)

Solution:

a) d = 1/2at2

(1) = 1/2(9.8)t2 Use quadratic formula to find t

t = 0.45 s

b) Since the ball travels 2.07 m horizontally in 0.45 seconds, divide distance by time to get vi

vi = 2.07÷0.45

vi = 4.6 m/s (6.00 marks)

18. A stunt vehicle leaves an incline with a speed of 35 m s at a height of 52 m above level ground.

Air resistance is negligible.

a) What are the vehicle’s vertical and horizontal velocity components as it leaves the incline?

(1 mark)

b) What is the vehicle’s time of flight? (4 marks)

c) What is the vehicle’s range, R? (2 marks)

Solution:

a)

(1 mark)

b) d = vot + 0.5 at2

-52 = 16.4 t + 0.5 x 9.81 t2 use the quadratic equation to find t

t = 5.3 s

(4 marks)

c) R = vxt

R = 30.9 x 5.3

R = 165 m (1.6 x 102 m)

(2 marks)