D.W. Poppy Secondary School
Physics 12
PHYS 100 Mid-Term #1
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
1. Which of the following statements correctly describes the motion of an object shown by the
graph?
a)
b)
c)
d)
e)
the object has non-uniform motion
the object has constant velocity
the object has increasing velocity
the object has constant non-zero acceleration
the object has increasing acceleration
Constant velocity is represented on a d-t graph by a sloping straight line.
2. Which one of the following terms is defined as "the slope of a line tangent to a
displacement-time graph"?
a) average acceleration
b) instantaneous acceleration
c) instantaneous velocity
d) average velocity
Instantaneous velocity = slope of a tangent line on a d-t graph.
3. Select from the following statements the one that is true for the graph below.
the acceleration during the first second is uniform and greater than 0.
the velocity between t=1 s and t=3 s is uniform and greater than 0.
the average speed is always equal to the magnitude of the average velocity
the distance travelled between t=1 s and t=2 s is greater than the distance travelled
between t= 0 s and t=1 s.
e) none of the above are true
a)
b)
c)
d)
The object moves with uniform motion between t = 0-1 sec.
The object has stopped between t = 1-3 sec.
The object moves with uniform motion between t = 3-5 sec back towards the origin.
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4. Which graph represents the motion of an object decreasing in velocity?
a)
b)
d)
e)
c)
Since slope of a d-t graph represents velocity, if an object's velocity is to decrease, then the slope
should be decreasing (going from a large slope to a small slope). The following graph represents
decreasing velocity.
5. From the graph shown below it can be seen that at time t = 2.0 s the velocity changes
from:
a) 3.0 m/s to zero
b) 6.0 m/s to zero
c) 12 m/s to zero
d) zero to 3 m/s
e) 3.0 m/s to 1.5 m/s
Solution:
The slope for the first two seconds is: 3 m/s and after has a value of 0.
6. What does the shaded area of this velocity versus time graph represent?
a) t2 – t1
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b) v2 – v1
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c) Displacement
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d) Average velocity
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Solution:
The area under a v-t graph gives displacement.
7. Which one of the following terms can be defined as "the slope of a line tangent to a
velocity-time graph"?
a) instantaneous velocity
b) velocity
c) average acceleration
d) instantaneous acceleration
Instantaneous acceleration = slope of tangent line on a v-t graph.
8. The graph shown below represents the velocity of an object moving in a straight line.
From the graph, find the velocity of the object at time = 5.0 seconds.
a) 0.0 m/s
b) 5.0 m/s
c) 6.5 m/s
d) 26 m/s
e) 40 m/s
Solution:
From a v-t graph, velocity is found by reading the value from the vertical axis. At t = 5 s, this is
26 m/s.
9. If a body is accelerating , it MUST be changing speed.
a) True
b) False
Solution:
Acceleration is a vector quantity. Velocity is a vector quantity (speed + direrction). We could
have an object changing direction only (not speed) and still have an acceleration.
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10. A locomotive has a mass of 1.0 × 102 kg. The engineer operates it when no cars are being
pulled in such a way that he exerts the maximum possible force at any speed. The force depends
on the speed. The speed of the locomotive, as function of time, under these conditions, is shown
in the graph below.
The acceleration of the locomotive when it is travelling at 7.0 m/s is closest to:
a) 0.0 m/s2
b) 0.16 m/s2
c) 0.35 m/s2
d) 0.63 m/s2
Solution:
The slope of the graph when the speed is 7.0 m/s is 0.35 m/s2.
e) 0.80 m/s2
11. This graph shows the speed of a 3 kg mass as a function of time.
The average acceleration over the period 0 to 5 s is:
a) 0 m/s2
b) 1.0 m/s2
c) 2.5 m/s2
d) 1.25 m/s2
e) 5.0 m/s2
Solution:
Acceleration is the slope between two points on a v-t graph. The start-finish points give zero
acceleration.
12. This graph shows the velocity of a 3 kg mass as a function of time.
What is the average velocity for the 5 second interval?
a) 0 m/s
b) 2.5 m/s
c) 3.5 m/s
d) 5.0 m/s
e) 7.5 m/s
Solution:
Average velocity is d/t. The area under the graph is 17.5m; v = 17.5m/5s = 3.5 m/s.
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Use the following information to answer the next 1 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.
13. Car Y is exceeding the speed limit by:
a) 5 m/s
b) 10 m/s
c) 15 m/s
d) 20 m/s
e) 50 m/s
Car X is traveling at the speed limit, so take the slope of car X = 15.4 m/s
Slope of car Y is 20 m/s. So car Y is traveling 5 m/s faster than car X.
14. A projectile is fired from ground level such that it has an initial vertical speed of 20 m/s and
initial horizontal speed of 30 m/s. How far from the point of firing does the projectile land?
Take g=10 m/s2, assume that the terrain is flat, and neglect air resistance.
A. 40 m
B. 60 m
C. 80 m
D. 120 m
E. 180 m
Solution:
vf = vi + at
(20) = (20) + (-10)t
t = 4 sec
dx = vx x t
dx = (30)(4)
dx = 120 m
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Use the following information to answer the next 2 questions.
A ball with a speed of 5.7 m/s rolls along a horizontal table 1.4 m high and falls off the edge.
(Air friction is insignificant.)
15. How long does the ball take to fall to the floor?
A. 0.14 s
B. 0.38 s
C. 0.53 s
D. 0.73 s
Solution:
Examine the motion in the vertical direction.
viy = 0, a = 9.81 m/s2 down, dy = -1.4 m
-1.4 = - 4.9 t2
t =√(1.4÷4.9)
t = 0.53 s
16. What is the magnitude of the vertical component of the ball's velocity 0.50 s after it starts to
fall?
A. 1.2 m/s
B. 4.9 m/s
C. 5.7 m/s
D. 7.5 m/s
Solution:
In the vertical direction we have:
viy = 0, a = 9.81 m/s2 down
vfy = viy + at
vfy = 0 + 9.81 x 0.5
vfy = 4.9 m/s
Use the following information to answer the next 2 questions.
A projectile is fired at 95 m/s at an angle of 22° above the horizontal.
17. Calculate the horizontal displacement of the projectile after 2.3 seconds.
A. 56 m
B. 83 m
C. 203 m
D. 301 m
Solution:
vx = v cos ø
dx = vxt = 202.6 ≈ 203 m
18. Calculate the vertical displacement of the projectile after 2.3 seconds.
A. 56 m
B. 83 m
C. 203 m
D. 301 m
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Solution:
dy = viyt - 0.5 at2
dy = 95 x sin 22 x 2.3 - 4.9 x 2.32
dy = 55.9 m
19. A missile is fired horizontally from the top of a 78.4 m building with an initial velocity of
50.0 m/s. Its horizontal range will be:
A. 392 m.
B. 158.8 m.
C. 200 m.
D. 98 m.
Solution:
time for missle to reach ground is found from the vertical values.
dy = viyt - 4.9 t2 where viy = 0 dy = -78.4 m.
-78.4 = -4.9t2
t = √(78.4 ÷ 4.9) = 4.0 s
Horizontal range is dx
dx = vxt
dx = 50 x 4.0
dx = 200 m
20. A police car moving 80 km/h is passed by a speeder moving at 120 km/h. The 1.5 seconds after the
speeder passes the police car, the police car begins to accelerate uniformly at 3.0 m/s/s until it overtakes
the speeding car.
a) how long does it take to overtake the speeding car?
b) how fast is the police car moving when it overtakes the speeding car?
21. a) A projectile of fired vertically upwards at 35 m/s from a point 300 m above the ground. What
is it's velocity when it strikes the ground?
b) How long loes it take to reach point X on the trajectory which is 210 meters above the
ground?
(6.00 marks)
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Solution:
(6.00 marks)
(a) vf2 = vi2 - 2ady
(b) dy = viyt - 0.5 at2
-300 = 35 t - 0.5 x 9.81 x t2
where dy = -300 m, vi = 35 m/s
vf =
vi2 − 2 ad y
use the quadratic equation to solve
t = 9.2 s
vf = 84.3 m/s
22. A spring gun inclined at an angle of 45° to the horizontal fires a ball which leaves the muzzle of
the gun 1.1 m above the floor. The initial velocity of the ball is 6.0 m/s.
a) Determine the total time that the ball is in the air.
b) Determine the horizontal range of the projectile.
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(7.00 marks)
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Solution:
a) Given
Calculations
viy = vi sin θ = (6.0)(sin 45° ) = 4.24 ms
vertical components only!
vi = 6.0 ms
q = 45°
d y = −11
.m
ay = −9.8 sm2
1
2
d = vi t -
at 2
-1.1 = 4.24 t - 4.9 t 2
4.9 t 2 - 4.24 t - 11
. = 0
Using the quadratic formula gives: t = 1.07 sec
OR
vix = vi cos θ = (6.0 )(cos 45° ) = 4.24 ms
b) Given
horizontal components only!
t = 1.07s
θ = 45°
Calculations
d = vi t -
vi = 6.0 ms
1
2
at 2
d = 4.24 (1.07) -
ax = 0 sm2
1
2
(0 )(1.07)2
d = 4.6 m
(7.00 marks)
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23. Construct a position vs. time graph, a velocity vs. time graph, and an acceleration vs.
time graph that represents the motion of the skydiver described below. Assume the
skydiver's initial position is at the origin on the graph.
(5.00 marks)
• the skydiver jumps from a stationary helicopter 3000 m above the ground.
• the skydiver's acceleration decreases until terminal velocity in achieved.
• after falling at terminal velocity for a while, the parachute is pulled.
• a new terminal velocity is achieved.
• after falling at terminal velocity for a while, the skydiver lands.
Solution:
1 mark for a horizontal straight line
2 marks for curved line
2 marks for straight line with slope of 15 m/s
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