Modified True/False
Indicate whether the sentence or statement is true or false. If false, change the identified word or phrase to make the
sentence or statement true.
____
1. The slopes of position-time, velocity-time, and acceleration-time graphs represent velocity, acceleration, and
displacement, respectively. _________________________
____
2. Two siblings leave their house and walk to the same school taking different routes. One of the routes is
farther, but the siblings take the same time to reach the school. It can be said that the two siblings have the
same average velocity but different average speeds. _________________________
____
3. For the following quantities, these are considered to be vectors: displacement, acceleration.
_________________________
____
4. During an interval of uniform acceleration, the average velocity over the entire interval is equivalent to the
instantaneous speed at the midpoint of the interval. ___________________________________
____
6. A rubber ball changes velocity from 6.5 m/s [down] to 6.5 m/s [up] in a time of 2.0 s. Its average acceleration
during this period can be considered to be zero. _________________________
____
7. The acceleration due to gravity on Earth varies according to both altitude and latitude.
_________________________
____
8. An object is thrown vertically upward. At the top of its flight, when its velocity is momentarily zero, its
acceleration is zero. _________________________
____
9. Aristotle recognized that all objects fall at the same rate of acceleration provided that air resistance is not a
factor. _________________________
____ 11. If air resistance is negligible, the acceleration of all projectiles is exactly the same.
_________________________
Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
____ 17. Which of the following measurements is a scalar quantity?
a. the tension in an elevator cable
b. the acceleration of a car along a test track
c. the instantaneous velocity of a parachutist in free fall
d. the displacement of a hiker from her base station
e. the measured distance between Toronto and Montreal
____ 18. A race car completes exactly 10 laps around an oval track. Which of the following pairs of quantities
concerning its motion would both have values of zero?
a. displacement, average velocity
b. average speed, average acceleration
c. distance, average speed
d. average speed, average velocity
e. displacement, average speed
____ 20. Two hikers set out from the same spot and arrive at the same destination but they take different routes. Which
of the following quantities must be the same for both hikers?
a. distance
d. displacement
b. average speed
e. acceleration
c. average velocity
____ 21. Which of the following graphs does NOT depict uniform motion?
a. A and B
d. B and D
b. C only
e. A and E
c. D and E
____ 22. Which of the following graphs depicts uniform motion?
____ 23.
____ 24.
____ 25.
____ 26.
____ 27.
____ 28.
a. A and B
d. B and D
b. C and D
e. E only
c. A and C
The slope of a position-time graph represents
a. displacement
d. acceleration
b. speed
e. distance
c. velocity
The slope of a velocity-time graph represents
a. displacement
d. acceleration
b. average velocity
e. distance
c. instantaneous velocity
The area under a velocity-time graph represents
a. acceleration
d. distance
b. displacement
e. average velocity
c. instantaneous velocity
The slope of a line drawn tangent to a curved position-time graph represents
a. displacement
d. acceleration
b. instantaneous velocity
e. distance
c. average velocity
An object is travelling north and slowing down. The directions associated with the object’s velocity and
acceleration, respectively, are
a. [N], [S]
d. [N], [N]
b. [N], [N]
e. [N], [S]
c. [S], [S]
Which of the following statements concerning motion graphs is NOT correct?
a. The slope of a position-time graph gives velocity.
b. The area under a velocity-time graph gives displacement.
____ 29.
____ 30.
____ 31.
____ 32.
____ 33.
____ 34.
____ 35.
____ 36.
____ 37.
c. The slope of a velocity-time graph gives acceleration.
d. The area under an acceleration-time graph gives velocity.
e. The slope of the tangent in a position-time graph gives instantaneous velocity.
Which of the following statements concerning motion graphs is correct?
a. The slope of a position-time graph gives acceleration.
b. The area under an acceleration-time graph gives instantaneous velocity.
c. The slope of a velocity-time graph gives displacement.
d. The area under a position-time graph gives velocity.
e. The area under a velocity-time graph gives displacement.
The area under an acceleration-time graph represents
a. displacement
d. instantaneous velocity
b. change in velocity
e. average speed
c. average velocity
The slope of an acceleration-time graph represents
a. change in velocity
d. displacement
b. instantaneous acceleration
e. instantaneous velocity
c. jerk
Which of the following descriptions best represents the acceleration-time graph of a car that pulls away from
a corner when the light turns green, reaches and maintains a constant velocity, then slows down until it stops?
Assume that all accelerations are uniform.
a. All three sections of the graph are comprised of horizontal lines.
b. Two sections of the graph are diagonal lines and one is horizontal.
c. Two sections of the graph are horizontal lines and one is diagonal.
d. All three sections of the graph are comprised of diagonal lines.
e. All three sections of the graph are comprised of curved lines.
In an emergency braking exercise, a student driver stops a car travelling at 83 km/h [W] in a time of 4.0 s.
What is the car’s acceleration during this time? (The answer is expressed in units of m/s2.)
a. 5.8 [W]
d. 21 [W]
b. 21 [E]
e. –5.8 [E]
c. 5.8 [E]
An object is thrown vertically upward with a speed of 25 m/s. How much time passes before it comes back
down at 15 m/s? (Air resistance is negligible.)
a. 1.0 s
d. 18 s
b. 4.1 s
e. 27 s
c. 9.8 s
An object is thrown vertically upward at 18 m/s from a window and hits the ground 1.6 s later. What is the
height of the window above the ground? (Air resistance is negligible.)
a. 3.7 m
d. 37 m
b. 16 m
e. 41 m
c. 21 m
How long does it take a car to slow down from a speed of 54 km/h to 32 km/h over a distance of 65 m?
Answer in seconds.
a. 21
d. 2.7
b. 5.9
e. 1.5
c. 5.4
A jogger is running at 4.2 m/s when she begins to accelerate uniformly. If she runs a distance of 14 m in the
next 3.0 s, what is her new speed?
a. 17 m/s
d. 5.1 m/s
b. 14 m/s
e. 4.9 m/s
____ 38.
____ 39.
____ 40.
____ 41.
____ 42.
c. 7.7 m/s
An object is thrown vertically downward at 3.2 m/s. How long will the object take to hit the ground 12 m
below?
a. 3.8 s
d. 1.0 s
b. 3.1 s
e. 1.3 s
c. 2.0 s
A bullet accelerates uniformly along a barrel, exiting the gun in 24 ms with a speed of 196 m/s. The
acceleration of the bullet, expressed in units of metre per second squared, is
a. 1.7  105
d. 1.7  103
4
b. 1.8  10
e. 3.6  102
3
c. 8.2  10
A car accelerates at 2.7 m/s2 for 5.4 s, reaching a speed of 18 m/s. During the period of acceleration, the car
travels a distance of
a. 1.8  102 m
d. 58 m
b. 1.4  102 m
e. 18 m
c. 9.0  101 m
A toy car is moving at 13 cm/s when it begins accelerating at 1.4 cm/s2. If the acceleration is uniform, what is
the speed of the car after it has travelled a distance of 27 cm?
a. 2.4  102 cm/s
d. 16 cm/s
b. 93 cm/s
e. 1.0  101 cm/s
c. 62 cm/s
What distance does an object travel during a period of uniform acceleration (a = 2.5 m/s2) when its speed
changes from 35 m/s to 45 m/s?
a. 6.5  102 m
d. 32 m
2
b. 3.2  10 m
e. 2.0 m
c. 1.6  102 m
____ 45. Ignoring air resistance, which of the following are exhibiting “free fall”?
a. an object, initially at rest, dropped out of a window
b. an object thrown vertically downward from a window
c. an object projected vertically upward from a window
d. an object thrown horizontally from a window
e. all of the above
____ 46. The acceleration due to gravity on Earth
a. is the same at all locations on the surface of Earth
b. is greater for heavier objects
c. is greater at the equator and less at the poles
d. is the same at any two locations provided that the distance to the centre of Earth at those
locations is the same
e. varies slightly with latitude
Completion
Complete each sentence or statement.
56. The slope of a velocity-time graph represents the ____________________, whereas the area under the same
graph represents the ____________________.
57. A toy car travels once around a circular track of radius 65 cm in a time of 2.8 s. The car’s average speed is
____________________ and the value of its average velocity is ____________________.
58. A car accelerates uniformly for 8.0 s. The car’s instantaneous velocity after a time of ____________________
is equivalent to its average velocity over the entire interval.
59. An object is travelling with a velocity directed 30 W of N. If the object is slowing, its acceleration is directed
______________________.
60. The locations where the acceleration due to gravity on the surface of Earth is greatest and least, respectively
are ____________________ and ____________________.
61. An object is thrown vertically upward. At the top of its flight, when its velocity is momentarily zero, the value
of its instantaneous acceleration is ____________________.
Matching
Match the calculation with the quantity it determines.
a. velocity
d. displacement
b. acceleration
e. change in velocity
c. jerk
____
____
____
____
____
66.
67.
68.
69.
70.
the slope of a velocity-time graph
the slope of an acceleration-time graph
the slope of a position-time graph
the area under an acceleration-time graph
the area under a velocity-time graph
Short Answer
81. Provide an example where an object’s average speed and average velocity for the same trip are different.
82. Under what condition is an object’s average velocity for a trip zero?
83. Consider the quantities displacement, velocity, change in velocity, acceleration, and jerk. Describe how these
quantities can be determined from position-time graphs, velocity-time graphs, and acceleration-time graphs.
84. With the aid of a position-time graph that illustrates uniform acceleration and making appropriate
constructions, show that the average velocity over the entire interval is equivalent to the instantaneous
velocity at the midpoint of the interval.
85. Describe a situation where an object is accelerating but its velocity is zero.
87. What is the conversion factor from mm/s2 to km/h2?
89. For the same initial upward velocities, how many times higher will an object travel above the lunar surface (g
= 1.6 m/s2 [down]) than above the surface of Earth? Assume negligible air resistance on Earth.
90. Provide a brief description of how the acceleration due to gravity across the surface of Earth varies with both
latitude and altitude.
91. Provide a brief historical account of the realization that the acceleration due to gravity is the same for all
objects in the absence of air resistance.
92. What is meant by the term terminal speed? Provide some examples to illustrate the difference in the terminal
speeds of various objects.
Problem
96. An object is pushed from rest across a sheet of ice, accelerating at 5.0 m/s2 over a distance of 80.0 cm. The
object then slides with a constant speed for 4.0 s until it reaches a rough section which causes it to stop in
2.5 s.
(a) What is the speed of the object when it reaches the rough section?
(b) At what rate does the object slow down once it reaches the rough section?
(c) What total distance does the object slide throughout its entire trip?
97. A ball is thrown vertically upward from a window that is 3.6 m above the ground. Its initial speed is 2.8 m/s.
(a) With what speed does the ball hit the ground?
(b) How long after the first ball is thrown should a second ball be simply dropped from the same window so
that both balls hit the ground at the same time?
98. An object is pushed along a rough horizontal surface and released. It slides for 10.0 s before coming to rest
and travels a distance of 20.0 cm during the last 1.0 s of its slide. Assuming the acceleration to be uniform
throughout
(a) How fast was the object travelling upon release?
(b) How fast was the object travelling when it reached the halfway position in its slide?
99. An arrow is shot vertically upward with an initial speed of 25 m/s. When it’s exactly halfway to the top of its
flight, a second arrow is launched vertically upward from the same spot. The second arrow reaches the first
arrow just as the first arrow reaches its highest point.
(a) What is the launch speed of the second arrow?
(b) What maximum height does the second arrow reach?
100. A pedestrian is running at his maximum speed of 6.0 m/s trying to catch a bus that is stopped at a traffic light.
When he is 16 m from the bus, the light changes and the bus pulls away from the pedestrian with an
acceleration of 1.0 m/s2.
(a) Does the pedestrian catch the bus and, if so, how far does he have to run? (If not, what is the pedestrian’s
distance of closest approach?)
(b) How fast is the bus moving when the pedestrian catches it? (or at the distance of closest approach)
(c) On a single set of axes, plot the corresponding position-time graphs of both the bus and pedestrian to
confirm your answer in (a).
101. A truck travels at a constant speed of 28.0 m/s in the fast lane of a two-lane highway. It approaches a
stationary car stopped at the side of the road. When the truck is still 1.2  102 m behind the car, the car pulls
out into the slow lane with an acceleration of 2.6 m/s2.
(a) How long will it take the truck to pass the car?
(b) How far will the car have travelled when the truck passes it?
(c) If the car were to maintain this acceleration, how fast would it be travelling when it overtakes the truck?
105. The graph below represents the motion of an object over a recorded time interval. Using methods of graphical
analysis wherever possible, determine
(a) the object’s displacement relative to its starting position at t = 6.0 s.
(b) the object’s average velocity between t = 0.0 s and t = 6.0 s.
(c) the object’s average speed between t = 0.0 s and t = 6.0 s.
(d) Including t = 0.0 s, how many times during the entire recorded time interval is the object at its starting
position?
(e) During which interval is the object’s acceleration the greatest? What is the value of the acceleration during
this interval?
(f) Plot the corresponding position-time graph.
(g) Plot the corresponding acceleration-time graph.
a
Answer Section
MODIFIED TRUE/FALSE
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F, jerk
FM1.02
T
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T
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F, instantaneous velocity
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F, south-east
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F, 6.5 m/s2 [up]
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T
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F, 9.8 m/s2 [down]
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F, Galileo
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F, vertical component
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T
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F, 45
FM1.03
T
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F, north-east
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F, slightly greater than 15
FM1.02
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MULTIPLE CHOICE
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COMPLETION
56. ANS:
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acceleration, displacement
FM1.02
1.5 m/s, 0 m/s
FM1.02
4.0 s
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30 E of S
FM1.02
at the poles, at the equator
FM1.05
9.8 m/s2 [down]
FM1.03
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constant, changing (at a rate of 9.8 m/s2 [down])
constant, changing
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63. ANS: acceleration, horizontal velocity component
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south-west
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distance across, speed of the boat in water, distance across/time to cross, speed of the current
K/U
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B
C
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MATCHING
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71. ANS: B
72. ANS: A
73. ANS: C
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C
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78. ANS: B
79. ANS: C
80. ANS: A
K/U
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SHORT ANSWER
81. ANS:
This is true for any object that doesn’t travel in a straight line during the trip. In this case, the distance
travelled by the object is different than its displacement, making its average speed and average velocity
different as well.
REF: C
OBJ: 1.1
LOC: FM1.02
82. ANS:
When the object returns to its starting position so that its displacement is zero, its average velocity will also be
zero.
REF: C
83. ANS:
OBJ: 1.1
LOC: FM1.02
position-time graphs: slope = velocity
velocity-time graphs: slope = acceleration, area under the graph = displacement
acceleration-time graphs: slope = jerk, area under the graph = change in velocity
REF: C
OBJ: 1.1
LOC: FM1.02
84. ANS:
The slope of the line joining the endpoints of the interval represents the average velocity over the entire
interval. This line is parallel and, therefore, has the same slope as the tangent to the graph at the interval
midpoint. This represents the instantaneous velocity at this point. The two quantities are equivalent.
REF: K/U
OBJ: 1.2
LOC: FM1.05
85. ANS:
This would occur at the instant when an object is reversing direction. For example, when an object is thrown
vertically upward its acceleration during the entire trip is 9.8 m/s2 [down], including its acceleration at the
peak of its flight when its velocity is momentarily zero.
REF: C
86. ANS:
north-west
OBJ: 1.2
LOC: FM1.02
REF: K/U
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87. ANS:
1.000 mm/s2 = 1.296 km/h2
LOC: FM1.05
REF: K/U
OBJ: 1.2
LOC: FM1.02
88. ANS:
A scale diagram approach to analyzing two dimensional motion is the least precise. Equally high precision is
achieved with either the trigonometric or component method approach. Generally, when only two vectors are
being considered, either a trigonometric or component approach is simplest. Once three or more vectors are
being considered, the component method of analysis is favorable.
REF: C
89. ANS:
From the expression
OBJ: 1.1
LOC: FM1.02
it can be seen that the distance an object will travel is inversely proportional to the acceleration due to gravity:
Therefore,
.
An object will travel 6.1 times higher on the moon than on Earth when projected vertically upward from the
two surfaces with the same initial velocity.
REF: C
90. ANS:
OBJ: 1.3
LOC: FM1.05
Latitude: The acceleration due to gravity increases as one approaches both poles because the distance to the
centre of Earth is decreasing. Similarly, the acceleration due to gravity is a minimum at the equator.
Altitude: The acceleration due to gravity decreases with altitude because the distance to the centre of Earth is
decreasing. The force of gravity between two objects is inversely proportional to the square of the distance
between the centres of the two objects.
REF: C
OBJ: 1.3
LOC: FM1.05
91. ANS:
Aristotle had suggested that heavier objects fall faster. This was disputed and proven to be false by Galileo in
a series of experiments where he rolled objects down inclined planes and showed that all objects “rolled” with
the same acceleration. For an inclined plane with a inclination of , the acceleration was gsin .
REF: C
OBJ: 1.3
LOC: FM1.05
92. ANS:
When the effects of air resistance are recognized, it is shown that objects in free fall will accelerate only to a
point beyond which they will fall no faster. When an object has reached this “terminal speed,” the
gravitational force acting downward on it is just balanced by the upward force exerted by the air the object is
falling through. For people, terminal speed is about 190 km/h without a parachute and anywhere from 18
km/h to 36 km/h with a parachute. This can be compared to the terminal speed of a dust particle which is
about 0.07 km/h.
REF: C
OBJ: 1.3
LOC: FM1.05
93. ANS:
The horizontal component of the motion would be uniform and the vertical component would show the
acceleration due to gravity.
REF: K/U
OBJ: 1.4
LOC: FM1.03
94. ANS:
All projectiles have an acceleration of 9.8 m/s2 [down] due to the effect of gravity. Upon launch, the vertical
component of the projectile’s motion is affected by this acceleration but the original horizontal component
remains fixed for the entire flight. Typically, the time of flight is found by considering the vertical motion.
The range is found using this time of flight and the horizontal component of the projectile’s velocity. This
horizontal motion is uniform.
REF: C
OBJ: 1.4
LOC: FM1.03
95. ANS:
The expression for horizontal range for a projectile returning to level ground is
where
is the angle above the horizontal for the launch.
For every angle less than 45, there will be a corresponding one greater than 45 (and less than 90), the two
of which will have the same value for sin2. For example, if a projectile is launched at 30 or at 60, the
values of sin2(30) and sin2(60) are both 0.87. These two launch angles will result in the same horizontal
range. Maximum range is achieved with a launch angle of 45. At this angle sin2(45) = 1 is maximized.
REF:
C
OBJ: 1.4
LOC: FM1.03
PROBLEM
96. ANS:
(a)
v1 = 0.0 m/s
a = 5.0 m/s2
d = 80.0 cm = 0.800 m
v2 = ?
The speed of the object upon reaching the rough section is 2.8 m/s.
(b)
v1 = 2.83 m/s
v2 = 0.0 m/s
t = 2.5 s
a=?
The object’s acceleration is 1.1 m/s2 and slowing.
(c)
During the period of acceleration:
d = 0.800 m
During the period of uniform motion:
v = 2.83 m/s
t = 4.0 s
d = vt = 2.83 m/s(4.0 s) = 11.32 m
During the period of deceleration:
v1 = 2.83 m/s
v2 = 0.0 m/s
t = 2.5 s
Total distance the object slides: 0.800 m + 11.32 m + 3.54 m = 16 m
The object slides a total distance of 16 m.
REF: K/U
OBJ: 1.2
LOC: FM1.02
97. ANS:
(a)
Using a sign convention of “down” as (+) and “up” as (–).
The speed of the object when it hits the ground is 8.9 m/s.
(b)
Time of flight for the first ball:
Time of flight for the second ball:
The difference in flight times is 1.19 s – 0.86 s = 0.33 s.
The second ball should be dropped 0.33 s after the first one is thrown so that both hit the ground at the
same time.
REF: K/U
OBJ: 1.3
LOC: FM1.02
98. ANS:
(a)
The object’s acceleration during the last 1.0 s:
t = 1.0 s
v1 = 40 cm/s
v2 = 0.0 cm/s
a=?
This is also the acceleration for the entire trip.
The speed upon release:
The object was travelling at 4.0 m/s upon release.
(b)
The distance travelled:
t = 10.0 s
v2 = 0.0 cm/s
a = –40 cm/s2
d = ?
At the halfway position:
d = 1.0  103 cm
v2 = 0.0 cm/s
a = –40 cm/s2
v1 = ?
The object is travelling at 2.8 m/s at the halfway position in its slide.
REF: K/U
OBJ: 1.2
LOC: FM1.02
99. ANS:
(a)
Using the sign convention that “up” is (–) and “down” is (+):
v1 = –25 m/s
v2 = 0.0 m/s
a = 9.8 m/s2
d = ?
The arrow travels 31.9 m upward to its highest point. The halfway position is 15.9 m.
The time to travel the last half of its flight:
d = –15.9 m
v2 = 0.0 m/s
a = 9.8 m/s2
t = ?
For the second arrow:
d = -31.9 m
a = 9.8 m/s2
t = 1.80 s
v1 = ?
The speed of the second arrow at launch is 27 m/s [upward].
(b)
Finding the maximum height of the second arrow:
v1 = –26.5 m/s
v2 = 0.0 m/s
a = 9.8 m/s2
d = ?
The second arrow reaches a maximum height of 36 m [upward].
REF: K/U
OBJ: 1.3
100. ANS:
(a)
Bus: v1B = 0.0 m/s, aB = 1.0 m/s2
Pedestrian: vP = 6.0 m/s
LOC: FM1.02
Bus:
Pedestrian: dP = vPt
dP = 6.0 t
dP = dB + 16 m
6.0 t = 0.5(t)2 + 16
solving the quadratic: t = 4.0 s, 8.0 s
The pedestrian does catch the bus after running for 4.0 s.
d
= 6.0 m/s(t)
= 6.0 m/s(4.0 s)
d = 24 m
The pedestrian runs 24 m to catch the bus.
(b)
v1B = 0.0 m/s
aB = 1.0 m/s2
t = 4.0 s
v2B = ?
v2B = v1B + aBt
= (1.0 m/s2)(4.0 s)
v2B = 4.0 m/s
The bus is travelling at 4.0 m/s when the pedestrian catches it.
(c) The position-time graph of the motion of the pedestrian is a straight diagonal line that begins at the origin.
The line that represents the motion of the bus is a curved line that begins at the 8.0 m mark and crosses that
pedestrian’s line twice, at 4.0 s and at 8.0 s.
REF:
K/U
OBJ: 1.3
LOC: FM1.02
101. ANS:
(a)
Car: v1C = 0.0 m/s, aC = 2.6 m/s2
Truck: vT = 28.0 m/s
Car:
Truck: dT = vTt
dT = 28.0 t
dT = dC + 1.2  102 m
28.0 t = 1.3(t)2 + 1.2  102
solving the quadratic: t = 5.9 s, 16 s
The truck passes the car after 5.9 s.
(b)
v1C = 0.0 m/s
aC = 2.6 m/s2
t = 5.9 s
dC = ?
The car travels 45 m by the time the truck passes it.
(c)
v1C = 0.0 m/s
aC = 2.6 m/s2
t = 15.6 s (the other root of the quadratic)
v2C
= v1C + aCt
= 2.6 m/s2(15.6 s)
v2C
= 41 m/s
The car will be travelling at 41 m/s when it passes the truck if it maintains its acceleration.
REF: K/U
OBJ: 1.2
LOC: FM1.02
102. ANS:
(a)
The car drives the following displacements:
= 80.0 km/h [E](1.50 h) = 120 km [E]
= 60.0 km/h [N](1.00 h) = 60.0 km [N]
= 100.0 km/h [30W of N](0.50 h) = 50 km [30W of N]
Using the component method for the displacement of A to B:
north-south components: 60 km [N] + 50 km(cos 30) [N] = 103.3 km [N]
east-west components: 120 km [E] + 50 km(sin 30) [W] = 95 km [E]
Using Pythagoras Theorem, the magnitude of the displacement is 140 km
Using trigonometry, the direction is: [43 E of N]
The displacement of point B from point A is 1.4  102 km [43E of N].
(b)
A vector diagram showing the relationship among the vectors is drawn:
vPG = velocity of plane with respect to the ground
vPA = velocity of plane with respect to the air
vAG = velocity of the air with respect to the ground
Since the car takes a total of 3.00 h to reach point B and the plane leaves 2.00 h later but arrives at the same
time, the time it takes the plane to make the flight is 1.00 h.
= 1.4  102 km/h [43 E of N]
Using cosine law:
The plane’s speed is 1.9  102 km/h.
(c)
Using sine law:
,  = 13
The plane must head in a direction of [30ºE of N]. (90 – 13 – 47)
(d)
If there is no wind:
The plane would take 0.74 h to fly from A to B if there was no wind.
REF: K/U
OBJ: 1.5
LOC: FM1.05
103. ANS:
(a) The triangle of velocity vectors appears as:
The plane must point [8.6 E of S].
(b) For the first leg of the trip of plane A:
time to fly from Toronto to Pittsburgh:


Layover time in Pittsburgh is 0.5 h.
Pittsburgh to Philadelphia:
Time to fly:
Total time Toronto to Philadelphia: 0.885 h + 0.5 h + 1.15 h = 2.5 h
The total time for plane A is 2.5 h.
(c) Distance from Toronto to Philadelphia:
Vector triangle of velocities:
Using sine law:  = 5.8, then  = 180 – 138 – 5.8 = 36
Using cosine law:
353 km/h
Time for plane B to fly from Toronto to Philadelphia:
Plane B must wait 2.5 h – 1.48 h = 1.0 h.
REF: K/U
104. ANS:
Canoeist B:
OBJ: 1.5
LOC: FM1.05
Using sine law:
.
The component of
across the river is: 2.4sin(56 + 16) = 2.28 m/s.
The time for B to cross to point X:
Canoeist A:
The time for A to reach point X:
Canoe A must wait 131.6 s – 125 s = 6.6 s.
REF: K/U
105. ANS:
(a)
displacement
displacement
OBJ: 1.5
LOC: FM1.05
= area under graph
= 23.75 m [S] + 18.75 m [N]
= 5.0 m [S]
(b)
The object’s average velocity during the first 6.0 s is 0.83 m/s [S].
(c)
The object’s average speed during the first 6.0 s is 7.1 m/s.
(d) The object is at its starting location 3 times throughout the motion.
(e) The object’s acceleration is greatest between t = 6.5 s and 7.0 s. (the greatest slope) acceleration = slope of
graph = 30 m/s2 [N]
(f)
(g)
REF: K/U
OBJ: 1.2
LOC: FM1.02
106. ANS:
(a)
Time of flight: let “up” be (–) and “down” be (+)
v1 = –80.0 m/s(sin 25) = –33.8 m/s
a = 9.8 m/s2
d = 36 m
t = ?
36 = (–33.8)t + 4.9(t)2
Solving the quadratic: t = 7.84 s
Horizontal range: d = vt = 80.0 m/s(cos 25)(7.84 s) = 5.7  102 m
The horizontal range of the shell is 5.7  102 m.
(b) Horizontal component of final velocity: 80.0 m/s(cos 25) = 72.5 m/s
Vertical component of final velocity: v2 = v1 + at = –33.8 m/s + 9.8 m/s2(7.84 s)
v2 = 43.0 m/s
Using Pythagoras:
=
The shell lands with a velocity of 84 m/s at an angle of 31 below the horizontal.
REF: K/U
OBJ: 1.4
LOC: FM1.03
107. ANS:
(a)
Time of flight: let “up” be (–) and “down” be (+)
v1 = –26 m/s(sin 60º) = –22.5 m/s
a = 9.8 m/s2
d = 2.0 m
t = ?
2.0 = (–22.5)t + 4.9(t)2
Solving the quadratic: t = 4.68 s
The time of flight is 4.7 s.
(b)
Horizontal range: d = vt = 26 m/s(cos 60)(4.68 s) = 60.8 m
The receiver must run: 60.8 m – 3.0 m = 57.8 m.
The time the receiver has to reach the football: 4.68 s + 2.0 s = 6.68 s.
The average speed of the receiver:
The receiver must run with an average speed of 8.7 m/s.
REF: K/U
OBJ: 1.4
LOC: FM1.03
108. ANS:
(a)
Time of flight: let “up” be (–) and “down” be (+)
v1 = –15 m/s(sin 50) = –11.5 m/s
a = 9.8 m/s2
d = –2.0 m
t = ?
–2.0 = (–11.5)t + 4.9(t)2
Solving the quadratic: t = 0.19 s (way up) and 2.16 s (way down)
Horizontal range: d = vt = 15 m/s(cos 50º)(2.16 s) = 21 m
The net must be placed 21 m away from the cannon.
(b) Horizontal component of final velocity: 15 m/s(cos 50) = 9.64 m/s
Vertical component of final velocity: v2 = v1 + at = –11.5 m/s + 9.8 m/s2(2.16 s)
v2 = 9.67 m/s
Using Pythagoras:
=
The shell lands with a velocity of 14 m/s at an angle of 45 below the horizontal.
REF: K/U
109. ANS:
(a)
OBJ: 1.4
LOC: FM1.03
Time of flight of shell:
Horizontal range of shell: 100 m
Horizontal component of shell’s velocity:
Angle of projection:
10 m/s = 40.0 m/s(cos )
 = 76º
The gun must be aimed at an angle of 76 to the horizontal.
(b)
Vertical component of shell’s velocity: 40.0 m/s(sin 75.5) = 38.8 m/s [up]
let “up” be (–) and “down” be (+)
v1 = –38.8 m/s
a = 9.8 m/s2
t = 10 s
d = ?
The cliff is 1.0 102 m high.
(c) Horizontal component of final velocity: 10 m/s
Vertical component of final velocity: v2 = v1 + at = –38.8 m/s + 9.8 m/s2(10 s)
v2 = 59.2 m/s
Using Pythagoras:
=
The shell lands with a velocity of 59.2 m/s at an angle of 9.6 to the vertical.
REF: K/U
OBJ: 1.4
LOC: FM1.03
110. ANS:
(a)
At maximum height: vertical component of velocity is zero:
let “up” be (–) and “down” be (+)
v1 = –40.0 m/s(sin 30.0º) = –20.0 m/s
a = 9.8 m/s2
v2 = 0.0 m/s
d = ?
The ball reaches a maximum height of 21 m above the ground.
(hit from 1.0 m above the ground)
(b)
Time of flight:
v1 = –40.0 m/s(sin 30.0º) = –20.0 m/s
a = 9.8 m/s2
d = –2.0 m
t = ?
–2.0 = (–20.0)t + 4.9(t)2
Solving the quadratic: t = 0.10 s (way up) and 3.98 s (way down)
The fielder must run for 4.0 s in order to catch the ball.
(c)
Horizontal range: d = vt = 40.0 m/s(cos 30)(3.98 s) = 138 m
The fielder must run a distance of: 138 m – 110.0 m = 28 m.
The speed of the fielder:
The fielder must run with an average speed of 7.0 m/s.
REF:
K/U
OBJ: 1.4
LOC: FM1.03
ESSAY
111. ANS:
Projectile motion: The time the ball is in the air will depend on both the vertical displacement the ball makes
during its flight and the vertical component of the velocity with which it is thrown. The horizontal range, in
turn, is determined by both the time of flight and the horizontal component of the original velocity.
Relative motion: The path the ball takes from the thrower to first base corresponds to the velocity of the ball
with respect to the ground. This is the vector sum of the velocity of the ball with respect to the thrower (the
velocity the thrower releases the ball with) and the velocity of the thrower with respect to the ground (the
velocity of the thrower as he or she runs along the ground). The direction the thrower must aim at is the
direction of the vector that represents the velocity of the ball with respect to the thrower. This is not directly at
first base (unless the thrower is running directly at or directly away from first base when the ball is released).
REF: MC
OBJ: 1.4, 1.5
LOC: FM3.02
112. ANS:
The pilot understands that the packages will fall with an acceleration of 9.8 m/s2 (if air resistance is
neglected). The time of flight will depend on the altitude from which they are dropped. The horizontal
distance the packages will travel while they fall will depend, in turn, on the time to fall and the horizontal
component of the airplane (and packages) when the packages are released.
REF: C
OBJ: 1.4
LOC: FM3.02
113. ANS:
A launching device that launches projectiles with a consistent speed is mounted to launch them horizontally.
Projectiles are launched from a variety of heights and the horizontal range is measured each time.
The time of flight depends on the vertical height in the following way:
v1 = 0.0 m/s
a = 9.8 m/s2
Solved for t, this reduces to
.
This shows that the time of flight is proportional to the square root of the vertical height.
From the expression for horizontal range, d = vt, the range is proportional to the time of flight. Therefore,
the horizontal range will be proportional to the square root of the vertical height.
A graph of horizontal range plotted against vertical height will yield the following curve:
If the data is then re-plotted as “range” versus “square root of vertical height,” a straight line will result,
thereby identifying the relationship that exists between them.
REF:
I
OBJ: 1.4
LOC: FM2.02