# Functions 11

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Unit 1 Forces and Motion

(Pages 2-3)

NOTE:
Answers may vary. Students can revisit their answers as they progress through the unit.
Knowledge and Understanding
1. The car accelerates to the right for the first 2.0 s, then moves with a constant velocity for 3.0 s, and then slows to a stop in
the final 2.0 s.
2. (a) Starting at A, the dog moves east, then south, then west, and finally north.
(b)

56
m
(c)

0.0
m
(d)

3.5
m/s
(e) 0.0 m/s
3. Set A represents the motion of the falling ball because the ball undergoes constant acceleration, so the distance it travels
increases with each successive time interval.
4. (a) At A, the force of gravity is pulling downward and the force of the board on the diver is pushing upward. At B, only
the downward force of gravity is acting on the diver.
(b) The required diagram is shown below.

(c) The force of gravity is the non-contact force because it acts on a body without needing to be in contact with the body.
5. For A:
For B:

A = lw
2
B = r
= (3.0 m)(1.8 m)
2

A = 5.4 m2
2.7 m

=

2

2
B = 5.7 m
6. The travel times are all equal (20 s). Car N has the highest velocity because it travels 80 m in 20 s (which is a velocity of
4 m/s). Car M has a velocity of 2 m/s, and car L has a velocity of 1 m/s.
7. Finding the slope of the line:
rise
slope = run
4 m/s [S] - 4 m/s [S]
=

8 s
slope = 0 m/s
The slope represents the acceleration, which in this case is zero. Finding the area under the line:

area
=
lw

= (4 m/s [S])(8 s)

= 32 m [S]
area
=
3
x 101 m [S]

The area represents the change of position or the displacement from t = 0 s to t = 8 s.
NOTE: Students can choose either one or two significant digits. The concepts of significant digits and rounding off
Copyright (c) 2002 Nelson Thomson Learning
Unit 1 Are You Ready? 1

Inquiry and Communication
8. (a) The density determined by Group A is the most reasonable.
(b) When rounded off to two significant digits, Group B has the same value as Group A. However, stating an experimental
measurement to six significant digits does not make sense. Group C is the only group with a density greater than
1.0 g/mL. Likely their value is the reciprocal of what it should be. Group D has the same basic numerical value as
Groups A and B, but their value would be the density of a gas. Likely they meant to write 0.76 g/mL.
9. To determine velocity, both the distance travelled and the time interval to travel that distance are needed. These
measurements can be determined using a metric measuring tape and a stopwatch as the jogger runs from one marked
distance
position to another. Then velocity =
.
time
10. (a) The motion is occurring at a constant velocity because in each 1.0 s time interval the distance covered is
4.0 m.
(b) See the graph below.

(c) Several sets of data points can be used to calculate the velocity. One example is:
distance
speed = time
4.0 m

=

1.0 s
speed = 4.0 m/s

CHAPTER 1 MOTION
Try This Activity: Comparing Velocities
(Page 5)

(a) Answers may vary, but a typical arrangement with estimated values is:
D: 10 000 km/h, B: 1000 km/h, A: 10 km/h, C: 1 km/h
(b) A: 3 km/h; B: 8.5 x 102 km/h; C: 1.0 km/h; D: 1.38 x 104 km/h
(c) Answers will depend on the predictions. To improve skills of estimating average velocities, students could learn more
about approximate distances between places (on maps or a globe), learn the approximate range of velocities of various
vehicles, and practise various estimations.

1.1 MOTION IN OUR LIVES
PRACTICE
(Pages 6-8)
Understanding Concepts
1. (a) The rubber stopper is accelerating under gravity as it falls.
(c) The rocket must accelerate from rest while rising from the launch pad.
(d) The motorcycle slows (decelerates) as the brakes are applied, bringing the motorcycle to rest.
2. (a) 12 ms, (b) 500 MHz, (e) 15 cm2, (f) 50 mL are all scalar quantities. None of these have directions associated with them.
3. (a) scalar quantities: velocity, distance, time, mass, grade (inclination), frequency, area
(b) other scalar quantities: volume, electrical current, sound intensity, luminosity
2
Unit 1 Forces and Motion
Copyright (c) 2002 Nelson Thomson Learning

Inquiry and Communication
8. (a) The density determined by Group A is the most reasonable.
(b) When rounded off to two significant digits, Group B has the same value as Group A. However, stating an experimental
measurement to six significant digits does not make sense. Group C is the only group with a density greater than
1.0 g/mL. Likely their value is the reciprocal of what it should be. Group D has the same basic numerical value as
Groups A and B, but their value would be the density of a gas. Likely they meant to write 0.76 g/mL.
9. To determine velocity, both the distance travelled and the time interval to travel that distance are needed. These
measurements can be determined using a metric measuring tape and a stopwatch as the jogger runs from one marked
distance
position to another. Then velocity =
.
time
10. (a) The motion is occurring at a constant velocity because in each 1.0 s time interval the distance covered is
4.0 m.
(b) See the graph below.

(c) Several sets of data points can be used to calculate the velocity. One example is:
distance
speed = time
4.0 m

=

1.0 s
speed = 4.0 m/s

CHAPTER 1 MOTION
Try This Activity: Comparing Velocities
(Page 5)

(a) Answers may vary, but a typical arrangement with estimated values is:
D: 10 000 km/h, B: 1000 km/h, A: 10 km/h, C: 1 km/h
(b) A: 3 km/h; B: 8.5 x 102 km/h; C: 1.0 km/h; D: 1.38 x 104 km/h
(c) Answers will depend on the predictions. To improve skills of estimating average velocities, students could learn more
about approximate distances between places (on maps or a globe), learn the approximate range of velocities of various
vehicles, and practise various estimations.

1.1 MOTION IN OUR LIVES
PRACTICE
(Pages 6-8)
Understanding Concepts
1. (a) The rubber stopper is accelerating under gravity as it falls.
(c) The rocket must accelerate from rest while rising from the launch pad.
(d) The motorcycle slows (decelerates) as the brakes are applied, bringing the motorcycle to rest.
2. (a) 12 ms, (b) 500 MHz, (e) 15 cm2, (f) 50 mL are all scalar quantities. None of these have directions associated with them.
3. (a) scalar quantities: velocity, distance, time, mass, grade (inclination), frequency, area
(b) other scalar quantities: volume, electrical current, sound intensity, luminosity
2
Unit 1 Forces and Motion
Copyright (c) 2002 Nelson Thomson Learning

4. The distance from the equator to the geographic North Pole (10 million metres) is imprecise because the exact location of
these positions cannot be known to absolute precision (and may change slightly with time). The second, as originally
defined (86 400 s for Earth to rotate once on its own axis), changes over the course of time. The length of the day is not
constant due to the gravitational drag the Sun places on Earth, slowing its rotation.
5. Surface area is measured in m2, derived from metres x metres; volume is measured in m3, derived from metres x metres x
metres.
6. d = 42.2 = 4.22 x 104 m
d
t = 3 h 53 min 17 s
v =
av
t
= 3(3600 s) + 53(60 s) + 17 s
4
4.22x10 m
t = 13 997 s
=

v
13 997 s
av = ?
v = 3.01 m/s
av
3.0 m = 3.0 x 10-3 km
1
1.0s = 3600h

3.0 m / s = 3.0 x 10-3 km
1
3600 h
=10.9 km / h

The average velocity is 3.01 m/s or 10.9 km/h.
7. d = 38 cm
d
d = 0.38 m
v =
av
t
t = 1.9 x 10-7 s
0.38 m
vav = ?
=
7
1.9 x10- s
6
v = 2.0x10 m/s
av

The average velocity of the electrons is 2.0 x 106 m/s.
d
8. d = vav(t), t =

vav
9. d
= 3.8 x 105 m
d
t =
vav = 9.5 x 10-3 m/s
v
t = ?
av
5
3.8x10 m
=

3
9.5x10- m/s
7
t = 4.0x10 s

t =
2.5 s
d = vav(t)
vav = 480 m/s
= 480 m/s(2.5 s)
d = ?
d = 1.2 x 103 m

d = 1800 m
d
v
t =
av = 24 m/s
v
t = ?
av
1800 m
=

24 m/s
t = 75 s
10. vav = 28 cm/s
d = vav(t)
t = 0.20 s
= 28 cm/s (0.20 s)
d = ?
d = 5.6 cm
Blood will travel a distance of 5.6 cm in the aorta in 0.20 s.
Copyright (c) 2002 Nelson Thomson Learning
Chapter 1 Motion 3

11. vav = 1.6 x 103 km/h
d
d
= 2r
t =
av
v
= 2(3.14)(6.5 x 103 km)
4
d = 4.08 x 104 km
4.08x10 km
=
t = ?
3
1.6x10 km/h
t = 26 h

It will take 26 h to complete the trip.

Activity 1.1.1 Calibrating a Ticker-Tape Timer
(Page 10)

Analysis
(a) Most timers operate at a frequency of 60.0 Hz, which corresponds to a period of 0.0167 s. Students' results should be
expressed to two significant digits and are acceptable if they are within about 15% of the accepted values.
(b) Answers will depend on the calculated values.
(c) The major source of error in this activity is human reaction time. The student operating the stopwatch will have a difficult
time starting and stopping the watch at the exact instants required. Another major source of error is the overlapping of
dots at the start of the motion where a few dots may be difficult or impossible to distinguish. Yet another source of error
could occur if the timer is not working properly; it could be double dotting or skipping dots.
PRACTICE
(Page 10)
Understanding Concepts
12. (a) f = 60.0 Hz
1

T = ?
T = f
1
= 60 Hz
-2
T = 1.67x10 s

The period of the spark timer is 1.67 x 10-2 s.
(b)
f = 30.0 Hz
1

T = ?
T = f
1
=

30.0 Hz
2
T = 3.33x10- s

The period of the spark timer is 3.33 x 10-2 s.
13. T = 0.10 s
1
f = ?
f = T
1
= 0.10s
1
f =1.0x10 Hz
The frequency of the spark timer is 1.0 x 101 Hz.
4
Unit 1 Forces and Motion
Copyright (c) 2002 Nelson Thomson Learning

Section 1.1 Questions
(Page 11)
Understanding Concepts
1. d =3.9 km + 180.2 km + 42.2 km
d
d
= 226.3 km (2.263 x 105 m)
av
v = t
t = 8 h 17 min 17 s
5
= 8 h (3600 s/h) + 17 min (60 s/min) + 17 s
2.263x10 m
=

t = 29 837 s
29 837 s
vav =?
= 7.584m/s
av
v = 27.30 km/h

(1.0 m/s = 3.6 km/h)

The winner's average velocity was 7.584 m/s or 27.30 km/h.
2. (a) v =3.00 x 108 m/s (velocity of light)
d = vav(t)

t
=1.00 s
=
3.00
x 108 m/s (1.00 s)

d =?
d = 3.00 x 108 m

Light travels a distance of 3.0 x 108 m in a time of 1.00 s.
(b)
v = 3.00 x 108 m/s
d = vav(t)

t
= 1.00 ms
=
3.00
x 108 m/s(1.00 x 10-3 s)

=
1.00
x 10-3 s
d = 3.00 x 105 m.

Light travels a distance of 3.00 x 105 m in a time of 1.00 ms.
3. Estimated distance: d = 5.0 x 103 km = 5.0 x 106 m
d
Estimated
average
walking
velocity:
v
t =
av = 1.5 m/s
v
t = ?
av
6
5.0x10 m
=

1.5 m/s
6
= 3.3x10 s
t = 38 d
Walking non-stop, it would take about 38 d to walk across Canada.
Applying Inquiry Skills
4. (a) One could measure the distance travelled by the tip of the golf club as it moves through the arc. This is the value of d.
By counting the number of separate images and knowing the period of the stroboscope's flash, the total time for the
swing is determined by the number of intervals (number of images minus one) multiplied by the period of the
d
stroboscope. This is the value of t. The average velocity is calculated by using the equation vav = .
t
(b) The slowest instantaneous velocity occurs when the distance between successive images is the least. The fastest
instantaneous velocity occurs when this distance is the greatest (as the club is striking the ball). In both cases, the
d
velocities can be calculated using vav = , where d = distance travelled and t = time taken (period of strobe).
t
5. (a) 138 dots represent 137 complete vibrations (cycles). The time interval is t = 2.50 s.
137

frequency:
f =
= 54.8 Hz
2.50s
(b) measured frequency: 54.8 Hz; true frequency: 60.0 Hz
experimental - a
ccepted
% error =
x100
accepted
54.8 Hz --60.0 Hz

=
x100

60.0 Hz
% error = 8.7%

The experimental error is 8.7%.
Copyright (c) 2002 Nelson Thomson Learning
Chapter 1 Motion 5

Making Connections
6. The car's odometer measures the distance travelled. Its speedometer measures the instantaneous velocity.
7. Sign (d) is the best because it indicates the "maximum" allowed velocity (as opposed to a possible interpretation of
"velocity" as average velocity). Second, the unit of km/h is proper SI form. The unit "kph" is not recognized and the
velocity of "60" omits units completely.

1.2 UNIFORM MOTION
PRACTICE
(Pages 12-14)
Understanding Concepts
1. An object that falls straight down experiences linear motion but not uniform motion because it is accelerating as it falls.

2. d 1 = 2.1 m [S]
d d
d

= 2 - 1
d
= 9.7 m [S] - 2.1 m [S]
2 = 9.7 m [S]

d = ?
d = 7.6 m [S]
The curling rock's displacement is 7.6 m [S].

3. d 1 = 2.8 m [W]
d d
d

= 2 - 1
d
= 12.6 m [E] - 2.8 m [W]
2 = 12.6 m [E]

d = ?
d = 15.4 m [E]
The dog's required displacement is 15.4 m [E].

4. (a) d 1 = 0.0 m
d = d
d

2 - 1

d

= 4.4 m [fwd] - 0.0 m
2 = 4.4 m [fwd]

d = ?
d = 4.4 m [fwd]

(b) d 1 = 4.4 m [fwd]
d = d
d

2 - 1

d

= 8.8 m [fwd] - 4.4 m [fwd]
2 = 8.8 m [fwd]

d = ?
d = 4.4 m [fwd]

(c)
d 1 = 4.4 m [fwd]
d d
d

= 2 - 1

d

= 13.2 m [fwd] - 4.4 m [fwd]
2 = 13.2 m [fwd]

d = ?
d = 8.8 m [fwd]
5. They are the same.

6. d = 50.0 m [fwd]
d
t = 16.9 s
av
v =

t
v av = ?
50.0 m [fwd]
=

16.9s
av
v = 2.96 m/s [fwd]
The athlete's average velocity was 2.96 m/s [fwd].

d
7. d = v av (t), t =

vav

8. v av = 2.4 mm/s [fwd]
d = v av (t)
t = 140 s

= 2.4 mm/s [fwd](140 s)
d = ?
= 3.4 x 102 mm [fwd]

d = 34 cm [fwd]
The snail's displacement was 34 cm [fwd].

9. v av = 20.8 m/s [fwd] (assumed)

d

t =
d = 178 m [fwd]
v
t = ?
av
178 m [fwd]
= 20.8 m/s [fwd]
t = 8.56 s
It would take the record holder 8.56 s.
6
Unit 1 Forces and Motion
Copyright (c) 2002 Nelson Thomson Learning

Making Connections
6. The car's odometer measures the distance travelled. Its speedometer measures the instantaneous velocity.
7. Sign (d) is the best because it indicates the "maximum" allowed velocity (as opposed to a possible interpretation of
"velocity" as average velocity). Second, the unit of km/h is proper SI form. The unit "kph" is not recognized and the
velocity of "60" omits units completely.

1.2 UNIFORM MOTION
PRACTICE
(Pages 12-14)
Understanding Concepts
1. An object that falls straight down experiences linear motion but not uniform motion because it is accelerating as it falls.

2. d 1 = 2.1 m [S]
d d
d

= 2 - 1
d
= 9.7 m [S] - 2.1 m [S]
2 = 9.7 m [S]

d = ?
d = 7.6 m [S]
The curling rock's displacement is 7.6 m [S].

3. d 1 = 2.8 m [W]
d d
d

= 2 - 1
d
= 12.6 m [E] - 2.8 m [W]
2 = 12.6 m [E]

d = ?
d = 15.4 m [E]
The dog's required displacement is 15.4 m [E].

4. (a) d 1 = 0.0 m
d = d
d

2 - 1

d

= 4.4 m [fwd] - 0.0 m
2 = 4.4 m [fwd]

d = ?
d = 4.4 m [fwd]

(b) d 1 = 4.4 m [fwd]
d = d
d

2 - 1

d

= 8.8 m [fwd] - 4.4 m [fwd]
2 = 8.8 m [fwd]

d = ?
d = 4.4 m [fwd]

(c)
d 1 = 4.4 m [fwd]
d d
d

= 2 - 1

d

= 13.2 m [fwd] - 4.4 m [fwd]
2 = 13.2 m [fwd]

d = ?
d = 8.8 m [fwd]
5. They are the same.

6. d = 50.0 m [fwd]
d
t = 16.9 s
av
v =

t
v av = ?
50.0 m [fwd]
=

16.9s
av
v = 2.96 m/s [fwd]
The athlete's average velocity was 2.96 m/s [fwd].

d
7. d = v av (t), t =

vav

8. v av = 2.4 mm/s [fwd]
d = v av (t)
t = 140 s

= 2.4 mm/s [fwd](140 s)
d = ?
= 3.4 x 102 mm [fwd]

d = 34 cm [fwd]
The snail's displacement was 34 cm [fwd].

9. v av = 20.8 m/s [fwd] (assumed)

d

t =
d = 178 m [fwd]
v
t = ?
av
178 m [fwd]
= 20.8 m/s [fwd]
t = 8.56 s
It would take the record holder 8.56 s.
6
Unit 1 Forces and Motion
Copyright (c) 2002 Nelson Thomson Learning

Try This Activity: Attempting Uniform Motion
(Page 14)

(a) If the motion is uniform, the position-time graph is a straight line with a positive or negative slope. (The corresponding
velocity-time graph is a straight horizontal line.)
(b) It is fairly easy to use a battery-powered toy vehicle or a glider on an air table to produce uniform velocity; however, in
both cases, the moving object should be on a level surface to prevent slowing down or speeding up. It is much more
difficult for a person to create uniform motion; in fact, it is impossible when starting or stopping. Walking backwards is
also a problem.
PRACTICE
(Pages 15-16)
Understanding Concepts

10. d = 108 m [W] - 72 m [W]

d
m =

d = 36 m [W]
t
t = 2.0 s
36m [W]
m = ?
=

2.0s

m = 18 m/s [W]
The slope from 2.0 s to 4.0 s is the same as the slope of the entire line.
11. (a)
Position (m [S])
Time (s)
0.0 0.0
9.3 x 102
1.0
1.86 x 103
2.0
2.79 x 103
3.0
3.72 x 103
4.0
4.65 x 103
5.0
5.58 x 103
6.0
6.51 x 103
7.0
7.44 x 103
8.0
8.37 x 103
9.0
9.30 x 103
10.0
1.023 x 104
11.0
1.116 x 104
12.0
(b)

d
(c)

slope:
m =

t
1.116x 104 m [S]

= 12.0s

m = 9.3 x 102 m/s [S]

All line segments have the same slope, indicating that the velocity is constant.
Copyright (c) 2002 Nelson Thomson Learning
Chapter 1 Motion 7

(d)

(e) area = length x width

= 9.3 x 102 m/s [S] (12.0 s)

area = 1.1 x 104 m [S]

This area represents the jet's displacement during the time interval.

d
12. (a) m =

t
15.0 m [E]
=

0.10s

m = 1.5 x 102 m/s [E]

d
(b)
m =

t
5.0 m [E]
=

0.10s

m = 5.0 x 101 m/s [E]

d
(c)
m =

t
10.0 m [W]
=

0.20s

m = 5.0 x 101 m/s [W]
13. (a) d = area under graph
= 40.0 m/s [N](3.0 s)

d = 1.2 x 102 m [N]

(b)
d = area under graph
= 30.0 m/s [N](4.0 s)

d = 1.2 x 102 m [N]

(c)
d = area under graph
= -15.0 m/s [N](8.0 s)
= -1.2 x 102 m [N]

d = 1.2 x 102 m [S]

Section 1.2 Questions
(Pages 17-18)
Understanding Concepts
1. (a) The slope of a line on a position-time graph represents "average velocity."
(b) The area under a line on a velocity-time graph represents "displacement."
2. The magnitude of the slope of the line on a position-time graph and the magnitude of the velocity of the motion are
equivalent.
3. (a) The runner's average velocity should be greater in the 60.0-m sprint than in the 50.0-m sprint since the athlete is able to
maintain the top velocity for a greater distance (and time). This assumes the period of acceleration from the start of the
race to the top velocity is the same for both sprints.
8
Unit 1 Forces and Motion
Copyright (c) 2002 Nelson Thomson Learning