-
Copyright 2003 Nelson Unit 1 Are You Ready? 1
Unit 1 Forces and Motion: Dynamics
ARE YOU READY? (Pages 23)
Knowledge and Understanding 1. Scalar quantities include
distance (metre, 5.0 m), time (second, 15 s), mass (kilogram, 65
kg), and frequency (hertz,
60 Hz). Vector quantities include velocity (metres per second,
15 m/s [E]), displacement (metre, 6.5 m [S]), acceleration (metres
per second squared, 9.8 m/s2 [down]), and force (newton, 25 N
[forward]).
2. (a) Both masses will hit the floor at the same time since the
speed at which an object falls is independent of mass, and is
related only to acceleration due to gravity (neglecting air
resistance).
(b)
(c) m = 20 g = 0.02 kg g ?F =
G
g
g
(0.02 kg)(9.8 N/kg [down])
= 0.2 N [down]
F mg
F
=
=
G G
G
The weight of the 20-g mass is 0.2 N [down]. (d) One example is
the force of Earth pulling downward on the 20-g mass and the force
of the 20-g mass pulling upward
on Earth.
3. E MoonG 2GM mF
r= The magnitude of the force of gravity between Earth and the
Moon depends linearly on the masses of
Earth and the Moon, and depends inversely as the square of the
distance between the centres of Earth and the Moon. 4. (a)
Kinematics is the study of motion (e.g., analyzing motion with
constant acceleration). Dynamics is the study of the
causes of motion (e.g., analyzing forces by applying Newtons
three laws of motion).
(b) Average speed is a scalar quantity, total distanceav total
time.v = Average velocity is a vector quantity, change of
positionav
time intervalv =G
.
(c) Static friction is a force that acts to prevent a stationary
object from starting to move. Kinetic friction is a force that acts
against a moving object. For a given situation, kinetic friction
tends to be less than maximum static friction.
(d) Helpful friction is needed in many cases (e.g., turning a
doorknob, walking, and travelling around a corner on a highway).
Unwanted friction usually increases the production of waste heat
(e.g., friction in the moving parts of an engine).
(e) Frequency is the number of cycles of a vibration per unit
time; it is measured in hertz (Hz) or s1. Period is the time for
one complete cycle of a vibration; it is measured in seconds
(s).
(f) Rotation is the spinning of an object on its own axis (e.g.,
Earth rotates daily). Revolution is the motion of one body around
another (e.g., Earth revolves around the Sun once per year).
Inquiry and Communication 5. (a) The units of acceleration are
m/s2, so the inspector could use a metre stick to measure the
distance in metres (m), and a
stopwatch to measure the time in seconds (s). (b) The
acceleration is the dependent variable, and the distance and time
are the independent variables. (Students will
discover in Chapter 3 that the distance is actually the radius
of the circle and the time is the period of rotation of the
ride.)
-
2 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
6. Error analysis can be reviewed by referring to page 755 of
the text. Note that possible error is also called uncertainty. (a)
The possible error is half of the smallest division of the
measurement, or 0.05 m/s2.
(b) possible error
% possible error = 100%measurement
2
2
0.05 m/s100%
9.4 m/s% possible error 0.5%
=
=
The percent possible error is 0.5%.
(c) measured value accepted value
% error = 100%accepted value
2 2
2
9.4 m/s 9.8 m/s100%
9.8 m/s% error = 4.0%
=
The percentage error is 4.0%.
(d) difference in values
% difference = 100%average of values
( )2 2
2 2
9.7 m/s 9.4 m/s100%
19.7 m/s 9.4 m/s
2% difference = 3.0%
=
+
The percentage difference is 3.0%. 7. A prediction is a stated
outcome expected from an experiment. A hypotenuse is a tentative
explanation of what is
expected in an experiment.
Making Connections 8. The rapid spinning in the centrifuge would
cause the liquids to separate according to their densities, with
the liquid of
greatest density moving farthest away, that is, toward the base
of the tube. 9. (a) Let gF
G = force of gravity on the truck
NFG
= normal force of the road on the truck
sFG
= force of static friction on the truck = angle of the banked
curve in (a) and (c) (a) (b) (c)
(b) Choice (c) would be the best because the force of the road
on the truck (the normal force) will help a lot in forcing the
truck to the right.
-
Copyright 2003 Nelson Unit 1 Are You Ready? 3
Math Skills 10. (a) The equation is rearranged as follows.
i
2i
i2
2
12
2
12
d v t
d v t a t
vda
tt
a t = +
=
=
G G
G G GG GG
G
(b) The quadratic formula is used to solve an equation in the
form 2 0ax bx c+ + =
As shown on page 750 of the text, the quadratic equation is
2 4
2b b ac
xa
=
Solving for t, we have:
( )
( ) ( )( )2
i
i2
2
i i
12
2
2
d v t a t
a t v t d
v v a dt
a
=
= +
+
=
G G GG GG
GG G GG
11. (a) Using the scale indicated in the question: 29.0 m/s [35
N of E]A =
G
The north and east components of this vector are, respectively:
A (sin 35) = 29.0 m/s (sin 35) = 17 m/s A (cos 35) = 29.0 m/s (cos
35) = 24 m/s
Notice that the answers are written to two significant digits
because the angle is stated to two significant digits. (b) The
vectors can be added by using a vector scale diagram (adding the
vectors head-to-tail), by using components, or by
applying trigonometry (the laws of sines and cosines). (c)
Scale: 1.0 cm = 5.0 cm
3.9 cm 5.0 cm 20 cm [65 N of E]B A+ = = GG
3.3 cm 5.0 cm 17 cm [2 S of E]A B = =
G G
-
4 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
Technical Skills and Safety 12. (a) The total time is 6(0.10 s)
= 0.60 s. (b)
As shown in the illustration, the x-component of each
displacement vector is about 1.0 cm in the diagram, or 5.0 cm
using the scale indicated. We can conclude that the motion in
the x direction is constant-speed motion. (c) The y-component of
the displacements constantly changes, becoming smaller as the puck
rises, and then becoming
larger as the puck descends. (d) The displacement (or change of
position) from the initial position to the final position is:
6.15 cm 5.0 cm/cm 31 cm [right]d = =G
t = 0.60 s
avvG = ?
av
av
31 cm [right]0.60 s
52 cm/s [right]
dv
t
v
=
=
=
GG
G
The average velocity of the puck is 52 cm/s [right]. 13. (a)
Using a stopwatch, determine the total time for a certain number of
complete revolutions of the stopper (e.g., 20
cycles). Then apply the following relationships:
frequency: number of cycles
total timef =
period: total time
number of cyclesT =
(b) The string should be strong, the stopper should be securely
attached to the string, and the lab partner should hold the string
securely while twirling the stopper a safe distance away from
objects or people.
(c) Typical sources of error are: starting and stopping the
stopwatch at the precise instant required (also called human
reaction time error.) choosing the same position to indicate both
the starting and finishing locations of the motion keeping the
stopper moving at a constant speed and/or in a horizontal circle
measuring the radius of the circle of motion
-
Copyright 2003 Nelson Chapter 1 Kinematics 5
CHAPTER 1 KINEMATICS
Reflect on Your Learning (Page 4)
1. (a)
(b) Ball B will land first because it has an initial downward
component of velocity. Balls A and C will land at the same
instant because, as students will discover in the Try This
Activity, page 5, the horizontal component of the velocity of ball
C does not affect its downward acceleration. Ball D will land last
because its initial velocity has an upward vertical component.
2. (a)
(b) The canoeists arrive at the north shore at the same time.
The motion of the canoeist in the river is perpendicular to the
flow of the river, so the motion is not affected by the flow of
the river. (c) The figure below shows that the canoeist must aim
upstream in order to arrive directly north of the starting
position.
This trip will take longer because the component of the velocity
perpendicular to the shore is less than it was previously.
Try This Activity: Choose the Winner It is important in setting
up this demonstration that the device be fixed horizontally. For an
alternative suggestion to the apparatus shown in the text, refer to
Section 1.4 Questions, page 51, question 10, and the corresponding
solution. (a) Predictions may vary. Refer to the Reflect on your
learning questions 1(b) above.
(b) Balls A and B will land simultaneously. The horizontal
motion of the ball projected horizontally is independent of its
downward acceleration.
-
6 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
1.1 SPEED AND VELOCITY IN ONE AND TWO DIMENSIONS
PRACTICE (Pages 78)
Understanding Concepts 1. (a) The motion of a tennis ball that
falls vertically downward is in one dimension. (b) The motion of a
tennis ball that falls vertically downward and then bounces is in
one dimension. (c) The motion of a basketball moving through the
air toward the hoop is in two dimensions. (d) The motion of a curve
ball is in three dimensions. (e) The motion of a passenger seat of
a Ferris wheel is in two dimensions. (f) The motion of a roller
coaster is in three dimensions. 2. (a) The force exerted by an
elevator cable is a vector measurement. (b) The reading on a cars
odometer is a scalar measurement. (c) The gravitational force of
Earth on you is a vector measurement. (d) The number of physics
students in your class is a scalar measurement. (e) Your age is a
scalar measurement. 3. A cars speedometer indicates instantaneous
speed, a scalar quantity. It does not indicate any direction. 4.
(a) t = 6.69 h d = 4.02 km 200 laps = 8.04 102 km vav = ?
av
2
2av
8.04 10 km6.69 h
1.20 10 km/h
dvt
v
=
=
=
The average speed in 1911 was 1.20 102 km/h. (b) t = 3.32 h vav
= ?
av
2
2av
8.04 10 km3.32 h
2.42 10 km/h
dvt
v
=
=
=
The average speed in 1965 was 2.42 102 km/h. (c) t = 2.69 h vav
= ?
av
2
2av
8.04 10 km2.69 h
2.99 10 km/h
dvt
v
=
=
=
The average speed in 1990 was 2.99 102 km/h. 5. (a) d = 16 m
t = 21 s vav = ?
av
av
16 m21s
0.76 m/s
dvt
v
=
=
=
The swimmers average speed is 0.76 m/s.
-
Copyright 2003 Nelson Chapter 1 Kinematics 7
(b) circumference = D = (16 m) = 50.26 m t = ?
av
av
50.26 m0.76 m/s66s
dvt
dtv
t
=
=
=
=
It would take the swimmer 66 s to swim around the edge of the
pool. 6. (a) vav = 342 m/s
t = 3.54 102 s d = ?
av
2(342 m/s) (3.54 10 s)12.1m
d v t
d
= =
=
The distance travelled is 12.1 m. (b) vav = 1.74 km/h
t = 60.0 h d = ?
av
(1.74 km/h) (60.0 h)104 km
d v t
d
= =
=
The distance travelled is 104 km, or 1.04 105 m.
PRACTICE (Page 10)
Understanding Concepts 7. Typical examples of different types of
vector quantities include a displacement while walking at a
constant velocity, an
acceleration while on a school bus, and the force of gravity. 8.
(a) It is possible for the total distance travelled to be equal to
the magnitude of the displacement if the motion is all in one
direction. (b) It is possible for the total distance travelled
to exceed the magnitude of the displacement if the motion is in
one
dimension along a path that turns back on itself, or if the
motion is in two or three dimensions. For example, if a car travels
50 km [N] and then 20 km [S], the distance travelled is 70 km, but
the displacement is 30 km [N].
(c) It is not possible for the magnitude of the displacement to
exceed the total distance travelled. The total distance is the sum
of the magnitudes of all of the vectors. Thus, the total distance
is always equal to or greater than the magnitude of the final
displacement vector.
9. Yes, the average speed can equal the magnitude of the average
velocity if the motion is all in one direction. 10. (a) d = 12 km +
12 km = 24 km
t = 24 min + 24 min = (48 min) 1 h60 min
= 0.80 h vav = ?
av
1av
24 km0.80 h
3.0 10 km/h
dvt
v
=
=
=
The average speed of the bus for the entire route is 3.0 101
km/h.
-
8 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
(b) dG
= 12 km [E]
t = (24 min) 1 h60 min
= 0.40 h av ?v =G
av
1av
12 km[E]0.40 h
3.0 10 km/h[E]
dvt
v
=
=
=
GG
G
The average velocity of the bus is 3.0 101 km/h [E]. (c) vav = ?
First we must calculate the total displacement:
12 km[E] 12 km[W]12 km [E] ( 12 km[E])
0.0 km
d
d
= += +
=
G
G
Since the total displacement is 0.0 km, the average velocity of
the bus for the entire route is 0.00 km/hr. (d) The answers in (b)
and (c) are different because the average velocity is determined by
the displacement, which is a
vector. In (b) the bus has reached its maximum displacement,
however in (c), the bus returns to its starting position, so its
displacement is zero and its average velocity is zero.
11. t = 0.32 s avvG
= 27 m/s [fwd]
?d =G
av
(27 m/s [fwd])(0.32s)
8.6 m [fwd]
d v t
d
= =
=
G G
G
The truck is displaced 8.6 m [fwd] during the time it takes for
the driver to react. 12. d
G= 1.6 104 km
avvG =21 km/h [S]
t = ?
av4
2
1.6 10 km[S]21km/h
1d 7.6 10 h24 h
32 d
dtv
t
t
=
=
= =
GG
The terns journey takes 7.6 102 h or 32 days.
Applying Inquiry Skills 13. (a) The windsock indicates both the
approximate speed and the approximate direction of the wind. Speed
with a direction
is velocity, a vector quantity. (b) Calibrating the windsock
involves finding how the angle of the sock above the vertical
depends on the speed of the
wind. (At a wind speed of zero, the sock hangs vertically
downward.) Thus, an experiment must be devised to measure the angle
at various increasing speeds (e.g., at 10 km/h, 20 km/h, etc.). The
easiest way to do this would be to hold the sock securely out of an
open window of a car as the car travels at different speeds on a
calm day. (Students are not expected to attempt such an
experiment.)
-
Copyright 2003 Nelson Chapter 1 Kinematics 9
Try This Activity: Graphing Linear Motion (Page 12) The required
graphs are drawn with the assumption that the zero displacement is
the location of the motion sensor, and the positive displacement
direction is away from the sensor.
PRACTICE (Pages 1314)
Understanding Concepts 14. (a) The motion, beginning north of
the zero displacement position, is southward with constant velocity
to a position south
of the zero displacement position. (b) The motion begins at the
zero displacement position and is upward but slowing down (downward
acceleration). (c) The motion begins west of the zero displacement
position and is eastward but slowing down (westward acceleration).
15. (a)
-
10 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
(b)
(c)
16. The area under the velocity-time graph represents the
displacement. Assume two significant digits.
1Area (15m/s [N])(0.20s) (15m/s [N])(0.40s 0.20s)2
3.0 m [N] 1.5m [N]Area 4.5m [N]
= +
= +
=
Thus, the area is 4.5 m [N]. 17. In each case, the initial
velocity is equal to the slope of the tangent at the time
indicated. The diagram with typical
tangents is shown below:
Using the equation slope dv mt
= = =
GG , students should calculate approximate answers of 7 m/s [E],
0 m/s, 7 m/s [W], 13 m/s [W], and 7 m/s [W].
-
Copyright 2003 Nelson Chapter 1 Kinematics 11
PRACTICE (Page 16)
Understanding Concepts 18. Using a vector scale diagram to solve
this problem, the vector sum of the displacements is found to be
5.6 m [24 E of S].
Using components would yield the same result.
19. (a)
(b) Solving Sample Problem 5 (c) using components, with +x
eastward and +y northward:
1
1,
1,
22 m
22 m (cos33 )18m
x
x
d
dd
=
= =
G2
2,
2,
11m
11m (cos 28 )10 m
x
x
d
dd
=
= =
G
1,1,
22 m (sin 33 )
12 my
y
d
d
= =
2,2,
11m (sin 28 )
5my
y
d
d
= =
1, 2,
18m 10 m28m
x x x
x
d d d
d
= + = +
=
1, 2,
12 m 5.0 m7.0 m
y y y
y
d d d
d
= =
=
-
12 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
To calculate the total displacement:
2 2
2 2(28m) (7.0 m)
29 m
x yd d d
d
= +
= +
=
G
G
1
tan
7.0 mtan28 m
14
y
x
dd
=
= =
The total displacement is 29 m [14 N of E]. 20. 1d
G= 8.5 102 m [25 N of E]
2dG
= 5.6 102 m [21 E of N]
t = 4.2 min 60 s1 min
= 252 s (a) Add the vectors together:
Use the cosine law to find the magnitude of the resultant
displacement:
2 2 2
1 2 1 2
2 2 2 2 2 2 2
3
2 cos
(8.5 10 m) (5.6 10 m) 2(8.5 10 m)(5.6 10 m)(cos136 )
1.3 10 m
d d d d d
d
d
= +
= +
=
G G G
GG
Use the sine law to solve for the angle of the displacements
direction:
2
2 3
21
3
sin sin
sin sin1365.6 10 m 1.3 10 m
(5.6 10 m)sin136sin1.3 10 m
17
d d
=
=
= =
G G
The angle of the displacement is 17 + 25 = 42 The skaters
displacement is 1.3 103 m [42 N of E].
-
Copyright 2003 Nelson Chapter 1 Kinematics 13
(b) vav = ?
av
2 2
av
(8.5 10 m) (5.6 10 m) 252s
5.6 m/s
dvt
v
=
+
=
=
To calculate the average velocity: av ?v =
G
av
3
av
1.3 10 m [42 N of E]252s
5.2 m/s [42 N of E]
dvt
v
=
=
=
GG
G
The skaters average speed is 5.6 m/s and average velocity is 5.2
m/s [42 N of E].
Section 1.1 Questions (Page 17)
Understanding Concepts 1. (a) scalar (b) scalar (c) scalar (d)
vector (e) vector (equal to the displacement) 2. (a) A car
travelling at constant speed in one direction is at constant
velocity. (b) A car travelling at a constant speed around a
circular track has a velocity that is constantly changing. (c) A
bus travelling from a station to a bus stop and then travelling
back along the same route. (d) In the example in (c) above, if the
bus returns to the station its average speed is greater than zero,
but its average
velocity is zero. (e) A car travelling around a circular track
has an average speed greater than zero, but when it reaches its
starting position,
its average velocity is zero. 3. Measurements with different
dimensions can be multiplied; for example, velocity (m/s) time (s)
= distance (m).
Measurements with different dimensions cannot be added; for
example, velocity cannot be added to time. 4. (a) vav = 3.00 108
m/s d = 1.49 1011 m t = ?
av11
8
2
1.49 10 m3.00 10 m/s
4.97 10 s
dtv
t
=
=
=
Light takes 4.97 102 s to travel from the Sun to Earth. (b) d =
3.84 105 km = 3.84 108 m t = ?
( )
av
8
8
2 3.84 10 m
3.00 10 m/s2.56 s
dtv
t
=
=
=
The laser light takes 2.56 s to travel to the Moon and back to
Earth.
-
14 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
5. (a) t = 4.0 s t = 8.0 s d = 0 m d = 40 m vav = ? vav = ?
av
av
0 m 4.0s
0.0 m/s
dvt
v
=
=
=
av
av
40 m 8.0s
5.0 m/s
dvt
v
=
=
=
The average speed between 4.0 s and 8.0 s is 0.0 m/s. The
average speed between 0.0 s and 8.0 s is 5.0 m/s. (b) t = 1.0 s t =
4.0 s t = 16 s d
G= 2.5 m [E] d
G=45 m [W] d
G= 5.0 [E]
av ?v =G av ?v =G av ?v =G
av
av
2.5m[E]1.0s
2.5m/s[E]
dvt
v
=
=
=
GG
G
av
av
45m[W]4.0s
11m/s[W]
dvt
v
=
=
=
GG
G
av
av
5.0 m[E]16s
0.31m/s[E]
dvt
v
=
=
=
GG
G
The average velocity between 8.0 s and 9.0 s is 2.5 m/s [E]; the
average velocity between 12 s and 16 s is 11 m/s [W]; and the
average velocity between 0.0 s and 16 s is 0.31 m/s [E].
(c) The instantaneous speed at t = 6.0 s and t = 9.0 s is the
slope of the line at that instant. Thus,
slope
0.0 m4.0 s
0.0 m/s
dv mt
v
= = =
=
=
10 m4.0 s2.5 m/s
dv mt
v
= =
=
=
The instantaneous speed at 6.0 s is 0.0 m/s and at 4.0 s is 2.5
m/s. (d) The instantaneous velocity at t = 14 s is the slope of the
line at that instant. Thus,
slope
45 m [W]4.0 s
11 m/s [W]
dv mt
v
= = =
=
=
GG
G
The instantaneous velocity at 14 s is 11 m/s [W]. 6. The slope
of the line can be calculated from a position-time graph to
indicate velocity. If the graph is curved, the slope of
the tangent to the curve indicates the instantaneous velocity.
7. The data to draw the position-time graph are found by
calculating the area under the line at several instances and
adding
8.0 m [E] to the area in each case. The table shows the
results.
t (s) dG
(m [E]) 0.0 8.0 1.5 9.5 3.0 14 4.0 18 5.0 22
The required graph:
-
Copyright 2003 Nelson Chapter 1 Kinematics 15
8. t = 2.0 min = 120 s d1 = 22 m [36 N of E] d2 = 65 m [25 E of
S]
(a) d = ?
1 2
22 m 65 m87 m
d d d
d
= +
= +
=
Thus, the total distance travelled is 87 m. (b) vav = ?
av
av
87 m 120s
0.73m/s
dvt
v
=
=
=
The average speed is 0.73 m/s. (c) Add the vectors together as
shown in the illustration:
Use the cosine law to find the magnitude of the resultant
displacement:
2 2 21 2 1 2
2 2 2
2 cos
(22 m) (65 m) 2(22 m)(65 m)(cos79 )
65m
d d d d d
d
d
= +
= +
=
G G G
GG
Use the sine law to solve for the angle of the displacements
direction:
2
2
1
sin sin
sinsin
(65m)sin 79sin65m
79
d d
d
d
=
=
= =
G GGG
The angle of the displacement is 79 36 = 43. Thus, the total
displacement is 65 m [43 S of E].
-
16 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
(d) avv =G ?
av
av
65 m[43 Sof E]120 s
0.54 m/s[43 S of E]
dvt
v
=
=
=
GG
G
The average velocity is 0.54 m/s [43 S of E].
Applying Inquiry Skills 9. (a) Students can refer to the
Learning Tip titled The Image of a Tangent Line on page 13 of the
text to understand how
to use a plane mirror to check the accuracy of their tangents.
(b) One way to draw tangents accurately is to use the plane mirror
method, as described in the Learning Tip. Another way
that is useful for a displacement-time graph of uniform
acceleration is to draw the tangent parallel to an imaginary line
joining two points that are equal times away from the tangent time
(e.g., at t = 3.5 s, draw the tangent parallel to the line joining
the points at t = 2.5 s and t = 4.5 s).
Making Connections 10. Use the equation d = vavt to complete the
table.
Reaction Distance Speed
no alcohol 4 bottles 5 bottles 17 m/s (60 km/h) 14 m 34 m 51 m
25 m/s (90 km/h) 20 m 50 m 75 m 33 m/s (120 km/h) 26 m 66 m 99
m
1.2 ACCELERATION IN ONE AND TWO DIMENSIONS
PRACTICE (Page 20)
Understanding Concepts 1. All five examples could be units of
acceleration. 2. (a) It is possible to have an eastward velocity
with a westward acceleration. For example, a truck moving eastward
whle
slowing down has a westward acceleration. (b) It is possible to
have acceleration when the velocity is zero. For example, at the
instant that a ball tossed vertically
upward comes to a stop, its acceleration is still downward. 3.
(a) When the flocks acceleration is positive, the flock is moving
south with increasing velocity. (b) When the flocks acceleration is
negative, the flock is moving south with decreasing velocity. (c)
When the flocks acceleration is zero, the flock is moving south
with constant velocity. 4. i 0v =
G fvG = 9.3 m/s [fwd] t = 3.9 s
avaG = ?
f iav
2av
9.3 m/s [fwd] 03.9 s
2.4 m/s [fwd]
v vat
a
=
=
=
G GG
G
The runners average acceleration is 2.4 m/s2 [fwd].
-
16 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
(d) avv =G ?
av
av
65 m[43 Sof E]120 s
0.54 m/s[43 S of E]
dvt
v
=
=
=
GG
G
The average velocity is 0.54 m/s [43 S of E].
Applying Inquiry Skills 9. (a) Students can refer to the
Learning Tip titled The Image of a Tangent Line on page 13 of the
text to understand how
to use a plane mirror to check the accuracy of their tangents.
(b) One way to draw tangents accurately is to use the plane mirror
method, as described in the Learning Tip. Another way
that is useful for a displacement-time graph of uniform
acceleration is to draw the tangent parallel to an imaginary line
joining two points that are equal times away from the tangent time
(e.g., at t = 3.5 s, draw the tangent parallel to the line joining
the points at t = 2.5 s and t = 4.5 s).
Making Connections 10. Use the equation d = vavt to complete the
table.
Reaction Distance Speed
no alcohol 4 bottles 5 bottles 17 m/s (60 km/h) 14 m 34 m 51 m
25 m/s (90 km/h) 20 m 50 m 75 m 33 m/s (120 km/h) 26 m 66 m 99
m
1.2 ACCELERATION IN ONE AND TWO DIMENSIONS
PRACTICE (Page 20)
Understanding Concepts 1. All five examples could be units of
acceleration. 2. (a) It is possible to have an eastward velocity
with a westward acceleration. For example, a truck moving eastward
whle
slowing down has a westward acceleration. (b) It is possible to
have acceleration when the velocity is zero. For example, at the
instant that a ball tossed vertically
upward comes to a stop, its acceleration is still downward. 3.
(a) When the flocks acceleration is positive, the flock is moving
south with increasing velocity. (b) When the flocks acceleration is
negative, the flock is moving south with decreasing velocity. (c)
When the flocks acceleration is zero, the flock is moving south
with constant velocity. 4. i 0v =
G fvG = 9.3 m/s [fwd] t = 3.9 s
avaG = ?
f iav
2av
9.3 m/s [fwd] 03.9 s
2.4 m/s [fwd]
v vat
a
=
=
=
G GG
G
The runners average acceleration is 2.4 m/s2 [fwd].
-
Copyright 2003 Nelson Chapter 1 Kinematics 17
5. i 0v =G
fvG = 26.7 m/s [fwd] aG = 9.52 m/s2
(a) t = ?
f i
av
2
26.7 m/s 09.52 m/s
2.80 s
v vt
a
t
=
=
=
G GG
The Espace takes 2.80 s to achieve a speed of 26.7 m/s. (b) v =
?
m 1 km 3600 s26.7s 1000 m 1 h
96.1 km/h
v
v
=
=
The speed is 96.1 km/h.
(c) ?
f i
av
v vta =G GG
?
2
2?
?
L LT TT
LT
L TTT L
T T
=
= =
Thus, the equation is dimensionally correct. 6. ava
G =14 (km/h)/s [fwd] t = 4.7 s
ivG = 42 km/h [fwd]
f ?v =G
[ ] [ ]( )( )[ ]
f iav
f i av
f
42 km/h fwd + 14(km/h)/s fwd 4.7s
=108 km/h fwd
v vat
v v a t
v
=
= + =
G GGG G G
G
Thus, the final velocity is 108 km/h [fwd]. 7. ava
G = 1.37 103 m/s2 t = 3.12 103 s
fvG =0 m/s
i ?v =G
[ ]( )( )
[ ]
f iav
i f av
3 2 2
i
0 m/s 1.37 10 m/s W 3.12 10 s
42.8m/s E
v vat
v v a t
v
=
=
=
=
G GGG G G
G
The velocity of the arrow as it hits the target is 42.8 m/s
[E].
-
18 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
Try This Activity: Graphing Motion with Acceleration (Page 23)
The required graphs are shown below, in which the position of zero
displacement is located where the cart is near the bottom of the
ramp but is not experiencing a push.
PRACTICE (Pages 2324)
Understanding Concepts 8. (a) To determine the average
acceleration from a velocity-time graph, determine the slope of the
line if the acceleration is
constant. (b) To determine the change in velocity from an
acceleration-time graph, determine the area under the line. 9. (a)
The motion starts with a westward velocity, but constant eastward
acceleration. The motion then slows down to zero
velocity, then accelerates westward with increasing westward
velocity. The magnitude of the westward acceleration is somewhat
less than the magnitude of the eastward acceleration.
(b) The motion is southward with northward acceleration slowing
down to zero velocity. (c) The motion is forward with constant
acceleration forward. After a period of time, the motion increases
to a higher
constant acceleration forward. (d) The motion starts with
northward acceleration then increases its northward acceleration.
It starts to slow down
(southward acceleration), and then decreases its southward
acceleration to zero. 10. (a)
-
Copyright 2003 Nelson Chapter 1 Kinematics 19
(b)
11.
12. The cars displacement is the area under the velocity-time
graph. It is determined by adding the areas of rectangles and
triangles contained in each time segment. Referring to the
figure in the text: displacement = total area = A4 (0 to 3 s) + A5
(3 s to 5 s) +A6 (5 s to 9 s)
4
5
6
1 (12 m/s [S])(3.0s) 18m [S]2
1(12 m/s [S])(2.0s) (18m/s [S] 12 m/s [S])(2.0s) 30 m
[S]21(18m/s [S])(4.0s) (24 m/s [S] 18m/s [S])(4.0s) 84 m [S]2
A
A
A
= =
= + =
= + =
displacement = A4 + A5 + A6 = 132 m [S] The cars displacement is
132 m [S].
Making Connections 13. (a) The word idealized means that the
acceleration changes instantaneously from one value to another. In
real situations,
changes from one acceleration value to another occur over a
finite time interval. (b) Calculations are much easier if idealized
examples are used. For example, to find the change in velocity for
an
idealized acceleration graph, we can find the area of a
rectangle on the graph. That is much easier than finding the area
under a curved line.
(c)
14. The solution to this question depends on the software,
calculator, or planimeter available. Each device is accompanied
by
a set of instructions that students can follow to analyze
graphs.
-
20 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
PRACTICE (Page 27)
Understanding Concepts
15. (a) ( )2
i 2a t
d v t
= +GG G
.
(b) ( )i f12d v v t = + G G G
.
16. 2 2f i 2v v a d= +
( )
2 2?
2
2 2 2?
L L L2 LT T T
L L L2T T T
= +
= +
Since the dimensions of each term are the same, the equation is
dimensionally correct.
17. (a) ( )i f12d v v t = + G G G
i f
2 dtv v =+
GG G
(b) ( )i f12d v v t = + G G G
i f
f i
2
2
dv vt
dv vt
+ =
=
GG GGG G
18. Start with the defining equation for constant acceleration
and the equation for displacement in terms of average velocity:
( )f i
vat
v va
t
=
=
GGG GG ( )
av
i f
2
d v tv v
d t
= +
=
G GG GG
(a) To derive the constant acceleration equation in which the
final velocity has been eliminated: first solve acceleration
equation for fv
G, then substitute into the equation for displacement.
f i
f i
v va
tv v a t
=
= +
G GGG G G
( )( )
i f
i i
2i
2i
2
22
212
v vd t
v v a td t
v t a t
d v t a t
+ =
+ + =
+ =
= +
G GG
G G GG
G G
G G G
-
Copyright 2003 Nelson Chapter 1 Kinematics 21
(b) To derive the constant acceleration equation in which the
initial velocity has been eliminated: first solve the acceleration
equation for iv
G, then substitute into the equation for displacement.
i fv v a t= G G G
( )( )
i f
f f
2f
2f
2
22
212
v vd t
v a t vd t
v t a t
d v t a t
+ =
+ =
=
=
G GG
G G GG
G G
G G G
19. aG =18 m/s2 [E] t = 1.6 s
ivG
=73 m/s [W]
f ?v =G
f i
f i2
f
73m/s[W] 18m/s [E](1.6s)44 m/s[W]
v vat
v v a t
v
=
= + = +
=
G GGG G G
G
The velocity of the birdie is 44 m/s [W]. 20. iv
G= 41 m/s [S]
fvG
= 47 m/s [N] t = 1.9 ms = 1.9 103 s
?a =G
f i
3
4 2
47 m/s[N] 41m/s[S]1.9 10 s
4.6 10 m/s [N]
v vat
a
=
=
=
G GG
G
The acceleration is 4.6 104 m/s2 [N]. 21. Note: As was described
in the text, pages 24 and 25, and applied in Sample Problem 6, page
26, we can omit the vector
notation when 2fvG or 2ivG terms are involved. However, in the
solution shown here as well as the solution for question 24,
we have kept the vector notation in order to show what the final
direction is. i 0v =G aG = 2.3 m/s2 [fwd] t = 3.6 s
(a) ?d =G
( )[ ]( )
2i
2 2
12
2.3m/s fwd (3.6s)0.0 m
215m [fwd]
d v t a t
d
= +
= +
=
G G G
G
The sprinters displacement is 15 m [fwd].
-
22 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
(b) f ?v =G
( )
f i
f i
2
f
0 m/s 2.3 m/s [fwd] (3.6 s)
8.3 m/s [fwd]
v vat
v v a t
v
=
= +
= +
=
G GGG G G
G
The sprinters final velocity is 8.3 m/s [fwd]. 22. iv
G= 7.72 107 m/s [E]
fvG
= 2.46 107 m/s [E]
dG
= 0.478 m [E] (a) ?a =G
( ) ( )
2 2f i
2 2f i
2 27 7
15 2
15 2
2
2
2.46 10 m/s [E] 7.72 10 m/s [E]
2(0.478m [E])
5.60 10 m/s [E]
5.60 10 m/s [W]
v v a d
v vad
a
a
= +
=
=
=
=
GG G GG GG G
GG
The electrons acceleration is 5.60 1015 m/s2 [W]. (b) t = ?
( )
( )( ) ( )
i f
i f
7 7
9
122
2 0.478m [E]
7.72 10 m/s [E] 2.46 10 m/s [E]
9.39 10 s
d v v t
dtv v
t
= +
=+
=
+
=
G G GG
G G
The acceleration occurs over 9.39 109 s.
Applying Inquiry Skills 23. The experiment can be simple.
Besides the book and the desk, the only equipment required is a
metre stick and a
stopwatch. Slide the book along the desk at a constant speed for
a predetermined distance that is long enough that the time interval
to cover the distance is at least 2.0 s (e.g., slide the book at a
constant speed of about 0.50 m/s for 2.0 s). Remove the pushing
force from the book and determine the displacement from that
instant to the stopping position. The acceleration can be found by
applying the equation 2 2f i 2v v a d= + , where vf = 0, vi is the
measured value of the speed while the book is being pushed, and d
is the distance the book slides after it is no longer pushed.
Making Connections 24. iv
G = 75.0 km/h [N] = ( ) 1000 m 1 h75.0 km/h [N]km 3600 s
= 20.8 m/s [N] aG = 4.80 m/s2 [S] reaction time = ?
-
Copyright 2003 Nelson Chapter 1 Kinematics 23
First we must calculate the change in displacement:
( ) ( )
2 2f i
2 2f i
2 2
2
2
20 m/s 20.8m/s[N]
2(4.80 m/s [S])
45.2 m[N]
v v a d
v vda
d
= +
=
=
=
GG G GG GGG
G
Calculate reaction distance (distance before stopping) = 48.0 m
45.2 m = 2.8 m
i
reaction distancereaction time
2.8m20.8 m/s
reaction time 0.13s
v=
=
=
Thus, the reaction time is 0.13 s.
PRACTICE (Page 29)
Understanding Concepts 25. iv
G = 25 m/s [E] fvG =25 m/s [S] t = 15 s
av ?a =G
Using components, with +x eastward and +y southward:
f i f i( ) ( )
25m/s 25 m/s x x x y y y
x y
v v v v v v
v v
= + = + = =
( ) ( )2 2
2 225 m/s 25 m/s
35.3m/s
x yv v v
v
= +
= +
=
G
G
1
tan =
25 m/stan25 m/s
45
y
x
vv
= =
So, [ ]35m/s 45 S of W .v = G Therefore,
[ ]
[ ]
av
2av
t35m/s 45 S of W
15 s2.4 m/s 45 S of W
va
a
=
=
=
GG
G
Thus, the cars average acceleration is 2.4 m/s2 [45 S of W].
-
24 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
26. ivG = 6.4 m/s [E]
avaG = 2.0 m/s2 [S] t = 2.5 s
f ?v =G
( )( )
f iav
f i av
2
f
tt
6.4 m/s [E] 2.0 m/s [S] 2.5s
6.4 m/s [E] 5.0 m/s [S]
v va
v v a
v
=
= +
= +
= +
G GGG G G
G
( ) ( )2 2
f f f
2 2
f
6.4 m/s 5.0 m/s
8.1m/s
x yv v v
v
= +
= +
=
G
G
f
fx
1
tan
5.0 m/stan6.4 m/s
38
yvv
=
= =
The final velocity of the watercraft is 8.1 m/s [38 S of E]. 27.
iv
G = 26 m/s [22 S of E] fvG = 21 m/s [22 N of E]
av ?a =G
Using +x eastward and +y northward:
( ) ( )
av,
3
3 2av,
21m/s cos 22 26 m/s cos 224.6 m/s
t4.6 m/s
2.5 10 s1.9 10 m/s
x
x
xx
x
vv
va
a
= =
=
=
=
( ) ( )
av,
3
3 2av,
21m/s sin 22 26 m/s sin 22
17.6 m/s
t17.6 m/s
2.5 10 s7.0 10 m/s
y
y
yy
y
v
v
va
a
= =
=
=
=
( ) ( )2 2
av av av
2 23 2 3 2
3 2av
av
av
3 21
3 2
1.9 10 m/s 7.0 10 m/s
7.3 10 m/s
tan
7.0 10 m/stan1.9 10 m/s
75
x y
y
x
a a a
a
aa
= +
= +
=
=
=
=
G
G
Thus, the average acceleration of the puck is 7.3 103 m/s2 [75 N
of W].
-
Copyright 2003 Nelson Chapter 1 Kinematics 25
28. avaG =9.8 m/s2 [down] t = 2.0 s
fvG =24 m/s [45 below horizontal]
i ?v =G
[ ] [ ]( )( )[ ] [ ]
f iav
i f av
2
i
t
24 m/s 45below the horizontal 9.8 m/s down 2.0s
24 m/s 45below the horizontal 19.6 m/s down
v va
v v a t
v
=
=
=
=
G GGG G G
G
( )i
i
24 m/s cos 4517 m/s
x
x
vv
=
=
( ) ( )i
i
24 m/s sin 45 20 m/s
2.6 m/sy
y
v
v
=
=
( ) ( )2 2
i i i
2 2
i
17 m/s 2.6 m/s
17 m/s
x yv v v
v
= +
= +
=
G
G
i
i
1
tan
2.6 m/stan17 m/s
10
y
x
vv
=
= =
The balls initial velocity is 17 m/s [10 above the
horizontal].
29. iv = 82.0 km/h = 82.0 km 1000 m 1 h
1 h 1 km 3600 s = 22.8 m/s
i f 22.8m/sv v= =
t = 15 min = 60 s15 minmin
= 9.00 103 s
As stated in the question, +x east and +y north.
( ) ( )av,
3
3 2av,
22.8m/s cos12.7 22.8m/s sin 38.29.00 10 s
9.0 10 m/s
xx
x
vat
a
=
=
=
( ) ( )
av,
3
2 2av,
22.8 m/s sin12.7 22.8 m/s cos38.29.00 10 s
2.5 10 m/s
yy
y
va
t
a
=
=
=
Thus, the x-component of the average acceleration is 9.0 103
m/s2 and the y-component is 2.5 102 m/s2.
Section 1.2 Questions (Pages 3031)
Understanding Concepts 1. Instantaneous acceleration equals
average acceleration during motion of constant acceleration. 2. It
is possible to have a northward velocity with westward acceleration
if there is a change in direction. For example, if a
truck is initially moving northward at 50 km/h and changes
direction to obtain a final velocity of 50 km/h [45 W of N], the
change in velocity and, thus, the acceleration just at the instant
of the initial velocity, is westward.
-
26 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
3. ivG
= 1.65 103 km/h [W]
fvG
= 1.12 103 km/h [W] t = 345 s
(a) ?a =G
[ ]
[ ]
f i
3 31.12 10 km/h [W] 1.65 10 km/h W345 s
1.54 (km/h)/s E
v va
t
a
=
=
=
G GG
G
The average acceleration of the aircraft is 1.54 (km/h)/s
[E].
(b) 1000 m 1 h1.54 (km/h)/skm 3600 s
a = G
20.427 m/s [E]a =G The average acceleration of the aircraft is
0.427 m/s2 [E]. 4. (a)
(b) To determine the instantaneous acceleration at t = 2.0 s,
calculate the slope of the tangent to the curve indicated on
the
graph. 5. (a) Students can determine the data for the
velocity-time graph by using the constant acceleration equation avd
v t =
G G (applied at the times indicated), or by drawing the
position-time graph and finding the tangents to the curve. The
table below gives the data. The velocity-time graph is a straight
line, and its slope indicates the acceleration.
t (s) 0 0.2 0.4 0.6 0.8
vG (m/s [W]) 0 2.6 5.2 7.8 10.4
-
Copyright 2003 Nelson Chapter 1 Kinematics 27
(b) ?a =G
( )
( )
( )
2
i i
2
2
2
(where 0)2
2
2(4.16 m [W])0.80 s
13 m/s
a td v t v
dat
a
= + =
=
=
=
GG G GGG
G
6. (a) The motion starts at a specific location at a high
velocity with a negative acceleration, reaching zero velocity
midway through the motion. The motion then undergoes increasing
velocity in the opposite direction until reaching the initial
position.
(b) This motion is increasing velocity in the negative
direction, followed by a decreasing velocity in the same direction,
eventually reaching zero velocity. This is followed by increasing
velocity in the positive direction. The magnitudes of the negative
and positive accelerations are equal.
(c) The motion in this graph starts with a constant velocity,
then accelerates at a high rate for a short time before slowing
down with negative acceleration at a lower rate to zero
velocity.
7. ivG
=26 m/s [E] aG = 5.5 m/s2 [W] = 5.5 m/s2 [E] t = 2.6 s
f ?v =G
f i
f i
2
f
26 m/s[E] ( 5.5m/s [E])(2.6s)12 m/s[E]
v vat
v v a t
v
=
= +
= +
=
G GGG G G
G
The cars velocity is 12 m/s [E]. 8. aG = 9.7 m/s2 [fwd]
t = 2.9 s i ?v =G
f i
i f2
i
0 ( 9.7 m/s [fwd])(2.9s)28 m/s[fwd]
v vat
v v a t
v
=
= =
=
G GGG G
G
The cars initial speed is 28 m/s. 9. The data points for the
position-time graph can be found by finding the total area on the
velocity-time graph up the each
second. The results are shown in the table.
t (s) 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0
dG
(m [W]) 0 2.5 10.0 22.5 37.5 52.5 67.5 78.8 82.5 78.8 67.5 56.2
52.5
-
28 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
The acceleration-time graph is generated by calculating the
slopes of the line segments on the velocity-time graph.
10. aG
= 4.4 m/s2 [fwd] t = 3.4 s
ivG
= 0 (a) f ?v =
G
[ ]( )
f i
f i
2
f
0 4.4 m/s fwd (3.4 s)
15 m/s[fwd]
v va
tv v a t
v
=
= +
= +
=
G GGG G G
G
The jumpers final velocity is 15 m/s [fwd]. (b) ?d =
G
( )( )
2i
2 2
12
10 4.4 m/s [fwd] (3.4 s)2
25 m[fwd]
d v t a t
d
= +
= +
=
G G G
G
The jumpers displacement is 25 m [fwd]. 11. iv
G= 0
fvG
= 2.0 107 m/s [E]
d =G
0.10 m [E] (a) ?a =G
2 2f i
2f
7 2
15 2
2
2(2.0 10 m/s[E])
2(0.10 m[E])
2.0 10 m/s [E]
v v a d
va
d
a
= +
=
=
=
GG G GGG G
G
The acceleration of the electron is 2.0 1015 m/s2 [E].
-
Copyright 2003 Nelson Chapter 1 Kinematics 29
(b) t = ?
f i
f i
7
15 2
8
2.0 10 m/s[E] 02.0 10 m/s [E]
1.0 10 s
v va
tv v
ta
t
=
=
=
=
G GGG GG
The electron takes 1.0 108 s to reach its final velocity. 12.
iv
G= 204 m/s [fwd]
fvG
= 508 m/s [fwd] t = 29.4 s
?d =G
( )( ) ( )
i f
4
12204 m/s [fwd] 508 m/s [fwd]
29.4 s2
1.05 10 m [fwd]
d v v t
d
= +
+=
=
G G G
G
The displacement of the rocket is 1.05 104 m [fwd]. 13. iv
G= 0
fvG
= 4.2 102 m/s [fwd]
dG
= 0.56 m [fwd] (a) av ?v =
G
i fav
2
2av
20 4.2 10 m/s[fwd]
22.1 10 m/s[fwd]
v vv
v
+=
+ =
=
G GG
G
The average velocity of the bullet is 2.1 102 m/s [fwd]. (b) t =
?
av
av
2
3
0.56 m[fwd]2.1 10 m/s [fwd]
2.7 10 s
d v t
dtv
t
=
=
=
=
G GGG
The uniform acceleration occurs over 2.7 103 s. 14. (a) After 45
s, the car and the van have the same velocity (from the graph). (b)
Let the subscript V represent the van and the subscript C represent
the car. The displacements of the two vehicles are
equal at some time t, and can be found by determining the areas
under the lines on the graphs.
( )
( )( )( )( )
V triangle,V rectangle,V
V1 V1 V2 V2av
V
20 m/s 0 m/s 60s + 20 m/s 60s2
600 m+ 20 m/s 60s
d A A
v t v t
t
d t
= +
= +
+ =
=
GG G
G
( )
( )( )( )( )
C triangle,C rectangle,C
C1 C1 C2 C2av
C
15m/s 0 m/s 30s + 15m/s 30s2
225m+ 15m/s 30s
d A A
v t v t
t
d t
= +
= +
+ =
=
GG G
G
-
30 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
Set V Cd d = G G
and solve for t:
( )( ) ( )( )( ) ( ) ( )( ) ( )( )
600 m+ 20 m/s 60s 225m+ 15m/s 30s
20 m/s 15m/s 225m 15m/s 30s 20 m/s 60s 600 m375m5 m/s75s
t t
t t
t
t
=
= +
=
=
Thus, the van V overtakes the car C at 75 s. (c) Using t = 75 s,
substitute into the equation for Vd
G or Cd
G.
( )( )( )
V V1 V1 V2 V2av
V
20 m/s 0 m/s 60s + 20 m/s 75s 60s2
600 m+300 m
= 900 m
d v t v t
d
= +
+ = =
G G G
G
The displacement from the intersection when V overtakes C is 900
m. 15. Av
G= 4.4 m/s [31 S of E]
BvG
= 7.8 m/s [25 N of E] t = 8.5 s
?a =G
Using +x east and +y north, we find the components of the
velocities and then the accelerations.
( )
( )
A
A
A
A
4.4 m/s cos31 3.8m/s
4.4 m/s sin 31
2.3m/s
x
x
y
y
vv
v
v
=
=
=
=
( )
( )
B
B
B
B
7.8m/s cos 25 7.1m/s
7.8m/s sin 25
3.3m/s
x
x
y
y
vv
v
v
=
=
=
=
B A
2
t7.1m/s 3.8m/s
8.5s
0.39 m/s
x xx
x
v va
a
=
=
=
( )
B A
2
3.3m/s 2.3m/s8.5s
0.66 m/s
y yy
y
v va
t
a
=
=
=
( ) ( )2 2
2 22 2
2
0.39 m/s 0.66 m/s
0.76 m/s
x ya a a
a
= +
= +
=
G
G
2
12
tan =
0.66 m/stan0.39 m/s
31
y
x
aa
= =
The birds average acceleration is 0.76 m/s2 [31 E of N].
-
Copyright 2003 Nelson Chapter 1 Kinematics 31
16. ivG
= 155 km/h [E]
fvG
= 118 km/h [S] t = 56.5 s
?a =G
Using components with +x east and +y south:
f i
t0 km/h 155km/h
56.5s2.74(km/h)/s
x xx
x
v va
a
=
=
=
f i
t118km/h 0 km/h
56.5s2.09(km/h)/s
y yy
y
v va
a
=
=
=
( ) ( )2 2
av
2 2
av
2.74 (km/h)/s 2.09 (km/h)/s
3.45(km/h)/s
x ya a a
a
= +
= +
=
G
G
1
tan =
2.74 (km/h)/stan2.09 (km/h)/s
52.7
y
x
aa
= =
Thus, the helicopters average acceleration is 3.45 (km/h)/s
[52.7 W of S].
Applying Inquiry Skills 17. In both cases, the variable most
difficult to measure is the time interval for the acceleration.
Since the final speed is zero
and the initial speed is given, it remains to find the
displacement the object undergoes during the acceleration. Then the
variables can be used in the equation 2 2f i 2v v a d= + .
Estimates of the acceleration will vary if they are just guesses,
but should be fairly close if they are determined by a quick
calculation with estimated quantities.
(a) The displacement of the bullet during stopping can be found
by measuring the penetration of the bullet plus the distance the
wood moves. Assuming the value is 15 cm [fwd], the acceleration
is:
2 2f i
av
2
5 2av
2
0 (175 m/s [fwd])
2(0.15 m [fwd])
6.8 10 m/s [fwd]
v va
d
a
=
=
=
G GG G
G
The average acceleration of the bullet is 6.8 105 m/s2 [fwd].
(b) The stopping distance can be measured by adding the total
crunch distance of the car plus any change of position of the
barrels. Assuming the value is 1.5 m [fwd], and changing the
initial speed to 24 m/s, the acceleration is:
2 2f i
av
2
2 2av
2
0 (24 m/s) [fwd])
2(1.5 m [fwd])
1.9 10 m/s [fwd]
v va
d
a
=
=
=
G GG G
G
The average acceleration of the test car is 1.9 102 m/s2
[fwd].
-
32 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
Making Connections 18. In the relay, the second, third, and
fourth runners have already accelerated to vf before they handoff
to the next runner.
Thus, the time is shorter than if you include acceleration time
for all runners. 1.3 ACCELERATION DUE TO GRAVITY
PRACTICE (Pages 3233)
Understanding Concepts 1. The skydivers velocity is much greater
than the divers velocity. Air resistance increases with velocity
and cannot be
neglected for the skydiver. 2. The disadvantage is that what
might seem logical or reasonable does not agree with what actually
happens. One reason for
this is some events occur too rapidly for our senses to be able
to observe slight differences. Aristotles reasoning that heavy
objects fell faster than lighter ones provides an example of the
disadvantage of not using experimentation to determine the
dependency of one variable on another, in this case the dependency
of the acceleration of a falling body on the mass of the body.
Applying Inquiry Skills 3. The experimental setup would require
a vacuum chamber in which the coin and the feather are released
simultaneously.
(This device is available commercially from scientific supply
companies.)
Making Connections 4. Since there is no atmosphere on the Moon,
falling objects do not experience air resistance.
Case Study: Predicting Earthquake Accelerations (Page 34)
(a) The map stretches from Northern California (40 north
latitude) to a few kilometres north of Vancouver (50 north
latitude), and from the middle of Vancouver Island (125 west
longitude) to just east of Trail, B.C. (about 117 west longitude).
The regions affected most severely, as depicted by deep reds, lie
near the west coast of the continent, especially in Northern
California and Southern Oregon, as well as areas fairly close to
Vancouver and Victoria. Areas affected moderately, as depicted by
yellows, stretch inland somewhat in British Columbia, Washington,
and Oregon, and a long way in California. Areas affected slightly,
as depicted by blue-greens, are in the northern and eastern parts
of British Columbia, the eastern parts of Washington and Oregon,
and the state of Idaho. There are no areas unaffected, as depicted
by white.
(b) Students have to combine the colours on the map with the
colours in the legend, and they have to compare the contours and
locations on the map to a conventional atlas of the same
region.
PRACTICE (Page 35)
Understanding Concepts 5. iv
G= 0
aG = 9.80 m/s2 [down] (a) d
G= 5.00 m
f ?v =G
-
32 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
Making Connections 18. In the relay, the second, third, and
fourth runners have already accelerated to vf before they handoff
to the next runner.
Thus, the time is shorter than if you include acceleration time
for all runners. 1.3 ACCELERATION DUE TO GRAVITY
PRACTICE (Pages 3233)
Understanding Concepts 1. The skydivers velocity is much greater
than the divers velocity. Air resistance increases with velocity
and cannot be
neglected for the skydiver. 2. The disadvantage is that what
might seem logical or reasonable does not agree with what actually
happens. One reason for
this is some events occur too rapidly for our senses to be able
to observe slight differences. Aristotles reasoning that heavy
objects fell faster than lighter ones provides an example of the
disadvantage of not using experimentation to determine the
dependency of one variable on another, in this case the dependency
of the acceleration of a falling body on the mass of the body.
Applying Inquiry Skills 3. The experimental setup would require
a vacuum chamber in which the coin and the feather are released
simultaneously.
(This device is available commercially from scientific supply
companies.)
Making Connections 4. Since there is no atmosphere on the Moon,
falling objects do not experience air resistance.
Case Study: Predicting Earthquake Accelerations (Page 34)
(a) The map stretches from Northern California (40 north
latitude) to a few kilometres north of Vancouver (50 north
latitude), and from the middle of Vancouver Island (125 west
longitude) to just east of Trail, B.C. (about 117 west longitude).
The regions affected most severely, as depicted by deep reds, lie
near the west coast of the continent, especially in Northern
California and Southern Oregon, as well as areas fairly close to
Vancouver and Victoria. Areas affected moderately, as depicted by
yellows, stretch inland somewhat in British Columbia, Washington,
and Oregon, and a long way in California. Areas affected slightly,
as depicted by blue-greens, are in the northern and eastern parts
of British Columbia, the eastern parts of Washington and Oregon,
and the state of Idaho. There are no areas unaffected, as depicted
by white.
(b) Students have to combine the colours on the map with the
colours in the legend, and they have to compare the contours and
locations on the map to a conventional atlas of the same
region.
PRACTICE (Page 35)
Understanding Concepts 5. iv
G= 0
aG = 9.80 m/s2 [down] (a) d
G= 5.00 m
f ?v =G
-
Copyright 2003 Nelson Chapter 1 Kinematics 33
2 2f i
2 2f
2f
f
f
2
0 2(9.80 m/s [down])(5.00 m[down])
2(9.80 m/s )(5.00 m)9.90 m/s[down]
1km9.90 m/s 3600s/h1000 m
35.6 km/h[down]
v v a d
v
vv
v
= + = +
=
=
=
=
GG G G
G
G
The divers velocity is 9.90 m/s [down] or 35.6 km/h [down]. (b)
d
G= 10.00 m
f ?v =G
2 2f i
2 2f
2f
f
f
2
0 2(9.80 m/s [down])(10.0 m[down])
2(9.80 m/s )(10.0 m)14.0 m/s [down]
1km14.0 m/s 3600s/h1000 m
50.4 km/h[down]
v v a d
v
vv
v
= + = +
=
=
=
=
GG G G
G
G
The divers velocity is 14.0 m/s [down] or 50.4 km/h [down].
Applying Inquiry Skills 6. (a) A good guess would be between 150
ms and 250 ms. The actual answer depends on the students hand span.
(b) i 0yv =
29.8 m/sya = 0.20 my = (a typical hand span) t = ?
2i
2
2
1
21
2
2
2(0.20 m)
9.8 m/s0.20 s
y y
y
y
y v t a t
y a t
yt
a
t
= +
=
=
=
=
For a hand span of 0.20 m, the time interval is 0.20 s, or 200
ms. (c) Comparisons will vary. Students can improve their skills by
trying to estimate answers to problems before performing
calculations, and by practising estimating quantities in
everyday transactions and activities. 7. (a) iv
G= 0
dG
= 1.55 m [down] t = 0.600 s
?a =G Rearranging the equation ( )2i 12d v t a t = +
G G G:
-
34 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
( )[ ]( )
( )[ ]
2
2
2
2
2 1.55 m down
0.600 s
8.61m/s down
dat
a
=
=
=
GG
G
Thus, the acceleration of the ball is 8.61 m/s2 [down]. (b) Plot
the position-time graph, find the slopes of the tangents at three
or more times, plot the corresponding velocity-time
graph, and calculate the slope of the line on that graph to find
the acceleration. (c) aG = 9.81 m/s2 [down]
2 2
2
measured value accepted value% error 100%
accepted value
8.61 m/s 9.81 m/s100%
9.81 m/s% error 12.2%
=
=
=
The percent error is 12.2 %. (d) The high percent error results
from a variety of possible errors, with air resistance likely the
most crucial influence.
(Since the ball is very light, air resistance has a greater
effect than if the ball were more massive but the same size.)
Another important source of error is measuring the time interval of
the fall.
Making Connections 8. Answers may vary. The change in the
gravitational field strength at various altitudes is quite small,
and deciding how
much adjustment would be required might be controversial.
Furthermore, if records were adjusted downward for events that are
easier at higher altitudes, then calls would be made for records to
be adjusted upward for events that are more difficult in the
rarified atmosphere at higher altitudes. Adjustments are not made
for the effects of low winds, so they would not likely be made for
changes in altitude.
PRACTICE (Pages 3738)
Understanding Concepts 9. (a) The time the ball takes to rise
equals the time the ball takes to fall since air resistance is
negligible. (b) The initial and final velocities would be equal in
magnitude but opposite in direction. (c) The balls velocity at the
top of the flight is zero. (d) The balls acceleration during the
entire motion is the acceleration due to gravity (9.8 m/s2 [down]),
even when it is at
the top of the flight. (e)
-
Copyright 2003 Nelson Chapter 1 Kinematics 35
10. We could solve Sample Problem 1(b) using the following
equations: i f( )
2y yv vy t+
= 2i1 ( )2y y
y v t a t = + 2f1 ( )2y y
y v t a t =
11. (a) iyv = 0
ya = 9.80 m/s2
y = 12.5 m vfy = ?
2 2f i
2 2f
2 2f
f
2
0 2(9.80 m/s )(12.5m)
245m /s
15.6 m/s
y y y
y
y
y
v v a y
v
vv
= +
= +
=
=
The shellfish has a speed of 15.6 m/s at impact. (b) iyv = 0
ya = 9.80 m/s2
t = 3.37 s vfy = ?
f i
f i
2
f
0 (9.80 m/s )(3.37s)33.0 m/s
y yy
y y y
y
v va
tv v a t
v
=
= +
= +
=
The steel ball has a speed of 33.0 m/s at impact. 12. iyv = 15.0
m/s
ya = 9.80 m/s2
y = 15.0 m
Using the equation 2
i( )2
yy
a ty v t
= + , rearrange as 2 i
1 ( ) 02 y y
a t v t y + = , which is a form of the quadratic equation.
Thus,
( )
2
2i i
2i i
42
42
22
2
yy y
y
y y y
y
b b acta
av v y
a
v v a yt
a
=
= +
=
(a) The initial velocity is up. Define +y as down. t = ? vfy =
?
( )2 22
15.0 m/s 15.0 m/s 2(9.80 m/s )(15.0 m)
9.80 m/s1.53s 2.33s (only the positive root is
applicable)3.86s
t
t
+ =
= =
-
36 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
( )( )f i
2
f
15.0 m/s+ 9.80 m/s 3.86s
22.8m/s
y y y
y
v v a t
v
= +
=
=
Thus, the total flight time is 3.86 s and the speed of impact is
22.8 m/s. (b) The initial velocity is down. Define +y as down.
t = ? vfy = ?
( )2 22
15.0 m/s 15.0 m/s 2(9.80 m/s )(15.0 m)
9.80 m/s1.53s 2.33s (only the positive root is applicable)
0.794s
t
t
+ =
= =
( )( )f i
2
f
15.0 m/s+ 9.80 m/s 0.794 s
22.8 m/s
y y y
y
v v a t
v
= +
=
=
Thus, the total flight time is 0.794 s and the speed of impact
is 22.8 m/s if the initial velocity is down. (Notice that the
answers are written to three significant digits in order to compare
them to the answers in (a). Rules for rounding off have not been
followed exactly here.)
(c) The final velocity is independent of whether the ball is
thrown up or down (at the same speed). 13. Let y1 represent the
distance the ball travels from t = 0.0 s to t = 1.0 s and let y2
represent the distance the ball travels
from t = 1.0 s to t = 2.0 s. For y1, the initial velocity viy,1
= 0
Using the equation 2i1 ( )2y y
y v t a t = + :
( )2
1 i ,1 1 1
2 2
1
1 ( )2
10 m/s (1.0 s 0.0 s) 9.8 m/s (1.0 s 0.0 s)2
4.9 m
y yy v t a t
y
= +
= +
=
For y2, the initial velocity viy,2 is the velocity at t = 1.0 s.
Thus,
i ,2 i ,1 1
2
i ,2
( )
0 m/s 9.8 m/s (1.0 s 0.0 s)9.8 m/s
y y y
y
v v a t
v
= +
= +
=
22 i ,2 2 2
2 2
2
1 ( )2
19.8 m/s (2.0s 1.0s) 9.8 m/s (2.0s 1.0s)2
14.7 m
y yy v t a t
y
= +
= +
=
( )( )2
1
14.7 m3.0
4.9 myy
= =
Therefore, a ball travels three times as far from 1.0 s to 2.0 s
as from 0.0 s to 1.0 s. 14. Define +y as up.
vfy = 0 m/s
t = 4.2s2
= 2.1 s
ay = 9.80 m/s2
-
Copyright 2003 Nelson Chapter 1 Kinematics 37
(a) viy = ?
f i
i f
2
i
0 m/s ( 9.80 m/s )(2.1s)21m/s
y yy
y y y
y
v va
tv v a t
v
=
=
=
=
Thus, the pitcher threw the ball with a velocity of 21 m/s [up].
(b) y = ?
2i
2 2
1 ( )2
121 m/s (2.1 s) + ( 9.80 m/s )(2.1 s)2
22 m
y yy v t a t
y
= +
=
=
The ball rises 22 m. 15. Define +y as down.
viy = 2.1 m/s t = 3.8 s ay = 9.80 m/s2
(a) y = ?
2
i
2 2
( )2
(9.80 m/s )(3.8s)( 2.1m/s)(3.8s)2
63m
yy
a ty v t
y
= +
= +
=
The balloon was 63 m when the ballast was released. (b) vfy =
?
f i
f i
2
f
2.1 m/s (9.80 m/s )(3.8 s)35 m/s
y yy
y y y
y
v va
tv v a t
v
=
= +
= +
=
The velocity of the ballast at impact was 35 m/s [down]. 16.
Define +y as down.
d = 2.3 m t = 1.7 s viy = 0
(a) ?g =G
2i
2
2
2
2
1 ( )2
1 ( )22
( )2(2.3 m)(1.7 s)
1.6 m/s [down]
yy v t g t
y g t
ygt
g
= +
=
=
=
=
G
G
The acceleration of gravity on the Moon is 1.6 m/s2 [down].
-
38 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
(b) 2
Earth2
Moon
9.8 m/s = 6.11.6 m/s
gg
=
GG
Thus, the ratio of EarthgG to MoongG is 6.1:1.
Applying Inquiry Skills 17. (a) Students can use graphing
techniques or uniform acceleration equations to determine the
acceleration. Sample
calculations using the final data points are shown below. Let +y
be down. For mass 1: viy = 0 y = 0.736 m t = 0.40 s ay1 = ?
( )
2i 1
21
1 2
2
21
1
21
22
2(0.736 m)
0.40 s
9.2 m/s
y y
y
y
y
y v t a t
y a t
ya
t
a
= +
=
=
=
=
For mass 2: viy = 0 y = 0.776 m t = 0.40 s ay2 = ?
( )
2i 2
22
2 2
2
22
1
21
22
2(0.776 m)
0.40 s
9.7 m/s
y y
y
y
y
y v t a t
y a t
ya
t
a
= +
=
=
=
=
The accelerations of the two masses are 9.2 m/s2 and 9.7 m/s2,
respectively. (b) To find the percent difference:
( )
2 2
2 2
difference in values% difference = 100%
average of values
9.2 m/s 9.7 m/s100%
19.2 m/s 9.7 m/s
2% difference = 5.3%
=
+
-
Copyright 2003 Nelson Chapter 1 Kinematics 39
(c) Likely, the lower experimental acceleration, 9.2 m/s2, is
attributable to the ticker-tape timer because friction between the
paper strip and the timer would reduce the acceleration.
Making Connections 18. Answers will vary. Walking, running, and
going up stairways are just a few examples of the activities that
would be more
difficult with a higher acceleration due to gravity. One
advantage is that friction would be increased when it might be
helpful, for example for a construction worker on a slanted roof.
Another advantage is that objects subject to toppling over might be
more stable.
PRACTICE (Page 39)
Understanding Concepts 19. The two main factors that affect
terminal speed are mass and surface area. (For solid spherical
objects, this is equivalent
to saying that the terminal speed depends on the objects
density.) For objects of the same mass, an increased surface area
contacting the air causes the terminal speed to become lower. For
objects with the same surface area, the greater the mass the
greater is the terminal speed.
20. There is no terminal speed related to air resistance on the
Moon because the Moon does not have an atmosphere. 21. There are
two terminal speeds in this case, a high
speed followed by a much lower terminal speed when the parachute
is opened. The graph is shown below. The slope of the line near the
beginning of the motion is equal to the magnitude of the
acceleration due to gravity at that location. The terminal speed
values are found on page 39 of the text (Table 5).
Applying Inquiry Skills 22. (a) Examples of factors to consider
in designing the package are the mass, dimensions, and contents of
the package, the
materials of the packets within the package, the material of the
package itself, the method of securing the outer layer of the
package, the height from which the package will be dropped, the
forward speed of the aircraft that drops the package, the type of
terrain on which the package will land, and the ease with which the
package can be opened when it is found.
(b) Since there are so many factors that could be tested, it
would be wise to predict what choices would be best for the design,
create a prototype, and then perform a controlled experiment by
dropping the sample of the design from various heights onto various
surfaces to determine how well the package can survive the fall.
Modifications and further testing are part of the process.
Making Connections 23. There are two main factors to consider,
terminal speed and stopping distance. Once the terminal speed of a
body is
reached, the speed remains constant, or it may even drop as the
air becomes more dense nearer the ground. Furthermore, a person
falling from a greater height can reduce the terminal speed by
using the spread-eagle orientation. Increasing the stopping or
deceleration distance reduces the chances of death or serious
injury. The distance can be increased if the landing occurs on the
down slope of a hill or if the fall is broken by trees, bushes, a
layer of snow, or water.
Section 1.3 Questions (Page 40)
Understanding Concepts 1. Air resistance increases dramatically
with increased speed, so air resistance is negligible for most
common objects that
fall for a short distance. For example, air resistance is
negligible for all but the lowest density objects that can fall in
a classroom setting.
2. Aristotle believed that the central region of Earth was made
up of four elements: earth, air, fire, and water. Each element had
its proper place, which was determined by its relative heaviness.
Each element moved in a straight line to its proper place where it
could be at rest. For example, similar objects were attracted to
each other. Thus, objects made from the
-
40 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
earth were attracted down to Earth. Fire tended to rise from
Earth. Aristotle also believed that heavier objects fell faster
than less massive objects of the same shape.
Galileo believed that all objects fall toward Earth at the same
acceleration, regardless of their mass, size, or shape, when
gravity is the only force acting on them. Evidently, he proved his
theory by dropping two objects of different mass from the top floor
of the Leaning Tower of Pisa.
3. In each case, we define +y as down. (a) viy = 0
ay = 9.8 m/s2 y = 36 m vfy = ?
2 2f i
2 2f
f
f
2
0 2(9.8 m/s )(36 m)
27 m 1 km 3600 ss 1000 m 1 h
97 km/h
y y y
y
y
y
v v a y
v
v
v
= +
= +
= =
Thus, the landing speed of the diver is 27 m/s or 97 km/h. (b)
viy = 0
ay = 9.8 m/s2 t = 3.2 s vfy = ?
f i
f i
2
2f
0 (9.8 m/s )(3.2 s)1 km 3600 s(31m/s)
1000 m 1 h1.1 10 km/h
y yy
y y y
y
v va
tv v a t
v
=
= +
= +
= =
Thus, the landing speed of the stone is 31 m/s or 1.1 102 km/h.
4. Define +y as up.
viy = 5.112 m/s vfy = 0 y = ? In Java, g = 9.782 m/s2:
2 2f i
2 2f i
2i
2
2
2
2
2
(5.112 m/s)2( 9.782 m/s )1.336 m
y y y
y y y
y
y
v v a y
a y v v
vy
a
y
= +
=
=
=
=
In London, g = 9.823 m/s2:
2i
2
2
2
(5.112 m/s)2( 9.823 m/s )1.330 m
y
y
vy
a
y
=
=
=
In Java, the jumper achieved a height of 1.336 m; in London, the
jumper achieved a height of 1.330 m.
-
Copyright 2003 Nelson Chapter 1 Kinematics 41
5. Let +y be the forward direction of the shuttle, assumed to be
a straight line. viy = 0 ay = 5(9.80 m/s2) t = 1.0 min = 60 s (two
significant digits) vfy = ?
f i
f
2
3
4f
5(9.80 m/s )(60 s)1 km 3600 s(2.9 10 m/s)
1000 m 1 h1.1 10 km/h
y yy
y y
y
v va
tv a t
v
=
=
=
= =
The shuttles speed is 2.9 m/s or 1.1 104 km/h. 6. Let +y be
up.
t = 2.6 s ay = g = 9.80 m/s2
(a) t = ? The time for the ball to rise will be half of the
total time.
2.6 s
21.3 s
t
t
=
=
Therefore, the time for the ball to rise is 1.3 s. (b) vfy =
0
t = 1.3 s viy = ?
f i
i f
2
i
0 ( 9.8m/s )(1.3s)13m/s
y yy
y y y
y
v va
tv v a t
v
=
=
=
=
The velocity is 13 m/s [up] when the golf ball leaves the
persons hand. (c) Again, +y is up.
viy = 13 m/s ay = g = 3.8 m/s2 t = ?
f i
f i
20 13m/s
3.8m/s3.4s
y yy
y y
y
v va
tv v
ta
t
=
=
=
=
The total time in flight would be 2(3.4 s) = 6.8 s. 7. Let +y be
down.
t = 0.087 s y = 0.80 m ay = 9.8 m/s2
-
42 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
( )
2f
f
2
f
1 ( )2
1 ( )2
0.80 m 1 9.8m/s (0.087s)0.087s 29.6 m/s
y y
y y
y
y v t a t
yv a tt
v
=
= +
= +
=
The velocity of the ball as it hits the floor is 9.6 m/s [down].
8. Let +y be down.
viy = 14 m/s ay = 9.8 m/s2
(a) y = 21 m t = ?
Using the equation 2i1 ( )2y y
y v t a t = + , rearrange as 2 i( ) 02y
ya
t v t y + = , which is a form of the quadratic
equation. Therefore,
( )
2
2i i
2i i
2 2
2
42
42
22
2
14 m/s (14 m/s) 2(9.8m/s )(21m)9.8m/s
1.1s or 3.9s
yy y
y
y y y
y
b b acta
av v y
a
v v a y
a
t
=
= +
=
+=
=
The stone will take 1.1 s to reach the water below. (b) The
positive root is the actual answer when the stone is thrown
vertically downward. The negative root is the time that
the stone would have travelled had the initial velocity been
upward rather than downward (i.e., if vi = 14 m/s in this
case).
9.
10. Note: To make the solution more compact, units are omitted
until the final answer is written. Let +y be down, let F represent
the flowerpot, and let B represent the ball. Since the ball is
thrown 1.00 s after the flowerpot is released, after some time F B
1.00,t t = let us assume the ball will
pass the flowerpot. At this instant, the flowerpot will have
travelled 28.5 m 26.0 m = 2.5 m farther than the ball. Thus, F B
2.5.y y = +
But with uniform acceleration, 2i12y
y v t ay t = + .
Combining these relationships, we have:
-
Copyright 2003 Nelson Chapter 1 Kinematics 43
( ) ( )( ) ( )
F B
2 2i F F F i B B B
22F F F
2 2F F F F
2 2F F F F
F F
F
F
2.5
1 12.5
2 2
0 4.9 12 1 4.9 1 2.5
4.9 12 1 4.9 2 1 2.5
4.9 12 12 4.9 9.8 4.9 2.5
0 12 12 9.8 4.9 2.5
0 2.2 4.6
4.62.2
y y y y
y y
v t a t v t a t
t t t
t t t t
t t t t
t t
t
t
= +
+ = + +
+ = + +
= + + +
= + + += + +=
=
F 2.09 st =
After this time interval, the flowerpot has fallen:
( )( )2
F F
22
F
121
9.80 m/s 2.09 s221.4 m
yy a t
y
=
=
=
The flowerpot is 28.5 m 21.4 m = 7.10 m above the ground when
the ball passes it. 11. The ranking from highest terminal speed to
lowest is: pollen; a ping-pong ball; a basketball; a skydiver
diving headfirst;
and a skydiver in spread-eagle orientation.
Applying Inquiry Skills 12. (a) 9.809 060 m/s2 has 7 significant
digits, a possible error of 0.000 000 5, and percent possible error
of
2
20.000 000 5m/s 100% 0.000 005%.9.809 060 m/s
=
(b) 9.8 m/s2 has 2 significant digits, a possible error of 0.05
m/s2, and percent possible error of 2
20.05m/s 100% 0.5%.9.8 m/s
=
(c) 9.80 m/s2 has 3 significant digits, a possible error of
0.005 m/s2, and percent possible error of 2
20.005m/s 100% 0.05%.9.80 m/s
=
(d) 9.801 m/s2 has 4 significant digits, a possible error of
0.0005 m/s2, and percent possible error of 2
20.0005m/s 100% 0.005%.9.801 m/s
=
(e) 9.8 106 m/s2 has 2 significant digits, a possible error of
0.05 106 m/s2, and percent possible error of 6 2
6 20.05 10 m/s 100% 0.5%.9.8 10 m/s
=
13. (a) Students may recall performing this activity in a
previous grade. Hold a metre stick vertically while your partner
holds his or her separated index finger and thumb ready to catch
the stick at a specific mark, such as the 50-cm mark. Without
warning, drop the stick and determine how far the stick falls
before the partner catches it. Repeat several times and take an
average of the distances, y. Use this value in the appropriate
equation, as shown below.
Let +y be down.
Assume y = 18 cm = 0.18 m ay = 9.8 m/s2 viy = 0 t = ?
-
44 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
2i
2
2
1
21
2
2
0.19 s
2(0.18 m)9.8 m/s
y y
y
y
y v t a t
y a t
yt
a
t
= +
=
=
=
=
(b) Talking on a cell phone would likely increase reaction time.
Students can simulate this situation by engaging in distracting
conversation with the lab partner whose reaction time is being
tested.
Making Connections 14. Aristotle and Galileo influenced the
philosophy and scientific thought of their respective eras, and in
both cases their
influence lasted long after they died. Students can find
information about these science giants in books and encyclopedias,
or on the Internet. For example, an advanced word search on the
Internet, entering only the words Aristotle and Galileo, found more
than 20 thousand hits, many of which featured discussions of the
same topics featured in the text.
15. Deductive reasoning involves using theories to account for
specific experimental results. Thus, deductive reasoning uses ideas
to explain observed phenomena. Inductive reasoning involves making
and collecting observations, and then developing general theories
or hypotheses to account for the observations.
(a) Aristotle and other ancient scientists used deductive
reasoning. (b) Galileo used the process of inductive reasoning.
When he observed that heavy objects fall with increasing speed,
he
formed the hypothesis that the speed of the object was directly
proportional to the distance the object fell. When this hypothesis
proved false, he hypothesized that the speed of a falling object is
directly proportional to the time, not the distance. Through
experiments, he was able to verify his hypothesis.
(c) Various ways are used to contrast these types of reasoning.
For example, in deductive reasoning, particular results are
inferred from a general law, whereas in inductive reasoning, a
general law is inferred from particular results. Stated another way
for deduction, conclusions follow from premises, that is the
reasoning goes from the general to the specific. In induction,
premises lead to the conclusions, or the reasoning goes from the
specific to the general. In mathematics, induction involves proving
a theorem using a process in which the theorem is verified for a
small value of an integer, and then extending the verification to
greater values of the integer.
16. Many sites can be found by doing an advanced word search on
the Internet. For example, in entering the words Luis Alvarez,
Yucatan, and dinosaurs, more than 400 sites were found, many of
them highly credible. Two examples are:
www.ceemast.csupomona.edu/nova/alvarez2.html
www.space.com/scienceastronomy/planetearth/deep_impact_991228.html
1.4 PROJECTILE MOTION
PRACTICE (Page 46)
Understanding Concepts 1. A projectile is an object that moves
through the air without a propulsion system and follows a curved
path. An airplane
has a propulsion system and does not follow a trajectory. Thus,
an airplane is not a projectile. 2. The projectile experiences
constant downward acceleration due to gravity (vertical
acceleration) and the horizontal
component of acceleration is zero. 3. Let +x be to the right and
+y be downward. The initial position is the position where the
marble leaves the table. (a) Horizontally (constant ixv ):
i 1.93 m/s
? ?
xvxt
=
= =
-
44 Unit 1 Forces and Motion: Dynamics Copyright 2003 Nelson
2i
2
2
1
21
2
2
0.19 s
2(0.18 m)9.8 m/s
y y
y
y
y v t a t
y a t
yt
a
t
= +
=
=
=
=
(b) Talking on a cell phone would likely increase reaction time.
Students can simulate this situation by engaging in distracting
conversation with the lab partner whose reaction time is being
tested.
Making Connections 14. Aristotle and Galileo influenced the
philosophy and scientific thought of their respective eras, and in
both cases their
influence lasted long after they died. Students can find
information about these science giants in books and encyclopedias,
or on the Internet. For example, an advanced word search on the
Internet, entering only the words Aristotle and Galileo, found more
than 20 thousand hits, many of which featured discussions of the
same topics featured in the text.
15. Deductive reasoning involves using theories to account for
specific experimental results. Thus, deductive reasoning uses ideas
to explain observed phenomena. Inductive reasoning involves making
and collecting observations, and then developing general theories
or hypotheses to account for the observations.
(a) Aristotle and other ancient scientists used deductive
reasoning. (b) Galileo used the process of inductive reasoning.
When he observed that heavy objects fall with increasing speed,
he
formed the hypothesis that the speed of the object was directly
proportional to the distance the object fell. When this hypothesis
proved false, he hypothesized that the speed of a falling object is
directly proportional to the time, not the distance. Through
experiments, he was able to verify his hypothesis.
(c) Various ways are used to contrast these types of reasoning.
For example, in deductive reasoning, particular results are
inferred from a general law, whereas in inductive reasoning, a
general law is inferred from particular results. Stated another way
for deduction, conclusions follow from premises, that is the
reasoning goes from the general to the specific. In induction,
premises lead to the conclusions, or the reasoning goes from the
specific to the general. In mathematics, induction involves proving
a theorem using a process in which the theorem is verified for a
small value of an integer, and then extending the verification to
greater values of the integer.
16. Many sites can be found by doing an advanced word search on
the Internet. For example, in entering the words Luis Alvarez,
Yucatan, and dinosaurs, more than 400 sites were found, many of
them highly credi