forces - nelson phys12 textbook solutions

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


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.


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.


(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.


(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.)


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)


(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].


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

= =

=


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].


(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.


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:


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].


(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].


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].


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.


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)


(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.


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


(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].


(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 = ?


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].


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].


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.


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


(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


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].


(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


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].


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].


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


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


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:


( )[ ]( )

( )[ ]

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.


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)


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

+ =

= =


( )( )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


(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].


(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%

=

+


(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


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.


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


( )

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:


( ) ( )( ) ( )

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 = ?


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

=

= =


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

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