Chapter 11 - HRSBSTAFF Home Page

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erating or moving with uniform motion. Answers: 1. (a) uniform motion ... 5. +1.5 m /s2. 6. –1.2 m/s2 (to the left). 7. +3.1 m/s2 (upward). BLM 11-4, Velocity Vectors ...
ANSWER KEY

CHAPTER 11

BLM 11-1, Build Your Own Accelerometer/Science Inquiry Goal: Students record their observations in the Starting Point Activity. Answers: Students’ answers will vary.

BLM 11-2, Recognizing Accelerated Motion/Science Inquiry

(d) accelerated motion Speed and direction change. (e) accelerated motion Direction changes.

Goal: Students determine whether an object is accelerating or moving with uniform motion. Answers:

(f) accelerated motion Direction changes.

1. (a) uniform motion Speed is constant, and motion is in a constant direction.

2. (a) The object moves in a constant direction with constant speed. (b) The object’s speed and/or direction of motion changes.

(b) accelerated motion Direction changes. (c) accelerated motion Speed changes.

BLM 11-3, Acceleration Problems/ Problem Solving Goal: Students calculate the acceleration of objects that are moving in a straight line. Answers: 1.

t i (s)

t f (s)

∆t (s)

10

25

15

0

40

12.5 9.70



0

⫹12

⫹12

⫹0.80

40

⫹50

⫹10

–40

–1

41.6

29.1

–10.1

⫹32.4

⫹42.5

⫹1.46

51.9

42.2

⫹43.7

–12.6

–56.3

–1.33

2

4. ⫹52.4 m/s2 5. ⫹1.5 m/s2 6. –1.2 m/s2 (to the left) 7. ⫹3.1 m/s2 (upward)

aav (m/s2)

BLM 11-4, Velocity Vectors and Acceleration/Reinforcement Goal: Students use velocity vectors to determine acceleration. Answers: 1. (a) →

∆t = 5.0 s

vi = 30 m/s →

vf = 15 m/s

292



∆v (m/s)

2. ⫹1.37 m/s2 3. –3.66 m/s



vf (m/s)

vi (m/s)

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∆v = –15 m/s



a = –3 m/s2

ANSWER KEY

CHAPTER 11

(b)

(f)





∆v = 25 m/s

vi = 10 m/s →



∆t = 12.5 s

vi = –10 m/s





a = 2.0 m/s2

vf = 35 m/s



∆t = 11.0 s

vf = 12 m/s →

∆v = 22 m/s

(c)

a = –2 m/s2

2. (a) the head of the initial velocity vector →



∆v = –10 m/s

∆t = 2.5 s

vi = –12 m/s →

(b) the head of the final velocity vector (c) ∆v divided by the time gives the acceleration.



vf = –22 m/s

a = –4.0 m/s2



∆t = 3.4 s

(d) vi = –26 m/s →



∆v = 19 m/s



a = 5.6 m/s2

vf = –7 m/s

(e) →

∆t = 6.0 s

vi = 17m/s



vf = –13 m/s → ∆v = –30 m/s



a = –5 m/s2

BLM 11-5, Accelerating Parallel to Motion/Science Inquiry Goal: Students examine the motion of objects along a straight line and determine their acceleration. Answers: 1. ∆t (s) 5 5 5 5 5 5 5 5 5 5 5



vi (m/s) ⫹20 ⫹20 ⫹20 ⫹20 –20 –20 –20 –20 0 0 0



vf (m/s) ⫹30 ⫹10 ⫹20 0 –10 –30 –20 0 ⫹20 0 –20













∆v (m/s)

aav (m/s2)

vi R/L

vf R/L

∆v R/L

aav R/L

⫹10 –10

⫹2 –2

R R

R R

R L

R L

speeding up slowing down

0 –20

0 –4

R R

R –

– L

– L

neither slowing down

⫹10

⫹2

L

L

R

R

slowing down

–10

–2

L

L

L

L

speeding up

0

0

L

L





neither

⫹20

⫹4

L



R

R

slowing down

⫹20

⫹4



R

R

R

speeding up

Speeding up/slowing down/neither

0

0









neither

–20

–4



L

L

L

speeding up

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ANSWER KEY

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2. (a) They are both zero. (b) The acceleration is in the same direction as the initial velocity, or the object starts from rest.

Goal: Students review and compare both positiontime and velocity-time graphs. Answers: 1. (a) position-time, position, velocity

(c) They are in opposite directions. (d) If they are in the same direction, or if the initial velocity is zero, the object will speed up. If they are in opposite directions, the object will slow down. 3. (a) The car’s speed reduces to zero, and then increases, but in the reverse direction.

(b) slope of straight line between initial and final data points (c) slope of tangent 2. (a) velocity-time, velocity, acceleration (b) slope of straight line from initial to final data point

(b) The car initially moves to the right, stops, and then moves to the left. (c) Initial and final velocities are in different directions.

(c) slope of tangent 3. (a) Slopes give rate of change. Average and instantaneous acceleration are calculated in the same way as average and instantaneous velocity.

BLM 11-6, Acceleration Quiz/ Assessment

1. ⫹1.3 m/s2 2. ⫹1.9 m/s2 (to the right) 2

3. –1.9 m/s (to the observer’s left) 4. (a) The velocity will still be to the left and will increase in magnitude. (b) The velocity will decrease. If the acceleration continues, the object will stop, and then speed will increase as the direction changes from left to right. →



vs = 3.1 m/s

vi = +6.8 m/s

BLM 11-9, Velocity-Time Graph from a Position-Time Graph/ Science Inquiry Goal: Students use a position-time graph to draw a velocity-time graph. Answers: 1. Answers may vary. 2.

v Velocity (m/s)

Goal: Students test their knowledge of acceleration. Answers:

(b) Slopes represent different physical quantities.

4 3 2 1 0 0 1 2 3 4 5 6 7 8 Time (s)



t

3. 0.5 m/s2

∆v = –9.9 m/s

4. (a) The slope increases.

BLM 11-7, Comparing Velocity and Acceleration/Overhead Master

(b) The velocity is increasing. There is an acceleration.

Answers: not applicable

BLM 11-8, Picturing Acceleration/ Science Inquiry

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ANSWER KEY

CHAPTER 11

BLM 11-10, Velocity-Time Graphs/ Science Inquiry Goal: Students examine the relationships between velocity-time graphs and acceleration. Answers: 1. (a) Velocity (m/s)

v

BLM 11-11, Average and Instantaneous Acceleration/ Reinforcement Goal: Students calculate both average and instantaneous acceleration. Answers: 1.

10 8 6 4 2 0





0

4

0

ⴙ5.0

ⴙ1.3

4

10

ⴙ5.0

–2.0

–1.2

10

12

–2.0

0

ⴙ1.0

Slope = 2.0 m/s2

2

7

ⴙ3.8

ⴙ2.2

–0.32

Slope and acceleration are the same.

5

11

ⴙ4.7

–0.5

–0.87

0

12

0

0

0

t

(b)

1

2 3 4 Time (s)

5

Velocity (m/s)

v 30 25 20 15 10 5 0

vf (m/s)

aav (m/s2)

t f (s)

0

v i (m/s)



t i (s)

2. Allow for variation based on errors in measurement. →

ainst (m/s2)

t (s) 2

1.25

4

0.00

Slope = –5.0 m/s2

5

–0.63

Slope and acceleration are the same.

9

–1.00

10

0.00

1

2 3 4 Time (s)

5

t

Velocity (m/s)

v 25 20 15 10 5 0

3. (a) The tangent is horizontal. (b) The graph crosses the time axis.

0

2

4 6 8 Time (s)

(b) t = 2.0 s, Slope = 1.0 m/s2 t = 4.0 s, Slope = 2.0 m/s2 t = 6.0 s, Slope = 3.0 m/s2 t = 8.0 s, Slope = 4.0 m/s2 (c) Slopes are the same.

10

t

BLM 11-12, Using Graphs to Determine Acceleration/ Reinforcement Goal: Students use a position-time graph to draw a velocity-time graph and determine acceleration. Answers: Displacement (m)

2. (a)

0

d 12 10 8 6 4 2 0 0

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1

2 3 4 Time (s)

5

t

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ANSWER KEY

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2.



t (s)

v (m/s)

1

–4

2

–3

3

–2

4

–1

Velocity (m/s)

v 1

0 -1 -2 -3 -4 -5

2

3

4

5 t

Time (s)

3. ⴙ1 m/s2 4. Find the slopes of tangents to the position-time graph at multiple points. Use these values (time, slope) to create a velocity-time graph. Then find the slope of the velocity-time graph, which is also the acceleration.

BLM 11-13, Slope of a VelocityTime Graph/Reinforcement Goal: Students find average acceleration and instantaneous acceleration from a velocity-time graph. Answers: 1. →



Uniform/average/ instantaneous acceleration

Time/time interval (s)

∆t (s)

∆v (m/s)

0–10

ⴙ10

ⴙ20

ⴙ2

uniform

10–16

ⴙ6

0

0

uniform

16–22

ⴙ6

–30

–5

uniform

22–30

ⴙ8

ⴙ30

ⴙ3.75

average

26





ⴙ5.0

a (m/s2)

4. (a) ⫹45 m/s

2. The line is straight. 3. (a) slope of line joining the initial and final data points for the time interval

(b) ⫹90 m/s

(b) slope of tangent

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instantaneous

ANSWER KEY

CHAPTER 11

Quiz/Assessment

BLM 11-14, Estimating Final Velocity/Problem Solving

Goal: Students assess their understanding of velocitytime graphs. Answers:

Goal: Students calculate average velocity and final velocity for an object moving in a straight line. Answers: 1. (a) vi = 0 m/s

(b) vi = 0 m/s

1. 3.3 m/s2 2. 0 m/s2

(c) vi = 0 m/s

vf = ⫹10 m/s

vf = ⫹25 m/s

vf = ⫹20 m/s

vav = ⫹5 m/s

vav = ⫹12.5 m/s vav = ⫹10 m/s

3. –5.3 m/s2 4. 6.0 m/s2 5. 4.7 m/s2

2. (a) zero (b) Average velocity is half the final velocity. 3. ⫹9.0 m/s

6. Answers will vary. 7. –1.1 m/s2

BLM 11-15, Velocity-Time Graph BLM 11-16, Free Fall/Science Inquiry Goal: Students record their data for Think & Link Investigation 11-C: Free Fall. Answers: Displacement, → ∆d (m)

Time, ∆t (s)

Interval

Average → velocity, vav (m/s)

1

0.0479

0.033

1.45

2

0.0587

0.033

1.78

3

0.0693

0.033

2.10

4

0.0799

0.033

2.42

5

0.0906

0.033

2.75

6

0.1013

0.033

3.07

7

0.1118

0.033

3.39

8

0.1225

0.033

3.71

9

0.1331

0.033

4.03

Change in velocity, → ∆v (m/s)

} } } } } } } }

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Acceleration, → a (m/s2)

0.327

9.92

0.321

9.73

0.321

9.73

0.324

9.83

0.324

9.83

0.318

9.64

0.324

9.83

0.321

9.73

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ANSWER KEY

CHAPTER 11

BLM 11-17, Displacement of an Accelerating Object/Reinforcement Goal: Students use a velocity-time graph to find displacement. Answers: 1. Velocity (m/s)

v 7 6 5 4 3 2 1 0 0

0.2 0.4 0.6 0.8 1.0 Time (s)

t

2. →











tf

∆t

0

0.1

0.1

ⴙ0.000

ⴙ0.225

ⴙ0.225

ⴙ2.0

ⴙ2.5

ⴙ2.25

ⴙ0.225

0.1

0.2

0.1

ⴙ0.225

ⴙ0.500

ⴙ0.275

ⴙ2.5

ⴙ3.0

ⴙ2.75

ⴙ0.275

0.2

0.3

0.1

ⴙ0.500

ⴙ0.825

ⴙ0.325

ⴙ3.0

ⴙ3.5

ⴙ3.25

ⴙ0.325

0.3

0.4

0.1

ⴙ0.825

ⴙ1.200

ⴙ0.375

ⴙ3.5

ⴙ4.0

ⴙ3.75

ⴙ0.375

0.4

0.5

0.1

ⴙ1.200

ⴙ1.625

ⴙ0.425

ⴙ4.0

ⴙ4.5

ⴙ4.25

ⴙ0.425

0.5

0.6

0.1

ⴙ1.625

ⴙ2.100

ⴙ0.475

ⴙ4.5

ⴙ5.0

ⴙ4.75

ⴙ0.475

0.6

0.7

0.1

ⴙ2.100

ⴙ2.625

ⴙ0.525

ⴙ5.0

ⴙ5.5

ⴙ5.25

ⴙ0.525

0.7

0.8

0.1

ⴙ2.625

ⴙ3.200

ⴙ0.575

ⴙ5.5

ⴙ6.0

ⴙ5.75

ⴙ0.575

0.8

0.9

0.1

ⴙ3.200

ⴙ3.825

ⴙ0.625

ⴙ6.0

ⴙ6.5

ⴙ6.25

ⴙ0.625

0.9

1.0

0.1

ⴙ3.825

ⴙ4.500

ⴙ0.675

ⴙ6.5

ⴙ7.0

ⴙ6.75

ⴙ0.675

di

∆d



ti

df

vi

3. They are the same. (vi ⫹ vf ) 4. (a) vav = ᎏ 2 (vi ⫹ vf )∆t (b) ∆d = ᎏᎏ 2 5. 5.0 m/s2 6. Add the initial and final velocities, divide by 2, and multiply the result by the time interval.

298

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vf

vav

vav∆t

ANSWER KEY

CHAPTER 11

BLM 11-18, Calculating Displacement/Problem Solving

BLM 11-20, Stunt Driving/ Reinforcement

Goal: Students practise using the formulas introduced in Chapter 11. Answers:

Goal: Students relate types of motion and their characteristics to a velocity-time graph. Answers:

1. (a) 4500 m

1. Velocity (m/s)

(b) 550 s; or, 9 min 10 s (2 s.d.)

v

2. (a) 2.00 m/s2 (b) 12.0 m/s

A

(c) 144 m 3. (a) –0.8 m/s2

(c) 440 m 4. 11.8 s

BLM 11-19, Displacement from Average Velocity Quiz /Assessment Goal: Students assess their ability to calculate displacement for accelerated motion. Answers:

2. (a) 6.0 m/s

t

2. The car starts from rest and speeds up with an increasing positive acceleration. Then it moves with constant positive acceleration. It continues moving in a positive direction with uniform motion. It slows down with constant negative acceleration, still moving in a positive direction. It comes to rest and then begins moving in a negative direction with increasing positive acceleration. Finally it moves in a negative direction with positive acceleration before coming to rest once more.

Goal: Students practise solving motion problems. Answers:

(b) 57 m

1. (a) ∆d = ⫹20 m, a = ⫹2.5 m/s2

v Velocity (m/s)

B C E F H C

H

BLM 11-21, Motion Problems and Displacement/Reinforcement

1. 0.57 m/s2

9 8 7 6 5 4 3 2 1 0

C

Time (s)

(b) 22.0 m/s

3.

B C E

(b) ∆d = ⫹34 m, a = ⫹2.4 m/s2 (c) ∆d = ⫹23.6 m, a = –1.3 m/s2 (d) ∆d = ⫹107 m, a = ⫹0.90 m/s2 0

2

4

6 8 10 12 Time (s)

t

2. (a) ti = 0 s

4. According to the graph at t = 0, when the student’s velocity should have been zero, the student was moving at 2.0 m/s. The student did not start the race in a fair manner.

tf = 20 s

vi = 0 m/s

a = ⫹0.5 m/s2

vf = ⫹10 m/s

∆d = ⫹100 m

vav = ⫹5 m/s (b) ti = 0 s tf = 15 s

vi = ⫹20 m/s

a = 0 m/s2

vf = ⫹20 m/s

∆d = ⫹300 m

vav = ⫹20 m/s (c) Students’ answers will vary somewhat.

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ANSWER KEY

CHAPTER 11

BLM 11-22, Accelerated Motion/ Vocabulary Check Goal: Students assess their knowledge of the vocabulary used in this chapter. Answers: 1

2

A

C

C

E

L

E

3

R

A

T

A

C

C

V

E

R

4

D R 6 A G

8

I

N

S

T

A

N

T

A

N

E

O U

S

A

C

T H E D M O T I O N O R E L E R O M E T E R T I C A G E A C C E L E R L P H Y 7 S T R O B O S C O P I C E L E R A T I O N S

BLM 11-23, Chapter 11 Review/ Reinforcement

11. (d)

Goal: Students assess their understanding of the concepts they studied in Chapter 11. Answers:

13. (b)

1. F: According to physicists, acceleration implies that an object is changing its motion.

12. (a)

14. (c) 15. (c) 16. (g)

2. F: Acceleration is the rate of change of velocity.

17. (j)

3. T

18. (i)

4. T

19. (c)

5. T

20. (h)

6. F: Objects tend to fall with a constant acceleration when dropped from rest.

21. (a)

7. T

23. (f)

8. F: Air brakes stop vehicles by increasing drag.

24. (e)

9. F: An object that is moving at a constant speed around a circle is accelerating.

25. (l)

22. (b)

10. F: The unit for acceleration is metres per second squared.

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5

A

E

T

A I R B R A K E

O N