vectors and motion in two dimensions - university of …n00006757/physicslectures/knight...

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PowerPoint® Lectures for

College Physics: A Strategic Approach, Second Edition

Chapter 3

Vectors and

Motion in Two

Dimensions

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3 Vectors and Motion in Two Dimensions

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Reading Quiz

1. Ax is the __________ of the vector A.

A. magnitude

B. x-component

C. direction

D. size

E. displacement

Slide 3-7

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Answer

1. Ax is the __________ of the vector A.

A. magnitude

B. x-component

C. direction

D. size

E. displacement

Slide 3-8

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Reading Quiz

2. The acceleration vector of a particle in projectile motion

A. points along the path of the particle.

B. is directed horizontally.

C. vanishes at the particle’s highest point.

D. is directed down at all times.

E. is zero.

Slide 3-9

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Answer

Slide 3-10

2. The acceleration vector of a particle in projectile motion

A. points along the path of the particle.

B. is directed horizontally.

C. vanishes at the particle’s highest point.

D. is directed down at all times.

E. is zero.

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3. The acceleration vector of a particle in uniform circular motion

A. points tangent to the circle, in the direction of motion.

B. points tangent to the circle, opposite the direction of motion.

C. is zero.

D. points toward the center of the circle.

E. points outward from the center of the circle.

Reading Quiz

Slide 3-11

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Answer

Slide 3-12

3. The acceleration vector of a particle in uniform circular motion

A. points tangent to the circle, in the direction of motion.

B. points tangent to the circle, opposite the direction of motion.

C. is zero.

D. points toward the center of the circle.

E. points outward from the center of the circle.

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Vectors

Slide 3-13

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Note that a is

v

t

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Which of the vectors below best represents

the vector sum P + Q?

Checking Understanding

Slide 3-16

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Which of the vectors below best represents

the vector sum P + Q?

Answer

Slide 3-17

A.

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Checking Understanding

Which of the vectors below best represents

the difference P – Q?

Slide 3-18

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Which of the vectors below best represents

the difference P – Q?

Answer

Slide 3-19

B.

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Checking Understanding

Which of the vectors below best represents

the difference Q – P?

Slide 3-20

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Which of the vectors below best represents

the difference Q – P?

Answer

Slide 3-21

C.

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Component Vectors and Components

Slide 3-22

“component” = “projection”

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h

hosin

h

hacos

a

o

h

htan

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h

ho1sin

h

ha1cos

a

o

h

h1tan

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Component Vectors and Components

Slide 3-22

cos

sin

x

y

B B

B B

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Component Vectors and Components

Slide 3-22

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What are the x- and y-components of this vector?

A. 3, 2

B. 2, 3

C. 3, 2

D. 2, 3

E. 3, 2

Checking Understanding

Slide 3-23

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What are the x- and y-components of this vector?

Answer

Slide 3-24

A. 3, 2

B. 2, 3

C. 3, 2

D. 2, 3

E. 3, 2

Ways to represent a vector:

2, 3

3.6, 56

ˆ ˆ2 3

ˆ ˆ2 3

A

A

A x y

A i j

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What are the x- and y-components of this vector?

A. 3, 1

B. 3, 4

C. 3, 3

D. 4, 3

E. 3, 4

Checking Understanding

Slide 3-25

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What are the x- and y-components of this vector?

Answer

Slide 3-26

A. 3, 1

B. 3, 4

C. 3, 3

D. 4, 3

E. 3, 4

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The following vector has length 4.0 units.

What are the x- and y-components of this vector?

A. 3.5, 2.0

B. 2.0, 3.5

C. 3.5, 2.0

D. 2.0, 3.5

E. 3.5, 2.0

Checking Understanding

Slide 3-27

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The following vector has length 4.0 units.

What are the x- and y-components of this vector?

Answer

Slide 3-28

A. 3.5, 2.0

B. 2.0, 3.5

C. 3.5, 2.0

D. 2.0, 3.5

E. 3.5, 2.0

Solve by using common sense…

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The following vector has length 4.0 units.

What are the x- and y-components of this vector?

Answer

Slide 3-28

A. 3.5, 2.0

B. 2.0, 3.5

C. 3.5, 2.0

D. 2.0, 3.5

E. 3.5, 2.0

4 0 60 2 0

4 0 30 2 0

4 0 120 2 0

x

x

x

A

A

A

Solve by calculation:

. cos .

. sin .

. cos .

4 0 30 3 5

4 0 60 3 5

4 0 120 3 5

. cos .

. sin .

. sin .

y

y

y

A

A

A

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The following vector has length 4.0 units.

What are the x- and y-components of this vector?

A. 3.5, 2.0

B. 2.0, 3.5

C. 3.5, 2.0

D. 2.0, 3.5

E. 3.5, 2.0

Checking Understanding

Slide 3-29

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The following vector has length 4.0 units.

What are the x- and y-components of this vector?

Answer

Slide 3-30

A. 3.5, 2.0

B. 2.0, 3.5

C. 3.5, 2.0

D. 2.0, 3.5

E. 3.5, 2.0

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Example Problem

The labeled vectors each have length 4 units. For each vector,

what is the component perpendicular to the ramp?

Slide 3-31

30

cos30PP

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Example Problem

The labeled vectors each have length 4 units. For each vector,

what is the component perpendicular to the ramp?

Slide 3-31

30

sin30Q

Q

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Example Problem

The labeled vectors each have length 4 units. For each vector,

what is the component perpendicular to the ramp?

Slide 3-31

30

cos30R

R

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Example Problem

The labeled vectors each have length 4 units. For each vector,

what is the component perpendicular to the ramp?

Slide 3-31

30sin30S

S

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Example Problem

The labeled vectors each have length 4 units. For each vector,

what is the component parallel to the ramp?

Slide 3-31

30

sin30P

P

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Example Problem

The labeled vectors each have length 4 units. For each vector,

what is the component parallel to the ramp?

Slide 3-31

30

cos30Q

Q

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Example Problem

The labeled vectors each have length 4 units. For each vector,

what is the component parallel to the ramp?

Slide 3-31

30

sin30R

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Example Problem

The labeled vectors each have length 4 units. For each vector,

what is the component parallel to the ramp?

Slide 3-31

S

cos30S30

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Example Problems

The Manitou Incline was an extremely steep cog railway in the

Colorado mountains; cars climbed at a typical angle of 22 with

respect to the horizontal. What was the vertical elevation

change for the one-mile run along the track?

Slide 3-32

22

1 mih?oh

sin

(1 mi)sin(22 )

0.4 mi

oh h

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Example Problems

The maximum grade of interstate highways in the United

States is 6.0%, meaning a 6.0 meter rise for 100 m of

horizontal travel.

a. What is the angle with respect to the horizontal of the

maximum grade?

Slide 3-32

?

100 mah

6.0 moh 1 1 6.0 mtan tan 3.4

100 m

o

a

h

h

h

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Example Problems

The maximum grade of interstate highways in the United

States is 6.0%, meaning a 6.0 meter rise for 100 m of

horizontal travel.

b. Suppose a car is driving up a 6.0% grade on a mountain

road at 67 mph (30 m/s). How many seconds does it take

the car to increase its height by 100 m?

Slide 3-32

100 msin 1700 m

sin sin(6 )

oo

hh h h

1700 m

mi67

x xv t

t v

h

1609 m

1 mi

1 h

56 s

3600 s

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The diagram below shows two successive positions of a

particle; it’s a segment of a full motion diagram. Which of the

acceleration vectors best represents the acceleration between

vi and vf?

Checking Understanding

Slide 3-33

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The diagram below shows two successive positions of a

particle; it’s a segment of a full motion diagram. Which of the

acceleration vectors best represents the acceleration between

vi and vf?

Answer

Slide 3-34

D.

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A new ski area has opened that emphasizes the extreme nature

of the skiing possible on its slopes. Suppose an ad intones

“Free fall skydiving is the greatest rush you can

experience…but we’ll take you as close as you can get on land.

When you tip your skis down the slope of our steepest runs, you

can accelerate at up to 75% of the acceleration you’d

experience in free fall.” What angle slope could give such an

acceleration?

Example Problems: Motion on a Ramp

Slide 3-35

?

90

sin(90 ) a g

g

1 1

1

1

0.75

0.75 cos(90 )

cos(90 ) 0.75

cos cos(90 ) cos (0.75)

90 cos (0.75)

90 cos (0.75)

49

a g

g g

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Ski jumpers go down a long slope on slippery skis, achieving a

high speed before launching into air. The “in-run” is essentially a

ramp, which jumpers slide down to achieve the necessary

speed. A particular ski jump has a ramp length of 120 m tipped

at 21 with respect to the horizontal. What is the highest speed

that a jumper could reach at the bottom of such a ramp?

Example Problems: Motion on a Ramp

Slide 3-35

21

g

sing

120 m x

2 2

2

2

2

0

2

2

2(9.8 m/s )(sin 21 )(120 m)

29 m/s

xf xi x

xi

xf x

xf x

xf

xf

v v a x

v

v a x

v a x

v

v

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Example Problems: Relative Motion

An airplane pilot wants to fly due west from Spokane to

Seattle. Her plane moves through the air at 200 mph, but the

wind is blowing 40 mph due north. In what direction should

she point the plane—that is, in what direction should she fly

relative to the air?

Slide 3-36

200 mi/h40 mi/h

1 1 40 mi/htan tan

o

a

h

h 200 mi/h11 S of W

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Example Problems: Relative Motion

A skydiver jumps out of an airplane 1000 m directly above his

desired landing spot. He quickly reaches a steady speed,

falling through the air at 35 m/s. There is a breeze blowing at 7

m/s to the west. At what angle with respect to vertical does he

fall? When he lands, what will be his displacement from his

desired landing spot?

Slide 3-36

35 m/s

7 m/s

1 1 7 m/ssin sin

oh

h 35 m/s11.31 10

1000 m

x

opptan opp adj tan

adj

(1000 m)tan(11.31 )

200 m to the west

x

x

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Projectile Motion

The horizontal and vertical

components of the motion are

independent.

The horizontal motion is

constant; the vertical motion

is free fall:

Slide 3-37

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© 2010 Pearson Education, Inc. Slide 3-39

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Example Problem: Projectile Motion

In the movie Road Trip, some students are seeking to jump a car

across a gap in a bridge. One student, who professes to know

what he is talking about (“Of course I’m sure—with physics, I’m

always sure.”), says that they can easily make the jump.

Continued next slide…

Slide 3-40

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Example Problem: Projectile Motion

The car weighs 2100 pounds, with passengers and luggage.

Right before the gap, there’s a ramp that will launch the car at an

angle of 30°. The gap is 10 feet wide. He then suggests that they

should drive the car at a speed of 50 mph in order to make the

jump.

a. If the car actually went airborne at a speed of 50 mph at an

angle of 30° with respect to the horizontal, how far would it

travel before landing?

Slide 3-40

50 mi/hiv

cos iv

sin iv

30

21

2

0 0 0, so

10

2

0 or

2 2 sin10

2

f i yi y

f i

yi y

yi i

yi y

y

y y v t a t

y y y

t v a t

t

v vv a t t

a g

solution continued next slide...

© 2010 Pearson Education, Inc.

Example Problem: Projectile Motion

The car weighs 2100 pounds, with passengers and luggage.

Right before the gap, there’s a ramp that will launch the car at an

angle of 30°. The gap is 10 feet wide. He then suggests that they

should drive the car at a speed of 50 mph in order to make the

jump.

a. If the car actually went airborne at a speed of 50 mph at an

angle of 30° with respect to the horizontal, how far would it

travel before landing?

Slide 3-40

50 mi/hiv

cos iv

sin iv

30

2

2

1

22 2 sin

0, 0, and from the previous slide, , so

2 sin( cos )

2sin cos the "range equation"

xi xif

yi ixi

y

ixi if

if

x x v t a t

v vx a t

a g

vx v t v

g

vx

g

solution continued next slide...

© 2010 Pearson Education, Inc.

Example Problem: Projectile Motion

The car weighs 2100 pounds, with passengers and luggage.

Right before the gap, there’s a ramp that will launch the car at an

angle of 30°. The gap is 10 feet wide. He then suggests that they

should drive the car at a speed of 50 mph in order to make the

jump.

a. If the car actually went airborne at a speed of 50 mph at an

angle of 30° with respect to the horizontal, how far would it

travel before landing?

Slide 3-40

50 mi/hiv

cos iv

sin iv

30

mi50 iv

h

1609 m

1 mi

1 h

2 2

2

22 m/s3600 s

2 2(22 m/s)sin cos sin 30 cos30

9.8 m/s

100 cm 1 in 1 ft44 m 140 ft

1 m 2.54 cm 12 in

So the car could easily make the jump.

i

f

f

vx

g

x

© 2010 Pearson Education, Inc.

Example Problem: Projectile Motion

The car weighs 2100 pounds, with passengers and luggage.

Right before the gap, there’s a ramp that will launch the car at an

angle of 30°. The gap is 10 feet wide. He then suggests that they

should drive the car at a speed of 50 mph in order to make the

jump.

a. If the car actually went airborne at a speed of 50 mph at an

angle of 30° with respect to the horizontal, how far would it

travel before landing?

b. Does the mass of the car make any difference in your

calculation?

Slide 3-40

The mass does not make any difference, because the flight time depends only on g.

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Example Problem: Broad Jumps

A grasshopper can jump a distance of 30 in (0.76 m) from a

standing start. If the grasshopper takes off at the optimal angle

for maximum distance of the jump, what is the initial speed of

the jump?

Slide 3-41

2

2

45

Use the range equation:

2sin cos

2sin cos

9.8 m/s(0.76 m)

2sin(45 )cos(45 )

2.7 m/s

optimal

i

f i

i

i

v gx v x

g

v

v

cos iv

ivsin iv

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Example Problem: Broad Jumps

A grasshopper can jump a distance of 30 in (0.76 m) from a

standing start. If the grasshopper takes off at the optimal angle

for maximum distance of the jump, what is the initial speed of

the jump? Most animals jump at a lower angle than 45°.

Suppose the grasshopper takes off at 30° from the horizontal.

What jump speed is necessary to reach the noted distance?

Slide 3-41

2

30

9.8 m/s(0.76 m)

2sin cos 2sin(30 )cos(30 )

2.9 m/s

i

i

gv x

v

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Example Problem

Alan Shepard took a golf ball to the moon during one of the

Apollo missions, and used a makeshift club to hit the ball a

great distance. He described the shot as going for “miles and

miles.” A reasonable golf tee shot leaves the club at a speed of

64 m/s. Suppose you hit the ball at this speed at an angle of

30 with the horizontal in the moon’s gravitational acceleration

of 1.6 m/s2. How long is the ball in the air?

Slide 3-42

2

2 sin 2(64 m/s)sin(30 )

1.6 m/s

40 s

ivt

g

t

© 2010 Pearson Education, Inc.

Example Problem

Alan Shepard took a golf ball to the moon during one of the

Apollo missions, and used a makeshift club to hit the ball a

great distance. He described the shot as going for “miles and

miles.” A reasonable golf tee shot leaves the club at a speed of

64 m/s. Suppose you hit the ball at this speed at an angle of

30 with the horizontal in the moon’s gravitational acceleration

of 1.6 m/s2. How long is the ball in the air? How far would the

shot go?

Slide 3-42

2 2

2

2 2(64 m/s)sin cos sin(30 )cos(30 )

1.6 m/s

1 mi2200 m 1.4 miles!!!

1609 m

if

f

vx

g

x

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Circular Motion

There is an acceleration

because the velocity is

changing direction.

Slide 3-43

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Example Problems: Circular Motion

Two friends are comparing the acceleration of their vehicles.

Josh owns a Ford Mustang, which he clocks as doing 0 to 60

mph in a time of 5.6 seconds. Josie has a Mini Cooper that

she claims is capable of higher acceleration. When Josh

laughs at her, she proceeds to drive her car in a tight circle at

13 mph. Which car experiences a higher acceleration?

Slide 3-44

2

22

miniCooper

2

Mustang

mini Coopers have a turning radius of 17.5 ft = 5.3 m, so let's use 6 m:

mi 1 h 1609 m13

h 3600 s 1 mi5.6 m/s

6.0 m

mi 1609 m 1 h60

h 1 mi 3600 s 4.8 m/s5.6 s

va

r

va

t

The miniCooper wins!

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Example Problems: Circular Motion

Turning a corner at a typical large intersection is a city means

driving your car through a circular arc with a radius of about

25 m. If the maximum advisable acceleration of your vehicle

through a turn on wet pavement is 0.40 times the free-fall

acceleration, what is the maximum speed at which you should

drive through this turn?

Slide 3-44

2 2

22

2

0.40 0.40 9.8 m/s 3.9 m/s

m 1 mi 3600 s(3.9 m/s )(25 m) 9.9 22 mph

s 1609 m 1 h

a g

va v a r v a r

r

v

© 2010 Pearson Education, Inc. Slide 3-45

Summary

© 2010 Pearson Education, Inc. Slide 3-46

Summary