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Chapter 6: Motion in Two Dimensions

PHYSICSPrinciples and Problems

You can use vectors and Newton’s laws to

describe projectile motion and circular motion.

BIG IDEA

CHAPTER

6 Motion in Two Dimensions

Section 6.1 Projectile Motion

Section 6.2 Circular Motion

Section 6.3 Relative Velocity

CHAPTER

6 Table Of Contents

Click a hyperlink to view the corresponding slides. Exit

MAIN IDEA

A projectile’s horizontal motion is independent of its vertical

motion.

Essential Questions

• How are the vertical and horizontal motions of a

projectile related?

• What are the relationships between a projectile’s height,

time in the air, initial velocity and horizontal distance

traveled?

SECTION

6.1 Projectile Motion

Review Vocabulary

• Motion diagram a series of images showing the

positions of a moving object taken at regular time

intervals.

New Vocabulary

• Projectile

• Trajectory

SECTION


• If you observed the movement of a golf ball being

hit from a tee, a frog hopping, or a free throw

being shot with a basketball, you would notice

that all of these objects move through the air

along similar paths, as do baseballs, and arrows.

• Each path rises and then falls, always curving

downward along a parabolic path.

Path of a Projectile

SECTION


• An object shot through the air is called a

projectile.

• A projectile can be a football or a drop of water.

• You can draw a free-body diagram of a launched

projectile and identify all the forces that are acting

on it.

Path of a Projectile (cont.)

SECTION


• No matter what the object is, after a projectile has

been given an initial thrust, if you ignore air

resistance, it moves through the air only under

the force of gravity.

• The force of gravity is what causes the object to

curve downward in a parabolic flight path. Its path

through space is called its trajectory.

SECTION



• You can determine an object’s trajectory if you

know its initial velocity.

• Two types of projectile motion are horizontal and

angled.

SECTION



Independence of Motion in Two

Dimensions

Click image to view movie.

SECTION


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• When a projectile is launched at an angle, the

initial velocity has a vertical component as well as

a horizontal component.

• If the object is launched upward, like a ball tossed

straight up in the air, it rises with slowing speed,

reaches the top of its path, and descends with

increasing speed.

Angled Launches

SECTION


• The adjoining figure

shows the separate

vertical- and

horizontal-motion

diagrams for the

trajectory of the ball.

Angled Launches (cont.)

SECTION


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• At each point in the vertical

direction, the velocity of the

object as it is moving upward

has the same magnitude as

when it is moving downward.

• The only difference is that the

directions of the two velocities

are opposite.

SECTION



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• The adjoining figure

defines two quantities

associated with a

trajectory.

SECTION



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• One is the maximum

height, which is the

height of the projectile

when the vertical

velocity is zero and

the projectile has only

its horizontal-velocity

component.

SECTION



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• The other quantity depicted is the range, R, which is the horizontal distance that the projectile travels when the initial and final heights are the same.

• Not shown is the flight time, which is how much time the projectile is in the air.

SECTION



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• For football punts,

flight time is often

called hang time.

SECTION



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ch6_scn11.swf

The Flight of a Ball

A ball is launched at 4.5 m/s at 66° above the

horizontal. What are the maximum height and flight

time of the ball?

SECTION


Step 1: Analyze and Sketch the Problem

The Flight of a Ball (cont.)

• Establish a coordinate system with the initial

position of the ball at the origin.

SECTION


• Show the positions of the ball at the beginning, at

the maximum height, and at the end of the flight.

SECTION



• Draw a motion diagram showing v, a, and Fnet.

SECTION



Known:

yi = 0.0 m

θi = 66°

vi = 4.5 m/s

ay = −9.8 m/s2

Vy, max = 0.0 m/s

Unknown:

ymax = ?

t = ?

Identify the known and unknown variables.

SECTION



Step 2: Solve for the Unknown

SECTION



Find the y-component of vi.

SECTION



Substitute vi = 4.5 m/s, θi = 66°

SECTION



Use symmetry to find the y-component of vf

vyf = - vyi = - 4.1 m/s

SECTION



Solve for the maximum height.

Vy,max2 = vyi

2 + 2a(ymax – yi)

(0.0 m/s2) = vyi + 2a(ymax – 0.0m/s)

ymax = - vyi / 2a

SECTION



Substitute vyi = 4.1 m/s, a = -9.8 m/s2

ymax = (4.1 m/s)2 = 0.86m

2(-9.8 m/s2)

SECTION



Solve for the time to return to the launching height.

Vyf = vyi + at

SECTION



Solve for t.

t = vyf – vyi

a

SECTION



Substitute vyf = - 4.1 m/s, vyi = 4.1 m/s,

a = - 9.80 m/s2

-4.1 m/s – 4.1 m/s = 0.84 s

-9.8 m/s2

SECTION



Step 3: Evaluate the Answer

SECTION



Are the units correct?

Dimensional analysis verifies that the units are

correct.

Do the signs make sense?

All of the signs should be positive.

Are the magnitudes realistic?

0.84 s is fast, but an initial velocity of 4.5 m/s makes

this time reasonable.

SECTION



The steps covered were:


Establish a coordinate system with the initial position of

the ball at the origin.

Show the positions of the ball at the beginning, at the

maximum height, and at the end of the flight.

Draw a motion diagram showing v, a, and Fnet.

SECTION





Find the y-component of vi.

Find an expression for time.

Solve for the maximum height.

Solve for the time to return to the launching height.


SECTION



• So far, air resistance has been ignored in the analysis

of projectile motion.

• Forces from the air can significantly change the motion

of an object.

– Ex. Flying a kite, parachute for a skydiver.

Forces from Air

SECTION


• Picture water flowing from a hose:

– If wind is blowing in the same direction as the water’s initial movement, the horizontal distance the water travels increases.

– If wind is blowing in the opposite direction as the water’s initial movement, the horizontal distance the water travels decreases.

• Even if the air is not moving, it can have a significant effect on some moving objects.

– Ex. Paper falling to the ground.

Forces from Air (cont.)

SECTION


A boy standing on a balcony drops one ball and throws

another with an initial horizontal velocity of 3 m/s. Which

of the following statements about the horizontal and

vertical motions of the balls is correct? (Neglect air

resistance.)

A. The balls fall with a constant vertical velocity and a constant

horizontal acceleration.

B. The balls fall with a constant vertical velocity as well as a constant

horizontal velocity.

C. The balls fall with a constant vertical acceleration and a constant


D. The balls fall with a constant vertical acceleration and an increasing


SECTION

6.1 Section Check

Answer

Reason: The vertical and horizontal motions of a

projectile are independent. The only force

acting on the two balls is the force of

gravity. Because it acts in the vertical

direction, the balls accelerates in the

vertical direction. The horizontal velocity

remains constant throughout the flight of

the balls.

SECTION

6.1 Section Check

Which of the following conditions is met when a

projectile reaches its maximum height?

A. The vertical component of the velocity is zero.

B. The vertical component of the velocity is maximum.

C. The horizontal component of the velocity is maximum.

D. The acceleration in the vertical direction is zero.

SECTION

6.1 Section Check

Answer

Reason: The maximum height is the height at which

the object stops its upward motion and

starts falling down, i.e. when the vertical

component of the velocity becomes zero.

SECTION

6.1 Section Check

Suppose you toss a ball up and catch it while

riding in a bus. Why does the ball fall in your

hands rather than falling at the place where you

tossed it?

SECTION

6.1 Section Check

Trajectory depends on the frame of reference.

For an observer on the ground, when the bus is

moving, your hand is also moving with the same

velocity as the bus, i.e. the bus, your hand, and the

ball will have the same horizontal velocity. Therefore,

the ball will follow a trajectory and fall back into your

hands.

Answer

SECTION

6.1 Section Check

MAIN IDEA

An object in circular motion has an acceleration toward the

circle’s center due to an unbalanced force toward the

circle’s center.

Essential Questions

• Why is an object moving in a circle at a constant speed

accelerating?

• How does centripetal acceleration depend upon the

object’s speed and the radius of the circle?

• What causes centripetal acceleration?

SECTION

6.2 Circular Motion

Review Vocabulary

• Average velocity the change in position divided by the

time during which the change occurred; the slope of an

object’s position-time graph.

New Vocabulary

• Uniform circular motion

• Centripetal acceleration

• Centripetal force

SECTION

6.2 Circular Motion

Describing Circular Motion

Click image to view movie.

SECTION

6.2 Circular Motion

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• The angle between position vectors r1 and r2 is

the same as that between velocity vectors v1 and

v2.

• Thus, ∆r/r = ∆v/v. The equation does not change

if both sides are divided by ∆t.

Centripetal Acceleration

SECTION

6.2 Circular Motion

Centripetal Acceleration (cont.)

• However, v = ∆r/∆t and a = ∆v/∆t

• Substituting v = ∆r/∆t in the left-hand side and

a = ∆v/∆t in the right-hand side gives the following

equation:

SECTION

6.2 Circular Motion

• Solve the equation for acceleration and give it the

special symbol ac, for centripetal acceleration.

• Centripetal acceleration always points to the

center of the circle. Its magnitude is equal to the

square of the speed, divided by the radius of

motion.

SECTION

6.2 Circular Motion


• One way of measuring the speed of an object moving in a

circle is to measure its period, T, the time needed for the

object to make one complete revolution.

• During this time, the object travels a distance equal to the

circumference of the circle, 2πr. The object’s speed, then,

is represented by v = 2πr/T.

SECTION

6.2 Circular Motion


• The acceleration of an object moving in a circle is

always in the direction of the net force acting on it,

there must be a net force toward the center of the

circle. This force can be provided by any number

of agents.

• When an Olympic hammer thrower swings the

hammer, the force is the tension in the chain

attached to the massive ball.

SECTION

6.2 Circular Motion


• When an object moves in a circle, the net force

toward the center of the circle is called the

centripetal force.

• To analyze centripetal acceleration situations

accurately, you must identify the agent of the force

that causes the acceleration. Then you can apply

Newton’s second law for the component in the

direction of the acceleration in the following way.

SECTION

6.2 Circular Motion


• The net centripetal force on an object moving in a

circle is equal to the object’s mass times the

centripetal acceleration.

• Newton’s Second Law for Circular Motion

SECTION

6.2 Circular Motion


• When solving problems, it is useful to choose a

coordinate system with one axis in the direction of

the acceleration.

• For circular motion, the direction of the

acceleration is always toward the center of the

circle.

SECTION

6.2 Circular Motion


• Rather than labeling this axis x or y, call it c, for

centripetal acceleration. The other axis is in the

direction of the velocity, tangent to the circle. It is

labeled tang for tangential.

• Centripetal force is just another name for the net

force in the centripetal direction. It is the sum of all

the real forces, those for which you can identify

agents that act along the centripetal axis.

SECTION

6.2 Circular Motion


• According to Newton’s first law, you will continue

moving with the same velocity unless there is a net

force acting on you.

Centrifugal “Force”

• The passenger in the

car would continue to

move straight ahead if it

were not for the force of

the car acting in the

direction of the

acceleration.

SECTION

6.2 Circular Motion

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• The so-called centrifugal, or outward force, is a

fictitious, nonexistent force.

Centrifugal “Force” (cont.)

• You feel as if you are being

pushed only because you

are accelerating relative to

your surroundings. There is

no real force because there

is no agent exerting a force.

SECTION

6.2 Circular Motion

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ch6_scn50.swf

ch6_scn50.swf

Explain why an object moving in a

circle at a constant speed is

accelerating.

SECTION

6.2 Section Check

Acceleration is the rate of change of velocity, the

object is accelerating due to its constant change in

the direction of its motion.

Answer

SECTION

6.2 Section Check

What is the relationship between the magnitude

of centripetal acceleration (ac) and an object’s

speed (v)?

A.

B.

C.

D.

SECTION

6.2 Section Check

Reason: From the equation for centripetal acceleration:

Centripetal acceleration always points to the

center of the circle. Its magnitude is equal to

the square of the speed divided by the radius

of the motion.

Answer

SECTION

6.2 Section Check

What is the direction of the velocity vector

of an accelerating object?

A. toward the center of the circle

B. away from the center of the circle

C. along the circular path

D. tangent to the circular path

SECTION

6.2 Section Check

Answer

Reason: While constantly changing, the velocity vector

for an object in uniform circular motion is

always tangent to the circle. Vectors are never

curved and therefore cannot be along a

circular path.

SECTION

6.2 Section Check

MAIN IDEA

An object’s velocity depends on the reference frame

chosen.

Essential Questions

• What is relative velocity?

• How do you find the velocities of an object in different

reference frames?

Relative VelocitySECTION

6.3

Review Vocabulary

• resultant a vector that results from the sum of two

other vectors.

New Vocabulary

• Reference frame


6.3

• Suppose you are in a school bus

that is traveling at a velocity of

8 m/s in a positive direction. You

walk with a velocity of 1 m/s

toward the front of the bus.

• If a friend of yours is standing on

the side of the road watching the

bus go by, how fast would your

friend say that you are moving?

Relative Motion in One Dimension


6.3

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ch6_scn58.swf

• If the bus is traveling at 8 m/s,

this means that the velocity of

the bus is 8 m/s, as measured

by your friend in a coordinate

system fixed to the road.

• When you are standing still,

your velocity relative to the

road is also 8 m/s, but your

velocity relative to the bus is

zero.

Relative Motion in One Dimension (cont.)


6.3

ch6_scn58.swf

ch6_scn58.swf

• In the previous example, your motion is viewed

from different coordinate systems.

• A coordinate system from which motion is

viewed is a reference frame.

Relative Velocity


SECTION

6.3

• Walking at 1m/s towards the front of the bus means your velocity is measured in the reference from of the bus.

• Your velocity in the road’s reference frame is different

• You can rephrase the problem as follows: given the velocity of the bus relative to the road and your velocity relative to the bus, what is your velocity relative to the road?

Relative Velocity


SECTION

6.3

• When a coordinate system is moving, two velocities are

added if both motions are in the same direction, and

one is subtracted from the other if the motions are in

opposite directions.

• In the given figure, you will find that your velocity relative to the street is 9 m/s, the sum of 8 m/s and 1 m/s.

Relative Velocity


SECTION

6.3

• You can see that when the velocities are along the

same line, simple addition or subtraction can be

used to determine the relative velocity.

Relative Velocity


SECTION

6.3

• Mathematically, relative velocity is represented as

vy/b + vb/r = vy/r.

• The more general form of this equation is:

Relative Velocity va/b + vb/c + va/c

• The relative velocity of object a to object c is the

vector sum of object a’s velocity relative to object b

and object b’s velocity relative to object c.

Relative Velocity


SECTION

6.3

• The method for adding relative velocities also

applies to motion in two dimensions.

• As with one-dimensional motion, you first draw a

vector diagram to describe the motion and then

you solve the problem mathematically.

Relative Motion in Two Dimensions


6.3

• For example, airline pilots

must take into account the

plane’s speed relative to

the air, and their direction

of flight relative to the air.

They also must consider

the velocity of the wind at

the altitude they are flying

relative to the ground.

Relative Motion in Two Dimensions (cont.)


6.3

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ch6_scn63.swf

• You can use the equations in the figure to solve problems for relative motion in two dimensions.

• Velocity of a reference frame moving relative to the ground is .

• Velocity of an object in the moving frame is .

Relative Velocity

Relative Motion in Two Dimensions (cont.)

SECTION

6.3

Relative Velocity of a Marble

Ana and Sandra are riding on a ferry boat that is

traveling east at a speed of 4.0 m/s. Sandra rolls a

marble with a velocity of 0.75 m/s north, straight

across the deck of the boat to Ana. What is the

velocity of the marble relative to the water?


6.3


Relative Velocity of a Marble (cont.)

• Establish a coordinate system.


6.3

• Draw vectors to represent the velocities of the boat

relative to the water and the marble relative to the

boat.



6.3

Identify known and unknown variables.

Known:

vb/w = 4.0 m/s

vm/b = 0.75 m/s

Unknown:

vm/w = ?

Relative Velocity


SECTION

6.3


Relative Velocity


SECTION

6.3

Because the two velocities are at right angles, use

the Pythagorean theorem.

Relative Velocity


SECTION

6.3

Substitute vb/w = 4.0 m/s, vm/b = 0.75 m/s

Relative Velocity


SECTION

6.3

Find the angle of the marble’s motion.

Relative Velocity


SECTION

6.3

Substitute vb/w = 4.0 m/s, vm/b = 0.75 m/s

= 11° north of east

The marble is traveling 4.1 m/s at 11° north of east.

Relative Velocity


SECTION

6.3


Relative Velocity


SECTION

6.3

Are the units correct?

Dimensional analysis verifies that the velocity is in m/s.

Do the signs make sense?

The signs should all be positive.

Are the magnitudes realistic?

The resulting velocity is of the same order of magnitude

as the velocities given in the problem.

SECTION

6.3 Relative Velocity




Establish a coordinate system.

Draw vectors to represent the velocities of the

boat relative to the water and the marble relative

to the boat.

SECTION





Use the Pythagorean theorem.


SECTION



Steven is walking on the top level of a double-decker bus

with a velocity of 2 m/s toward the rear end of the bus. The

bus is moving with a velocity of 10 m/s. What is the velocity

of Steven with respect to Anudja, who is sitting on the top

level of the bus and to Mark, who is standing on the street?

A. The velocity of Steven with respect to Anudja is 2 m/s and is 12 m/s

with respect to Mark.

B. The velocity of Steven with respect to Anudja is 2 m/s and is 8 m/s


C. The velocity of Steven with respect to Anudja is 10 m/s and is 12

m/s with respect to Mark.

D. The velocity of Steven with respect to Anudja is 10 m/s and is 8 m/s


SECTION

6.3 Section Check

Answer

Reason: The velocity of Steven with respect to Anudja is 2 m/s since Steven is moving with a velocity of 2 m/s with respect to the bus, and Anudja is at rest with respect to the bus.

The velocity of Steven with respect to Mark can be understood with the help of the following vector representation.

SECTION

6.3 Section Check

Which of the following formulas correctly relates the

relative velocities of objects a, b, and c to each other?

A. va/b + va/c = vb/c

B. va/b vb/c = va/c

C. va/b + vb/c = va/c

D. va/b va/c = vb/c

SECTION

6.3 Section Check

Answer

Reason: The relative velocity equation is

va/b + vb/c = va/c.

The relative velocity of object a to object c

is the vector sum of object a’s velocity

relative to object b and object b’s velocity

relative to object c.

SECTION

6.3 Section Check

An airplane flies due south at 100 km/hr relative to the air.

Wind is blowing at 20 km/hr to the west relative to the

ground. What is the plane’s speed with respect to the

ground?

A. (100 + 20) km/hr

B. (100 − 20) km/hr

C.

D.

SECTION

6.3 Section Check

Answer

Reason: Since the two velocities are at right angles, we

can apply the Pythagorean theorem. By using

relative velocity law, we can write:

vp/a2 + va/g

2 = vp/g2

SECTION

6.3 Section Check

CHAPTER

6

Resources

Physics Online

Study Guide

Chapter Assessment Questions

Standardized Test Practice

Motion in Two Dimensions

http://www.connected.mcgraw-hill.com/

• The vertical and horizontal motions of a projectile are

independent. When there is no air resistance, the

horizontal motion component does not experience an

acceleration and has constant velocity; the vertical

motion component of a projectile experiences a

constant acceleration under these same conditions.

Projectile MotionSECTION

6.1

Study Guide

• The curved flight path a projectile follows is called a

trajectory and is a parabola. The height, time of flight,

initial velocity and horizontal distance of this path are

related by the equations of motion. The horizontal

distance a projectile travels before returning to its initial

height depends on the acceleration due to gravity an on

both components on the initial velocity.

Study Guide

Projectile MotionSECTION

6.1

• An object moving in a circle at a constant speed has an

acceleration toward the center of the circle because the

direction of its velocity is constantly changing.

• Acceleration toward the center of the circle is called

centripetal acceleration. It depends directly on the

square of the object’s speed and inversely on the radius

of the circle.

Study Guide

Circular MotionSECTION

6.2

• A net force must be exerted by external agents toward

the circle’s center to cause centripetal acceleration.

Study Guide

Circular MotionSECTION

6.2

• A coordinate system from which you view motion is

called a reference frame. Relative velocity is the velocity

of an object observed in a different, moving reference

frame.

• You can use vector addition to solve motion problems of

an object in a moving reference frame.

Study Guide


6.3

What is the range of a projectile?

A. the total trajectory that the projectile travels

B. the vertical distance that the projectile travels

C. the horizontal distance that the projectile

travels

D. twice the maximum height of the projectile

Chapter Assessment

CHAPTER


Reason: When a projectile is launched at an angle, the

straight (horizontal) distance the projectile

travels is known as the range of the projectile.

Chapter Assessment

CHAPTER


Define the flight time of a trajectory.

A. time taken by the projectile to reach the

maximum height

B. the maximum height reached by the projectile

divided by the magnitude of the vertical velocity

C. the total time the projectile was in the air

D. half the total time the projectile was in the air

Chapter Assessment

CHAPTER


Reason: The flight time of a trajectory is defined as

the total time the projectile was in the air.

Chapter Assessment

CHAPTER


What is centripetal force?

Answer: When an object moves in a circle, the net

force toward the center of the circle is called

centripetal force.

Chapter Assessment

CHAPTER


Donna is traveling in a train due north at 30 m/s.

What is the magnitude of the velocity of Donna

with respect to another train which is running

due south at 30 m/s?

A. 30 m/s + 30 m/s

B. 30 m/s − 30 m/s

C. 302 m/s + 302 m/s

D. 302 m/s − 302 m/s

Chapter Assessment

CHAPTER


Reason: The magnitude of the velocity of Donna

relative to the ground is 30 m/s and the

magnitude of velocity of the other train

relative to the ground is also 30 m/s.

Now, with this speed, if the trains move in

the same direction, then the relative

speed of one train to another will be zero.

Chapter Assessment

CHAPTER


Reason: In this case, since the two trains are

moving in opposite directions, the relative

speed (magnitude of the velocity) of Donna

relative to another train is 30 m/s + 30 m/s.

Chapter Assessment

CHAPTER


What is the relationship between the magnitude

of centripetal acceleration and the radius of a

circle?

A.

B.

C.

D.

Chapter Assessment

CHAPTER


Reason: Centripetal acceleration always points to

the center of the circle. Its magnitude is

equal to the square of the speed, divided

by the radius of motion. That is,

Therefore,

Chapter Assessment

CHAPTER


Which of the following formulas can be used to

calculate the period (T) of a rotating object if the

centripetal acceleration (ac) and radius (r) are

given?

A.

B.

C.

D.

Chapter Assessment

CHAPTER


Reason: We know that the centripetal acceleration always

points toward the center of the circle. Its

magnitude is equal to the square of the speed

divided by the radius of motion. That is, ac = v2/r.

To measure the speed of an object moving in a

circle, we measure the period, T, the time needed

for the object to make one complete revolution.

That is, v = 2πr/T, where 2πr is the

circumference of the circle.

Chapter Assessment

CHAPTER


Reason: Substituting for v in the equation

v = 2πr/T, we get,

Chapter Assessment

CHAPTER


A 1.60-m tall girl throws a football at an angle of

41.0° from the horizontal and at an initial

velocity of 9.40 m/s. How far away from the girl

will it land?

A. 4.55 m

B. 5.90 m

C. 8.90 m

D. 10.5 m


CHAPTER


A dragonfly is sitting on a merry-go-round, 2.8

m from the center. If the tangential velocity of

the ride is 0.89 m/s, what is the centripetal

acceleration of the dragonfly?

A. 0.11 m/s2

B. 0.28 m/s2

C. 0.32 m/s2

D. 2.2 m/s2


CHAPTER


The centripetal force on a 0.82-kg object on the

end of a 2.0-m massless string being swung in a

horizontal circle is 4.0 N. What is the tangential

velocity of the object?

A. 2.8 m/s2

B. 3.1 m/s2

C. 4.9 m/s2

D. 9.8 m/s2


CHAPTER


A 1000-kg car enters an 80-m radius curve at 20

m/s. What centripetal force must be supplied by

friction so the car does not skid?

A. 5.0 N

B. 2.5×102 N

C. 5.0×103 N

D. 1.0×103 N


CHAPTER


A jogger on a riverside path sees a rowing team

coming toward him. If the jogger is moving at 10

km/h, and the boat is moving at 20 km/h, how

quickly does the jogger approach the boat?

A. 3 m/s

B. 8 m/s

C. 40 m/s

D. 100 m/s


CHAPTER


Practice Under Testlike Conditions

Test-Taking Tip

Answer all of the questions in the time provided

without referring to your book. Did you complete the

test? Could you have made better use of your time?

What topics do you need to review?


CHAPTER


Another example of combined relative velocities is

the navigation of migrating neotropical songbirds.

In addition to knowing in which direction to fly, a

bird must account for its speed relative to the air

and its direction relative to the ground.

Relative Velocity

CHAPTER

6

Chapter Resources


If a bird tries to fly over the Gulf of Mexico into a

headwind that is too strong, it will run out of energy

before it reaches the other shore and will perish.

Similarly, the bird must account for crosswinds or it

will not reach its destination.

Relative Velocity

CHAPTER

6

Chapter Resources



A ball is launched at 4.5 m/s at 66° above the

horizontal. What are the maximum height and flight

time of the ball?

CHAPTER

6

Chapter Resources



Ana and Sandra are riding on a ferry boat that is

traveling east at a speed of 4.0 m/s. Sandra rolls a

marble with a velocity of 0.75 m/s north, straight

across the deck of the boat to Ana. What is the

velocity of the marble relative to the water?

CHAPTER

6

Chapter Resources


Trajectories of Two Softballs

CHAPTER

6

Chapter Resources


Motion Diagrams for Horizontal and

Vertical Motions

CHAPTER

6

Chapter Resources


Projectiles Launched at an Angle

CHAPTER

6

Chapter Resources



CHAPTER

6

Chapter Resources


A Player Kicking a Football

CHAPTER

6

Chapter Resources


The Displacement of an Object in Circular

Motion

CHAPTER

6

Chapter Resources


Vectors at the Beginning and End of a Time

Interval

CHAPTER

6

Chapter Resources


Uniform Circular Motion

CHAPTER

6

Chapter Resources


A Nonexistent Force

CHAPTER

6

Chapter Resources


Calculating Relative Velocity

CHAPTER

6

Chapter Resources


The Plane’s Velocity Relative to the

Ground

CHAPTER

6

Chapter Resources



CHAPTER

6

Chapter Resources


A Hammer Thrower Swings a Hammer

CHAPTER

6

Chapter Resources


End of Custom Shows

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