My writer delivered an awesome dissertation on the Creek War, I definitely didn’t have the time to do all the research she did on my own.

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HW_Week2

Due: 11:59pm on Friday, September 12, 2014

You will receive no credit for items you complete after the assignment is due. Grading Policy

Speed of a Bullet

A bullet is shot through two cardboard disks attached a distance apart to a shaft turning with a rotational period , as

shown.

Part A

Derive a formula for the bullet speed in terms of , , and a measured angle between the position of the hole in

the first disk and that of the hole in the second. If required, use , not its numeric equivalent. Both of the holes lie at

the same radial distance from the shaft. measures the angular displacement between the two holes; for instance,

means that the holes are in a line and means that when one hole is up, the other is down. Assume that

the bullet must travel through the set of disks within a single revolution.

**Hint 1. **Consider hole positions

The relative position of the holes can be used to find the bullet’s speed. Remember, the shaft will have

rotated while the bullet travels between the disks.

**Hint 2. **How long does it take for the disks to rotate by an angle ?

The disks rotate by 2 in time . How long will it take them to rotate by ?

Give your answer in terms of , , and constants such as .

**Hint 1. **Checking your formula

If your formula is correct, when you plug 2 in for , your answer will be .

ANSWER:

2 J

R

J

J – J – R

J

R J

J R

R J

11/10/2014 HW_Week2

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ANSWER:

Correct

Exercise 2.14

A race car starts from rest and travels east along a straight and level track. For the first 5.0 of the car’s motion, the

eastward component of the car’s velocity is given by .

Part A

What is the acceleration of the car when = 14.8 ?

Express your answer with the appropriate units.

ANSWER:

Correct

Motion of Two Rockets

Learning Goal:

To learn to use images of an object in motion to determine velocity and acceleration.

Two toy rockets are traveling in the same direction (taken to be the *x *axis). A diagram is shown of a timeexposure

image

where a stroboscope has illuminated the rockets at the uniform time intervals indicated.

J =

J

R

2 = R

J

T

W40 – NT U

W4 NT

4 = 7.58

NT

11/10/2014 HW_Week2

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Part A

At what time(s) do the rockets have the same velocity?

**Hint 1. **How to determine the velocity

The diagram shows position, not velocity. You can’t find instantaneous velocity from this diagram, but you

can determine the average velocity between two times and :

.

Note that no position values are given in the diagram; you will need to estimate these based on the distance

between successive positions of the rockets.

ANSWER:

Correct

Part B

At what time(s) do the rockets have the same *x *position?

ANSWER:

0 0

2BWH<00>-

40Ã40

0Ã0

at time only

at time only

at times and

at some instant in time between and

at no time shown in the figure

0 –

0 –

0 – 0 –

0 – 0 –

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Correct

Part C

At what time(s) do the two rockets have the same acceleration?

**Hint 1. **How to determine the acceleration

The velocity is related to the spacing between images in a stroboscopic diagram. Since acceleration is the

rate at which velocity changes, the acceleration is related to the how much this spacing changes from one

interval to the next.

ANSWER:

Correct

Part D

The motion of the rocket labeled A is an example of motion with uniform (i.e., constant) __________.

ANSWER:

at time only

at time only

at times and

at some instant in time between and

at no time shown in the figure

0 –

0 –

0 – 0 –

0 – 0 –

at time only

at time only

at times and

at some instant in time between and

at no time shown in the figure

0 –

0 –

0 – 0 –

0 – 0 –

and nonzero acceleration

velocity

displacement

time

11/10/2014 HW_Week2

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Correct

Part E

The motion of the rocket labeled B is an example of motion with uniform (i.e., constant) __________.

ANSWER:

Correct

Part F

At what time(s) is rocket A ahead of rocket B?

**Hint 1. **Use the diagram

You can answer this question by looking at the diagram and identifying the time(s) when rocket A is to the

right of rocket B.

ANSWER:

Correct

Velocity from Graphs of Position versus Time

An object moves along the *x *axis during four separate trials. Graphs of position versus time for each trial are shown in

the figure.

and nonzero acceleration

velocity

displacement

time

before only

after only

before and after

between and

at no time(s) shown in the figure

0 –

0 –

0 – 0 –

0 – 0 –

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Part A

During which trial or trials is the object’s velocity not constant?

Check all that apply.

**Hint 1. **Finding velocity from a position versus time graph

On a graph of coordinate *x *as a function of time , the instantaneous velocity at any point is equal to the

slope of the curve at that point.

**Hint 2. **Equation for slope

The slope of a line is its rise divided by the run:

.

ANSWER:

Correct

The graph of the motion during Trial B has a changing slope and therefore is not constant. The other trials all

have graphs with constant slope and thus correspond to motion with constant velocity.

Part B

During which trial or trials is the magnitude of the average velocity the largest?

0

TMPQF- Y

U

Trial A

Trial B

Trial C

Trial D

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Check all that apply.

**Hint 1. **Definition of average velocity

Recall that

.

Then note that the question asks only about the magnitude of the velocity.

ANSWER:

Correct

While Trial B and Trial D do not have the same average velocity, the only difference is the direction! The

magnitudes are the same. Neither one is “larger” than the other, and it is only because of how we chose our

axes that Trial B has a positive average velocity while Trial D has a negative average velocity. In Trial C the

object does not move, so it has an average velocity of zero. During Trial A the object has a positive average

velocity but its magnitude is less than that in Trial B and Trial D.

± Average Velocity from a Position vs. Time Graph

Learning Goal:

To learn to read a graph of position versus time and to calculate average velocity.

In this problem you will determine the average velocity of a

moving object from the graph of its position as a function

of time . A traveling object might move at different speeds

and in different directions during an interval of time, but if we

ask at what *constant *velocity the object would have to travel to

achieve the same displacement over the given time interval,

that is what we call the object’s *average velocity*. We will use

the notation to indicate average velocity over the

time interval from to . For instance, is the

average velocity over the time interval from to .

Part A

BWFSBHF WFMPDJUZ- – QPTJUJPO

UJNF

Y

U

Trial A

Trial B

Trial C

Trial D

40

0

2BWF <0 0 >

0 0 2BWF <>

0 – 0 –

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Consulting the graph shown in the figure, find the object’s average velocity over the time interval from 0 to 1 second.

Answer to the nearest integer.

**Hint 1. **Definition of average velocity

Average velocity is defined as the constant velocity at which an object would have to travel to achieve a given

displacement (difference between final and initial positions, which can be negative) over a given time interval,

from the initial time to the final time . The average velocity is therefore equal to the displacement divided

by the given time interval. In symbolic form, average velocity is given by

.

ANSWER:

Correct

Part B

Find the average velocity over the time interval from 1 to 3 seconds.

Express your answer in meters per second to the nearest integer.

**Hint 1. **Find the change in position

The final and initial positions can be read off the *y *axis of the graph. What is the displacement during the time

interval from 1 to 3 seconds?

Express your answer numerically, in meters

ANSWER:

**Hint 2. **Definition of average velocity

Average velocity is defined as the constant velocity at which an object would have to travel to achieve a given

displacement (difference between final and initial positions, which can be negative) over a given time interval,

from the initial time to the final time . The average velocity is therefore equal to the displacement divided

by the given time interval. In symbolic form, average velocity is given by

.

ANSWER:

0J 0G

2BWF<0J 0G >-

40G Ã40J

0G Ã0J

2BWF <> = 0 NT

4G Ã4J = 40 N

0J 0G

2BWF<0J 0G >-

40G Ã40J

0G Ã0J

11/10/2014 HW_Week2

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Correct

A note about instantaneous velocity. The instantaneous velocity at a certain moment in time is represented by

the slope of the graph at that moment. For straightline

graphs, the (instantaneous) velocity remains constant

over the interval, so the instantaneous velocity at any time during an interval is the same as the average

velocity over that interval. For instance, in this case, the instantaneous velocity at any time from 1 to 3 seconds

is the same as the average velocity of .

Part C

Now find .

Give your answer to three significant figures.

**Hint 1. **A note on the displacement

Since the object’s position remains constant from time 0 to time 1, the object’s displacement from 0 to 3 is the

same as in Part B. However, the time interval has changed.

ANSWER:

Correct

Note that is not equal to the simple arithmetic average of and , i.e.,

, because they are averages for time intervals of different lengths.

Part D

Find the average velocity over the time interval from 3 to 6 seconds.

Express your answer to three significant figures.

**Hint 1. **Determine the displacement

What is the displacement?

Answer to the nearest integer.

ANSWER:

2BWF <> = 20 NT

NT

2BWF <>

2BWF <> = 13.3 NT

2BWF <> 2BWF <> 2BWF <>

2BWF<>2BWF<>

4 Ã4 = 40

N

11/10/2014 HW_Week2

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**Hint 2. **Determine the time interval

What is the time interval?

Answer to two significant figures.

ANSWER:

ANSWER:

Correct

Part E

Finally, find the average velocity over the whole time interval shown in the graph.

Express your answer to three significant figures.

**Hint 1. **Determine the displacement

What is the displacement?

Answer to the nearest integer.

ANSWER:

ANSWER:

Correct

Note that though the *average *velocity is zero for this time interval, the *instantaneous *velocity (i.e., the slope of

the graph) has several different values (positive, negative, zero) during this time interval.

Note as well that since average velocity over a time interval is defined as the change in position (displacement)

in the given interval divided by the time, the object can travel a great distance (here 80 meters) and still have

zero average velocity, since it ended up exactly where it started. Therefore, zero average velocity does not

necessarily mean that the object was standing still the entire time!

Given Positions, Find Velocity and Acceleration

0G Ã0J = 3.0 T

2BWF <> = 13.3

NT

4 Ã4 = 0 N

2BWF <> = 0 NT

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Learning Goal:

To understand how to graph position, velocity, and acceleration of an object starting with a table of positions vs. time.

The table shows the *x *coordinate of a moving object. The position is tabulated at 1s

intervals. The *x *coordinate is

indicated below each time. You should make the simplification that the acceleration of the object is bounded and

contains no spikes.

time (s) 0 1 2 3 4 5 6 7 8 9

*x *(m) 0 1 4 9 16 24 32 40 46 48

Part A

Which graph best represents the function , describing

the object’s position vs. time?

**Hint 1. **Meaning of a bounded and nonspiky acceleration

A bounded and nonspiky acceleration results in a smooth graph of vs. .

ANSWER:

Correct

Part B

Which of the following graphs best represents the function , describing the object’s velocity as a function of

40

4 0

1

2

3

4

20

11/10/2014 HW_Week2

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time?

**Hint 1. **Find the velocity toward the end of the motion

Velocity is the time derivative of displacement. Given this, the velocity toward the end of the motion is

__________.

ANSWER:

**Hint 2. **What are the implications of zero velocity?

Two of the possible velocity vs. time graphs indicate zero velocity between and . What would

the corresponding position vs. time graph look like in this region?

ANSWER:

**Hint 3. **Specify the characteristics of the velocity function

The problem states that “the acceleration of the object is bounded and contains no spikes.” This means that

the velocity ___________.

ANSWER:

positive and increasing

positive and decreasing

negative and increasing

negative and decreasing

0 – 0 – T

a horizontal line

straight but sloping up to the right

straight but sloping down to the right

curved upward

curved downward

11/10/2014 HW_Week2

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ANSWER:

Correct

In principle, you could also just compute and plot the average velocity. The expression for the average velocity

is

.

The notation emphasizes that this is not an instantaneous velocity, but rather an average over an

interval. After you compute this, you must put a single point on the graph of velocity vs. time. The most accurate

place to plot the average velocity is at the middle of the time interval over which the average was computed.

Also, you could work back and find the position from the velocity graph. The position of an object is the integral

of its velocity. That is, the area under the graph of velocity vs. time from up to time must equal the

position of the object at time . Check that the correct velocity vs. time graph gives you the correct position

according to this method.

Part C

Which of the following graphs best represents the function

, describing the acceleration of this object?

has spikes

has no discontinuities

has no abrupt changes of slope

is constant

1

2

3

4

2BWH<00>-

40Ã40

0Ã0

2BWH <0 0 >

0 – 0

0

0

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**Hint 1. **Find the acceleration toward the end of the motion

Acceleration is the time derivative of velocity. Toward the end of the motion the acceleration is __________.

ANSWER:

**Hint 2. **Calculate the acceleration in the region of constant velocity

What is the acceleration over the interval during which the object travels at constant speed?

Answer numerically in meters per second squared.

ANSWER:

**Hint 3. **Find the initial acceleration

Acceleration is the time derivative of velocity. Initially the acceleration is _________.

ANSWER:

ANSWER:

zero

positive

negative

= 0 NT

zero

positive

negative

1

2

3

4

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Correct

In one dimension, a linear increase or decrease in the velocity of an object over a given time interval implies

constant acceleration over that particular time interval. You can find the magnitude of the acceleration using the

formula for average acceleration over a time interval:

.

When the acceleration is constant over an extended interval, you can choose any value of and within the

interval to compute the average.

Velocity and Acceleration of a Power Ball

Learning Goal:

To understand the distinction between velocity and acceleration with the use of motion diagrams.

In common usage, *velocity *and *acceleration *both can imply having considerable speed. In physics, they are sharply

defined concepts that are not at all synonymous. Distinguishing clearly between them is a prerequisite to understanding

motion. Moreover, an easy way to study motion is to draw a *motion diagram*, in which the position of the object in motion

is sketched at several equally spaced instants of time, and these sketches (or snapshots) are combined into one single

picture.

In this problem, we make use of these concepts to study the motion of a power ball. This discussion assumes that we

have already agreed on a coordinate system from which to measure the position (also called the position vector) of

objects as a function of time. Let and be velocity and acceleration, respectively.

Consider the motion of a power ball that is dropped on the floor and bounces back. In the following questions, you will

describe its motion at various points in its fall in terms of its velocity and acceleration.

Part A

You drop a power ball on the floor. The motion diagram of the ball is sketched in the figure . Indicate whether the

magnitude of the velocity of the ball is increasing,

decreasing, or not changing.

**Hint 1. **Velocity and displacement vectors

BWH<00>-

20Ã20

0Ã0

0 0

. . 0

2 . 0 . 0

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By definition, the velocity is the ratio of the distance traveled to the interval of time taken. If you interpret the

vector displacement as the distance traveled by the ball, the length of is directly proportional to the

length of . Since the length of displacement vectors is increasing, so is the length of velocity vectors.

ANSWER:

Correct

While the ball is in free fall, the magnitude of its velocity is increasing, so the ball is accelerating.

Part B

Since the length of is directly proportional to the length of , the vector connecting each dot to the next could

represent velocity vectors as well as displacement vectors, as shown in the figure here . Indicate whether the

velocity and acceleration of the ball are, respectively,

positive (upward), negative, or zero.

Use P, N, and Z for positive (upward), negative, and

zero, respectively. Separate the letters for velocity

and acceleration with a comma.

**Hint 1. **Acceleration vector

The acceleration is defined as the ratio of the change in velocity to the interval of time, and its direction is

given by the quantity , which represents the change in velocity that occurs in the

interval of time .

ANSWER:

Correct

.. 2.

..

increasing

decreasing

not changing

2. ..

2.- 2 . 0 Ã2 . 0

0 – 0 Ã0

N,N

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Part C

Now, consider the motion of the power ball once it bounces upward. Its motion diagram is shown in the figure here .

Indicate whether the magnitude of the velocity of the ball

is increasing, decreasing, or not changing.

**Hint 1. **Velocity and displacement vectors

By definition, the velocity is the ratio of the distance traveled to the interval of time taken. If you interpret the

vector displacement as the distance traveled by the ball, the length of is directly proportional to the

length of . Since the length of displacement vectors is decreasing, so is the length of velocity vectors.

ANSWER:

Correct

Since the magnitude of the velocity of the ball is decreasing, the ball must be accelerating (specifically, slowing

down).

Part D

The next figure shows the velocity vectors corresponding to the upward motion of the power ball. Indicate whether

its velocity and acceleration, respectively, are positive (upward), negative, or zero.

Use P, N, and Z for positive (upward), negative, and zero, respectively. Separate the letters for velocity and

acceleration with a comma.

.. 2.

..

increasing

decreasing

not changing

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**Hint 1. **Acceleration vector

The acceleration is defined as the ratio of the change in velocity to the interval of time, and its direction is

given by the quantity , which represents the change in velocity that occurs in the

interval of time .

ANSWER:

Correct

Part E

The power ball has now reached its highest point above the ground and starts to descend again. The motion

diagram representing the velocity vectors is the same as that after the initial release, as shown in the figure of Part

B. Indicate whether the velocity and acceleration of the ball at its highest point are positive (upward), negative, or

zero.

Use P, N, and Z for positive (upward), negative, and zero, respectively. Separate the letters for velocity and

acceleration with a comma.

**Hint 1. **Velocity as a continuous function of time

In Part D you found that the velocity of the ball is positive during the upward motion. Once the ball starts its

descent, its velocity is negative, as you found in Part B. Since velocity changes continuously in time, it has to

be zero at some point along the path of the ball.

**Hint 2. **Acceleration as a continuous function of time

In Part D, you found that the acceleration of the ball is negative and constant during the upward motion, as

well as once the ball has started its descent, which you found in Part B. Since acceleration is a continuous

function of time, it has to be negative at the highest point along the path as well.

2.- 2 . 0 Ã2 . 0

0 – 0 Ã0

P,N

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ANSWER:

Correct

These examples should show you that the velocity and acceleration can have opposite or similar signs or that

one of them can be zero while the other has either sign. Try hard to think carefully about them as *distinct*

physical quantities when working with kinematics.

Analyzing Position versus Time Graphs: Conceptual Question

Two cars travel on the parallel lanes of a twolane

road. The

cars’ motions are represented by the position versus time

graph shown in the figure. Answer the questions using the

times from the graph indicated by letters.

Part A

At which of the times do the two cars pass each other?

**Hint 1. **Two cars passing

Two objects can pass each other only if they have the same position at the same time.

ANSWER:

Z,N

11/10/2014 HW_Week2

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Correct

Part B

Are the two cars traveling in the same direction when they pass each other?

ANSWER:

Correct

Part C

At which of the lettered times, if any, does car #1 momentarily stop?

**Hint 1. **Determining velocity from a position versus time graph

The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In

physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a

position versus time graph is the velocity of the object being graphed.

ANSWER:

A

B

C

D

E

None

Cannot be determined

yes

no

A

B

C

D

E

none

cannot be determined

11/10/2014 HW_Week2

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Correct

Part D

At which of the lettered times, if any, does car #2 momentarily stop?

**Hint 1. **Determining velocity from a position versus time graph

The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In

physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a

position versus time graph is the velocity of the object being graphed.

ANSWER:

Correct

Part E

At which of the lettered times are the cars moving with nearly identical velocity?

**Hint 1. **Determining Velocity from a Position versus Time Graph

The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In

physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a

position versus time graph is the velocity of the object being graphed.

ANSWER:

A

B

C

D

E

none

cannot be determined

11/10/2014 HW_Week2

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Correct

Conceptual Question 2.05

Part A

Suppose that an object is moving with constant nonzero acceleration. Which of the following is an accurate

statement concerning its motion?

ANSWER:

Correct

Conceptual Question 2.09

Part A

The graph in the figure shows the position of an object as a function of time. The letters HL

represent particular

moments of time. At which moments shown (H, I, etc.) is the speed of the object

A

B

C

D

E

None

Cannot be determined

In equal times its speed changes by equal amounts.

In equal times its velocity changes by equal amounts.

A graph of its position as a function of time has a constant slope.

In equal times it moves equal distances.

A graph of its velocity as a function of time is a horizontal line.

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(a) the greatest?

ANSWER:

Correct

Part B

(b) the smallest?

ANSWER:

Correct

Conceptual Question 2.04

Part A

H

I

J

K

L

H

I

J

K

L

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When can we be certain that the average velocity of an object is always equal to its instantaneous velocity?

ANSWER:

Correct

Conceptual Question 2.18

Part A

Two objects are thrown from the top of a tall building and experience no appreciable air resistance. One is thrown

up, and the other is thrown down, both with the same initial speed. What are their speeds when they hit the street?

ANSWER:

Correct

Conceptual Question 2.16

Part A

The figure represents the velocity of a particle as it travels along the *x*axis.

At what value (or values) of *t *is the

instantaneous acceleration equal to zero?

always

only when the acceleration is constant

never

only when the acceleration is changing at a constant rate

only when the velocity is constant

The one thrown down is traveling faster.

The one thrown up is traveling faster.

They are traveling at the same speed.

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ANSWER:

Correct

Problem 2.27

Part A

A package is dropped from a helicopter moving upward at 15 m/s. If it takes 8 s before the package strikes the

ground, how high above the ground was the package when it was released if air resistance is negligible?

ANSWER:

Correct

Problem 2.10

Part A

*t *= 1 s

*t *= 0.5 s and *t *= 2 s

*t *= 0

190 m

114 m

152 m

228 m

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The position of an object as a function of time is given by *x = bt2 ct*,

where *b *= 2.0 m/s2 and *c *= 6.7 m/s, and *x *and

*t *are in SI units. What is the instantaneous velocity of the object when *t *= 2.3?

ANSWER:

Correct

Conceptual Question 2.02

Part A

If the graph of the position as a function of time for an object is a horizontal line, that object cannot be accelerating.

ANSWER:

Correct

Prelecture Concept Question 2.01

Part A

Which of the following best describes how to calculate the average acceleration of any object?

ANSWER:

2.0 m/s

3.5 m/s

2.5 m/s

3.0 m/s

True

False

Average acceleration is always equal to the change in velocity of an object divided by the time interval.

Average acceleration is always halfway between the initial acceleration of an object and its final

acceleration.

Average acceleration is always equal to the displacement of an object divided by the time interval.

Average acceleration is always equal to the change in speed of an object divided by the time interval.

11/10/2014 HW_Week2

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Correct

Conceptual Question 2.13

Part A

An object is moving in a straight line along the *x*axis.

A plot of its velocity in the *x*direction

as a function of time is

shown in the figure. Which graph represents its acceleration in the *x*direction

as a function of time?

ANSWER:

11/10/2014 HW_Week2

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2992507 28/29

11/10/2014 HW_Week2

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2992507 29/29

Correct

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Your score on this assignment is 97.1%.

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