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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 xaxis.
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.
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Correct
Conceptual Question 2.13
Part A
An object is moving in a straight line along the xaxis.
A plot of its velocity in the xdirection
as a function of time is
shown in the figure. Which graph represents its acceleration in the xdirection
as a function of time?
ANSWER:
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Correct
Score Summary:
Your score on this assignment is 97.1%.
You received 17.48 out of a possible total of 18 points.