Ch 02 HW
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Aerospace Engineering
Date
Apr 3, 2024
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58
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Ch 02 HW
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Ch 02 HW
Due: 10:59pm on Tuesday, March 7, 2023
You will receive no credit for items you complete after the assignment is due. Grading Policy
PhET Tutorial: Moving Man
Learning Goal:
To understand the relationships between position, velocity, and acceleration.
NOTE: These activities use Java, and are therefore not screen-reader accessible and may not work on a mobile device. If the
browser you're using no longer supports Java, try a different browser and download the Java plugin for this content.
For this tutorial, use the PhET simulation The Moving Man
. This simulation allows you to drag a person back and forth and look
at the resulting position, velocity, and acceleration. You can also enter a position as a function of time mathematically and look at
the resulting motion.
Start the simulation.
When you click the simulation link, you may be asked whether to run, keep, or save the file. If no prompt
appears, you can also right-click the link and select "Copy link address", then manually paste this URL into a new tab (or
window).
If you are still unable to open this PhET in your browser, select the following link to open a separate compatible version of this
simulation. This version may not perform as well, therefore it is offered as a secondary option: Moving Man (backup link)
.
Under the Charts
tab you can click and drag the person left and right, or enter a numeric value in the boxes on the left panel to
see plots for the person’s position, velocity, and acceleration as a function of time. Click the Play
button to start a simulation and
the Pause
button to stop a simulation. You can also watch a playback by selecting the Playback
radio button instead of the
default Record
radio button. You can click Clear
to remove the current plot while maintaining your settings for position, velocity
and acceleration or click Reset All
to start over. In the Playback
mode, the grey bar can be dragged over the plot to any value in
time, and the digital readouts will show the corresponding values of the position, velocity, and acceleration.
Under the Special Features menu, the Expression Evaluator option produces a second window in which you can mathematically
type in any function for the position as a function of time, x
(
t
)
. After typing in a function, click the Play
button to start the
simulation. To zoom in vertically, click any of the three +
buttons to the top right of each plot. To zoom in horizontally, click the
+
button to the bottom right of the acceleration plot. Feel free to play around with the simulation. When you are done, click
Reset All
on the Charts
tab before beginning Part A.
Part A
First, you will focus on the relationship between velocity and position. Recall that velocity is the rate of change of position.
This means that the velocity is equal to the slope of the Position vs. Time graph.
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Move the person to the position x
= − 6 m
or enter –6.00 in the position box. If you dragged the person to position, click the
Pause
button and then the Clear
button. Next, drag the person to the right to roughly x
= 6 m
and reverse his direction,
returning him to the original position, at x
= − 6 m
. Move the person relatively quickly, about a few seconds for the round trip.
Your plots should look something like those shown below.
Look at the Position vs. Time and Velocity vs. Time plots. What is the person's velocity when his position is at its maximum
value (around 6 m
)?
ANSWER:
Correct
When the person’s position is a maximum, the slope of the position with respect to time is zero, which means the
velocity is also zero. However, due to the person’s acceleration, the velocity does not remain zero; he eventually
moves to the left.
Part B
Similarly, acceleration is the rate of change of the velocity, so it is the slope of the Velocity vs. Time graph.
Because it is difficult to drag the person in a consistent and reproducible way, use the Expression Evaluator under the
Special Features menu for this question.
Click Reset All
and type in the function x
(
t
) = 8
t
− 2
t
t
in the Expression Evaluator
. Click the Play
button and let the
simulation run roughly 5 simulation seconds before pressing the Pause
button. Use the zoom buttons to adjust the plots so
they fit in the screen. You should see a plot similar to what you got in the previous question, but much smoother.
Look at the Position vs. Time, Velocity vs. Time, and Acceleration vs. Time plots.
Hint 1. How to approach the problem
The person's velocity is zero.
positive.
negative.
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In Playback
mode, use the grey vertical bar. Slide the bar until the value x
= 8 m
is displayed in the position box on
the left panel. What are the values of velocity and acceleration when x
= 8 m
?
ANSWER:
Correct
At x
= 8 m
, the person turns to go back in the opposite direction. His velocity is zero, but his acceleration is
negative since the velocity is decreasing with time. This is similar to throwing a ball straight up into the air; at its
highest point, the velocity is zero but the acceleration is still directed downward.
Part C
Keep the function x
(
t
) = 8
t
− 2
t
t
in the Expression Evaluator. What is the value of the person’s acceleration a
x
at
t
= 2 s
?
Hint 1. How to approach the problem
Use the grey vertical bar. Slide the bar until it coincides with t
= 2 s
on the horizontal axis. What is the value displayed
in the acceleration box on the left panel?
ANSWER:
Correct
This is an example of one-dimensional motion with constant acceleration. The position of an object undergoing
this type of motion obeys the kinematic equation x
(
t
) =
x
0
+
v
x
, 0
t
+ 1/2
a
x
t
2
. In this case, the initial velocity is
v
x
, 0
= 8 m/s
and the acceleration is a
x
= − 4 m/s
2
(since 1/2
a
x
= − 2 m/s
2
).
Part D
In the previous question, the person had an initial velocity of 8 m/s
and a constant acceleration of −4 m/s
2
. How would the
maximum distance he travels to the right of the origin change if instead his initial velocity were doubled (
v
x
, 0
= 16 m/s
)?
When the person is 8 m
to the right of the origin,
both the velocity and the acceleration are zero.
the velocity is zero but the acceleration is negative.
both the velocity and the acceleration are nonzero.
the velocity is zero but the acceleration is positive.
a
x
= 4 m/s
2
−2 m/s
2
0
−4 m/s
2
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Hint 1. How to approach the problem
Go to the Introduction
tab to run the simulation using the new initial velocity v
x
, 0
= 16 m/s
and the same
acceleration of −4 m/s
2
, and read the value for position when the velocity equals zero. Remember to remove the
walls from the simulation by clicking on the red close button on the walls. In Playback
mode the simulation can be
run slowly and paused when the velocity is zero.
ANSWER:
Correct
Because it takes twice as much time to momentarily stop, and because his average velocity will be twice as fast,
the distance he travels will be four times greater. Using the kinematic equation,
x
(4 s) = (16 m/s)(4 s) − (1/2)(4 m/s
2
)(4 s)
2
= 32 m
. Alternatively, since his final velocity (i.e. the point he turns around)
is zero, you can use the other kinematic equation v
2
x
, f
=
v
2
x
, 0
+ 2
a
x
(
x
f
−
x
0
)
, or x
f
= −
v
2
x
, 0
2
a
x
to directly observe that
doubling the initial velocity results in the distance increasing by a factor of 4.
Now, assume going forward that the position is given by the equation x
(
t
) = 0.2
t
3
− 2.4
t
2
+ 7.2
t
− 5
.
Enter this function in the Expression Evaluator as x
(
t
) = 0.2
t
t
t
− 2.4
t
t
+ 7.2
t
− 5
and run the simulation by clicking
the Play
button in the Record
mode, waiting until the person collides with the wall to select the Pause
button. Consider the
person's motion from the beginning until the moment just before colliding with the brick wall.
Part E
Which of the following statements is true of the position at t
= 1.0 s
?
ANSWER:
Correct
The slope of the Position vs. Time graph is positive (but decreasing in magnitude) at this time.
Part F
Which of the following statements is true of the velocity at t
= 1.0 s
?
The maximum distance would not change.
The maximum distance would double.
The maximum distance would increase by a factor of four.
The position is decreasing at a rate increasing in magnitude.
The position is decreasing at a rate decreasing in magnitude.
The position is increasing at a rate increasing in magnitude.
The position is increasing at a rate decreasing in magnitude.
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ANSWER:
Correct
Since the slope of the Position vs. Time graph is positive, the velocity is positive (i.e. the person is moving to the
right), although since this slope is becoming smaller in magnitude, this implies the magnitude of the person's
velocity, or speed, is decreasing (i.e. the person is slowing down).
Part G
How many times are the person's position, velocity, and acceleration equal to zero?
Sort each item into the bin corresponding to the number of times it is equal to zero.
ANSWER:
Correct
Notice that since the position is given by x
= 4 t
3
, when the time is t
= 1 s
, the position is x
= 4(1)
3
m = 4 m.
Part H
At approximately what times does the person change direction?
Select all that apply.
Hint 1. How to approach the problem
The velocity is positive and the person is speeding up.
The velocity is negative and the person is speeding up.
The velocity is positive and the person is slowing down.
The velocity is negative and the person is slowing down.
Reset
Help
acceleration
velocity
position
never
1
2
3
4
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When a person moving in one dimension changes direction, the sign of their velocity changes (either from positive to
negative or negative to positive).
ANSWER:
Correct
The person changes direction when his velocity changes from positive to negative (at t
= 2.0 s
) or negative to
positive (at t
= 6.0 s
). In other words, this occurs when the person's velocity is equal to zero, which also
corresponds to when the slope of the Position vs. Time graph is zero as well.
Part I
Which of the following statements is true about the acceleration during the entire course of this motion?
ANSWER:
Correct
The graph showing Acceleration vs. Time is linear and increases at a constant rate. Since the acceleration is not
constant, the familiar kinematic equations can't be used to describe this motion, due to the fact that there is a term
in the position equation that is proportional to t
3
. However, it is still true that the slope of a Position vs. Time graph
gives velocity, and the slope of a Velocity vs. Time graph gives acceleration. In this case, you should be able to
observe that the slope of the Velocity vs. Time graph increases the entire duration of the motion (i.e. from
negative, to zero, to positive), which is captured by the Acceleration vs. Time graph. Using calculus methods, it is
possible to derive the equations for velocity and acceleration too if desired.
PhET Interactive Simulations
University of Colorado
http://phet.colorado.edu
Speed of a Bullet
1.0 s
4.0 s
6.0 s
3.2 s
2.0 s
7.8 s
The acceleration is constant (and non-zero) in time.
The acceleration is constant (and zero) in time.
The acceleration increases linearly.
The acceleration decreases linearly.
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A bullet is shot through two cardboard disks attached a distance D
D
apart to a shaft turning with a rotational period T
T
, as
shown.
Part A
Derive a formula for the bullet speed v
v
in terms of D
D
, T
T
, and a measured angle θ
theta
between the position of the hole in
the first disk and that of the hole in the second. If required, use π
pi
, not its numeric equivalent. Both of the holes lie at the
same radial distance from the shaft. θ
theta
measures the angular displacement between the two holes; for instance, θ
= 0
means that the holes are in a line and θ
=
π
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 θ
?
The disks rotate by 2 π
pi
in time T
T
. How long will it take them to rotate by θ
theta
?
Give your answer in terms of T
T
, θ
theta
, and constants such as π
pi
.
Hint 1. Checking your formula
If your formula is correct, when you plug 2 π
pi
in for θ
theta
, your answer will be T
T
.
ANSWER:
ANSWER:
T
θ
=
Tθ
2
π
v
v
=
2
πD
θT
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Correct
Motion of Two Rockets
Learning Goal:
To learn to use images of an object in motion to determine position, velocity, and acceleration.
Two toy rockets are traveling in the same direction (taken to be the x
axis). A diagram is shown of a time-exposure image where
a stroboscope has illuminated the rockets at the uniform time intervals indicated.
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 t
1
t_1
and t
2
t_2
:
v
avg
[
t
1
,
t
2
] =
x
(
t
2
) −
x
(
t
1
)
t
2
−
t
1
.
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:
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Correct
Part B
At what time(s) do the rockets have the same x
position?
ANSWER:
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:
at time t
= 1
only
at time t
= 4
only
at times t
= 1
and t
= 4
at some instant in time between t
= 1
and t
= 4
at no time shown in the figure
at time t
= 1
only
at time t
= 4
only
at times t
= 1
and t
= 4
at some instant in time between t
= 1
and t
= 4
at no time shown in the figure
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Correct
Part D
The motion of the rocket labeled A is an example of motion with uniform (i.e., constant) __________.
ANSWER:
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?
at time t
= 1
only
at time t
= 4
only
at times t
= 1
and t
= 4
at some instant in time between t
= 1
and t
= 4
at no time shown in the figure
and nonzero acceleration
velocity
position
time
and nonzero acceleration
velocity
position
time
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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
Overcoming a Head Start
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a
distance D
A
D_A
beyond the starting line at t
= 0
. The starting line is at x
= 0
. Car A travels at a constant speed v
A
v_A
. Car B
starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed v
B
v_B
, which is greater
than v
A
v_A
.
Part A
How long after Car B started the race will Car B catch up with Car A?
Express the time in terms of given quantities.
Hint 1. Consider the kinematics relation
Write an expression for the displacement of Car A from the starting line at a time t
t
after Car B starts. (Note that we
are taking this time to be t
= 0
.)
Answer in terms of v
A
v_A
, v
B
v_B
, D
A
D_A
, and t
t
for time, and take x
= 0
at the starting line.
Hint 1. What is the acceleration of Car A?
The acceleration of Car A is zero, so the general formula x
(
t
) =
x
0
+
v
0
t
+ (1/2)
at
2
has at least one term
equal to zero.
ANSWER:
before t
= 1
only
after t
= 4
only
before t
= 1
and after t
= 4
between t
= 1
and t
= 4
at no time(s) shown in the figure
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Hint 2. What is the relation between the positions of the two cars?
The positions of the two cars are equal at time t
catch
t_catch
.
Hint 3. Consider Car B's position as a function of time
Write down an expression for the position of Car B at time t
t
after starting.
Give your answer in terms of any variables needed (use t
t
for time).
ANSWER:
ANSWER:
Correct
Part B
How far from Car B's starting line will the cars be when Car B passes Car A?
Express your answer in terms of known quantities. (You may use t
catch
t_catch
as well.)
Hint 1. Which expression should you use?
Just use your expression for the position of either car after time t
= 0
, and substitute in the correct value for
t
catch
t_catch
(found in the previous part).
ANSWER:
Correct
± Average Velocity from a Position vs. Time Graph
Learning Goal:
x
A
(
t
)
=
D
A
+
v
A
t
x
B
(
t
)
=
v
B
t
t
catch
t_catch
=
D
A
v
B
−
v
A
d
pass
d_pass
=
v
B
t
catch
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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 x
(
t
)
as a function of time t
t
. 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 v
ave
[
t
1
,
t
2
]
to
indicate average velocity over the time interval from t
1
t_1
to t
2
t_2
. For
instance, v
ave
[1, 3]
is the average velocity over the time interval from
t
= 1
to t
= 3
.
Part A
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 t
i
t_i
to the final time t
f
t_f
. The average velocity is therefore equal to the displacement divided by the
given time interval. In symbolic form, average velocity is given by
v
ave
[
t
i
,
t
f
] =
x
(
t
f
) −
x
(
t
i
)
t
f
−
t
i
.
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:
v
ave
[0, 1]
=
0
m/s
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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 t
i
t_i
to the final time t
f
t_f
. The average velocity is therefore equal to the displacement divided by the
given time interval. In symbolic form, average velocity is given by
v
ave
[
t
i
,
t
f
] =
x
(
t
f
) −
x
(
t
i
)
t
f
−
t
i
.
ANSWER:
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 straight-line 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 20 m/s
.
Part C
Now find v
ave
[0, 3]
.
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 v
ave
[0, 3]
is not equal to the simple arithmetic average of v
ave
[0, 1]
and v
ave
[1, 3]
, i.e., v
ave
[ 0 , 1 ] +
v
ave
[ 1 , 3 ]
2
,
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.
x
f
−
x
i
= 40
m
v
ave
[1, 3]
=
20
m/s
v
ave
[0, 3]
=
13.3
m/s
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Hint 1. Determine the displacement
What is the displacement?
Answer to the nearest integer.
ANSWER:
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:
x
(6.0) −
x
(3.0)
= -40
m
t
f
−
t
i
=
3.0
s
v
ave
[3.0, 6.0]
=
-13.3
m/s
x
(6.0) −
x
(0.0)
= 0
m
v
ave
[0.0, 6.0]
=
0
m/s
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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!
Analyzing Position versus Time Graphs: Conceptual Question
Two cars travel on the parallel lanes of a two-lane 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:
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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
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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
A
B
C
D
E
none
cannot be determined
A
B
C
D
E
none
cannot be determined
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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
Given Positions, Find Velocity and Acceleration
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 1-s 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 in best represents the function x
(
t
)x(t)
, describing the object's position vs. time?
A
B
C
D
E
None
Cannot be determined
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Hint 1. Meaning of a bounded and nonspiky acceleration
A bounded and nonspiky acceleration results in a smooth graph of x
x
vs. t
t
.
ANSWER:
Correct
Part B
Which of the following graphs in best represents the function
v
(
t
)v(t)
, describing the object's velocity as a function of time?
Hint 1. Find the velocity toward the end of the motion
Graph 1
Graph 2
Graph 3
Graph 4
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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 t
= 4
and t
= 7 s
. 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:
ANSWER:
positive and increasing
positive and decreasing
negative and increasing
negative and decreasing
a horizontal line
straight but sloping up to the right
straight but sloping down to the right
curved upward
curved downward
has spikes
has no discontinuities
has no abrupt changes of slope
is constant
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Correct
In principle, you could also just compute and plot the average velocity. The expression for the average velocity is
v
avg
[
t
1
,
t
2
] =
x
(
t
2
) −
x
(
t
1
)
t
2
−
t
1
.
The notation v
avg
[
t
1
,
t
2
]
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 t
= 0
up to time t
t
must equal the position of
the object at time t
t
. Check that the correct velocity vs. time graph gives you the correct position according to this
method.
Part C
Which of the following graphs in best represents the function
a
(
t
)a(t)
, describing the acceleration of this object?
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:
Graph 1
Graph 2
Graph 3
Graph 4
zero
positive
negative
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Hint 2. Calculate the acceleration in the region of constant velocity
What is the acceleration a
a
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:
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:
a
avg
[
t
1
,
t
2
] =
v
(
t
2
) −
v
(
t
1
)
t
2
−
t
1
.
When the acceleration is constant over an extended interval, you can choose any value of t
1
t_1
and t
2
t_2
within
the interval to compute the average.
Kinematic Vocabulary
One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings,
among them velocity
, acceleration
, position
, speed
, and displacement
. The series of questions in this problem is designed to get
you to try to think of these quantities like a physicist.
a
a
= 0
m/s
2
zero
positive
negative
Graph 1
Graph 2
Graph 3
Graph 4
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Answer the questions in this problem using words from the following list:
A. position
B. direction
C. displacement
D. coordinates
E. velocity
F. acceleration
G. distance
H. magnitude
I. vector
J. scalar
K. components
Part A
Velocity
differs from speed
in that velocity
indicates a particle's __________ of motion.
Enter the letter from the list given in the problem introduction that best completes the sentence.
ANSWER:
Correct
Part B
Unlike speed
, velocity
is a __________ quantity.
Enter the letter from the list given in the problem introduction that best completes the sentence.
ANSWER:
Correct
Part C
A vector, by definition, has both __________ and direction.
Enter the letter from the list given in the problem introduction that best completes the sentence.
ANSWER:
Correct
B
I
H
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Part D
Once you have selected a coordinate system, you can express a two-dimensional vector using a pair of quantities known
collectively as __________.
Enter the letter from the list given in the problem introduction that best completes the sentence.
ANSWER:
Correct
Part E
Speed
differs from velocity
in the same way that __________ differs from displacement
.
Enter the letter from the list given in the problem introduction that best completes the sentence.
Hint 1. Definition of displacement
Displacement is the vector that indicates the difference of two positions (e.g., the final position from the initial
position). Being a vector, it is independent of the coordinate system used to describe it (although its vector
components depend on the coordinate system).
ANSWER:
Correct
Part F
Consider a physical situation in which a particle moves from point A to point B. This process is described from two
coordinate systems that are identical except that they have different origins.
The __________ of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the
__________ from A to B is/are the same as expressed in both coordinate systems.
Type the letters from the list given in the problem introduction that best complete the sentence. Separate the letters
with commas. There is more than one correct answer, but you should only enter one pair of comma-separated
letters. For example, if the words "vector" and "scalar" fit best in the blanks, enter I,J
.
ANSWER:
K
G
A,C
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The coordinates of a point will depend on the coordinate system that is chosen, but there are several other
quantities that are independent of the choice of origin for a coordinate system: in particular, distance,
displacement, direction, and velocity. In working physics problems, unless you are interested in the position of an
object or event relative to a specific origin, you can usually choose the coordinate system origin to be wherever is
most convenient or intuitive.
Note that the vector indicating a displacement from A to B is usually represented as →
r
BA
=
→
r
B
−
→
r
A
.
Part G
Identify the following physical quantities as scalars or vectors.
ANSWER:
Correct
Exercise 2.16 - Enhanced - with Feedback
An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity
changes, each taking place in a 11 s
interval. What is the average acceleration in each interval? Assume that the positive
direction is to the right.
Part A
At the beginning of the interval, the astronaut is moving toward the right along the x
-axis at 14.3 m/s
, and at the end of the
interval she is moving toward the right at 4.9 m/s
.
Express your answer in meters per second squared.
Reset
Help
distance
speed
position
velocity
average velocity
acceleration
displacement
Scalar quantity
Vector quantity
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ANSWER:
Correct
Part B
At the beginning she is moving toward the left at 4.9 m/s
, and at the end she is moving toward the left at 14.3 m/s
.
Express your answer in meters per second squared.
ANSWER:
Correct
Part C
At the beginning she is moving toward the right at 14.3 m/s
, and at the end she is moving toward the left at 14.3 m/s
.
Express your answer in meters per second squared.
ANSWER:
Correct
PSS 2.1 Motion with Constant Acceleration
Learning Goal:
To practice Problem-Solving Strategy 2.1 Motion with Constant Acceleration.
Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour
in 2.22 s
starting from rest. Assuming that they have constant
acceleration throughout that time, find their acceleration in meters per second squared.
Problem-Solving Strategy: Motion with constant acceleration
IDENTIFY the relevant concepts
:
In most straight-line motion problems, you can use the constant-acceleration equations. Occasionally, however, you will
encounter a situation in which the acceleration isn’t constant. In such a case, you’ll need a different approach.
SET UP the problem using the following steps
:
1. First, decide where the origin of coordinates is and which axis direction is positive. It is often easiest to place the
particle at the origin at time t
= 0
; then x
0
= 0
. It helps to make a motion diagram showing the coordinates and
a
= -0.85
m/s
2
a
= -0.85
m/s
2
a
= -2.6
m/s
2
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some later positions of the particle.
2. Keep in mind that your choice of the positive axis direction automatically determines the positive direction for x
velocity and x
acceleration. If x
x
is positive to the right of the origin, then v
x
v_x
and a
x
a_x
are also positive toward
the right.
3. Restate the problem in words, and then translate it into symbols and equations. 4. Make a list of known and unknown quantities such as x
x
, x
0
x_0
, v
x
v_x
, v
0
x
v_0x
, a
x
a_x
, and t
t
. Write down the values
of the known quantities, and decide which of the unknowns are the target variables. Look for implicit information.
EXECUTE the solution as follows
:
Choose an equation from the following list
v
x
=
v
0
x
+
a
x
t
x
=
x
0
+
v
0
x
t
+
1
2
a
x
t
2
v
2
x
=
v
2
0
x
+ 2
a
x
(
x
−
x
0
)
x
−
x
0
=
v
0
x
+
v
x
2
t
that contains only one of the target variables. Solve this equation for the target variable, using symbols only. Then, substitute the
known values and compute the value of the target variable. Sometimes you will have to solve two simultaneous equations for
two unknown quantities.
EVALUATE your answer
:
Take a hard look at your results to see whether they make sense. Are they within the general range of values you expected?
IDENTIFY the relevant concepts
This problem involves the motion of an object, the cheetah, whose acceleration is assumed constant. Thus, the equations given
in this strategy apply.
SET UP the problem using the following steps
Part A
Which of the following sketches and choice of coordinate axis best describe the physical situation presented in this
problem?
ANSWER:
(
)
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Part B
The next step is to translate the problem statement from words into symbols. Which of the following is an appropriate
restatement of the problem, "Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour
in 2.22 s
starting from rest.
Assuming that they have constant acceleration throughout that time, find their acceleration in meters per second squared."
Hint 1. Find the initial velocity using implicit information
The problem states that the cheetah starts running from rest. What is the initial velocity v
0
x
v_0x
of the cheetah?
Enter your answer in meters per second.
ANSWER:
Hint 2. The condition for the equations of motion presented in this problem
The equations presented in the strategy above only apply to situations involving motion under constant acceleration.
ANSWER:
A
B
C
D
v
0
x
v_0x
=
0
m/s
Cheetahs can reach v
0
x
=
50.0 miles/hour
in t
t
= 2.22 s
starting from v
x
= 0
. What is a
x
a_x
?
Cheetahs can reach v
x
v_x
= 50.0 miles/hour
in t
t
= 2.22 s
starting from v
0
x
= 0
. What is a
x
a_x
?
Cheetahs can reach v
x
v_x
= 50.0 miles/hour
in t
t
= 2.22 s
starting from v
0
x
= 0
. Assuming a
= constant
, what is
a
x
a_x
?
Cheetahs can reach v
0
x
=
50.0 miles/hour
in t
t
= 2.22 s
starting from v
x
= 0
. Assuming a
= constant
, what is a
x
a_x
?
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Now you compile a list of known and unknown quantities. You can organize this information in a table as shown
below.
Known
Unknown
x
0
= 0 m
x
x
v
0
x
= 0 m/s
a
x
a_x
v
x
v_x
= 50.0 miles/hour
_
t
t
= 2.22 s
_
Keep in mind that your target variable is a
x
a_x
.
EXECUTE the solution as follows
Part C
Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour
in
2.22 s
starting from rest. Assuming that they have constant acceleration throughout that time, find their acceleration in
meters per second squared.
Enter your answer in meters per second squared to three significant figures.
Hint 1. Identify what equation to use
Which of the following equations would be the best
to use when solving for a
x
a_x
?
ANSWER:
Hint 2. Convert to SI units
How many meters per second are equivalent to 50.0 miles/hour
?
Enter your answer in meters per second to three significant figures.
Hint 1. The conversion factor from miles to meters
To convert miles to meters, use
1 mile = 1609 m
.
x
=
x
0
+
v
0
x
t
+
1
2
a
x
t
2
v
2
x
=
v
2
0
x
+ 2
a
x
(
x
−
x
0
)
v
x
=
v
0
x
+
a
x
t
x
−
x
0
=
v
0
x
+
v
x
2
t
(
)
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ANSWER:
ANSWER:
Correct
EVALUATE your answer
Part D
Imagine you looked up the accelerations of the following objects: snails, humans, Thomson's gazelles, the space shuttle,
Formula One race cars, and F-16 fighter jets. Which of the following statements about the acceleration of a cheetah would
you expect to be true?
ANSWER:
Correct
The acceleration of the space shuttle on takeoff is 29.4 m/s
2
. Thomson's gazelles can accelerate at approximately
half the rate of a cheetah, which is why they often become tasty snacks for the fast cats.
If you had solved for the acceleration of a cheetah and calculated a number greater than 29.4 m/s
2
or smaller than
30 cm/s
2
(the acceleration of a snail), you most likely made an error and would want to review your work.
Exercise 2.30 - Enhanced - with Feedback
A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the
incline, and its speed after it has traveled 7.00 m
to the bottom of the incline is 3.80 m/s
.
Part A
What is the speed of the block when it is 5.00 m
from the top of the incline?
Express your answer with the appropriate units.
50.0 miles/hour
= 22.3
m/s
a
x
a_x
= 10.1
m/s
2
The acceleration of a cheetah is greater than the acceleration of a Thomson's gazelle but less than the
acceleration of the space shuttle during liftoff.
The acceleration of a cheetah is greater than the acceleration of a Formula One race car but less than the
acceleration of an F-16 fighter jet.
The acceleration of a cheetah is greater than the acceleration of a snail but less than the acceleration of a
human.
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ANSWER:
Correct
Exercise 2.47 - Enhanced - with Feedback
Part A
You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way
down, 7.00 s
after it was thrown. What is the speed of the rock just before it reaches the water 21.0 m
below the point where
the rock left your hand? Ignore air resistance.
Express your answer with the appropriate units.
ANSWER:
Correct
Exercise 2.50 - Enhanced - with Feedback
A small object moves along the x
-axis with acceleration a
x
(
t
)
= −(0.0320 m/s
3
)(15.0 s −
t
)
. At t
= 0 the object is at x
= -14.0 m
and
has velocity v
0
x
v_0x
= 9.00 m/s
.
Part A
What is the x
-coordinate of the object when t
= 10.0 s
?
Express your answer with the appropriate units.
ANSWER:
Correct
v
=
3.21 m
s
v
=
39.9 m
s
x
= 57.3 m
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Problem 2.65 - Enhanced - with Feedback
A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a
constant acceleration of 2.10 m/s
2
and the car an acceleration of 3.40 m/s
2
. The automobile overtakes the truck after the truck
has moved 60.0 m
.
Part A
How much time does it take the car to overtake the truck?
Express your answer with the appropriate units.
ANSWER:
Correct
Part B
How far was the car behind the truck initially?
Express your answer with the appropriate units.
ANSWER:
Correct
Part C
What is the speed of the truck when they are abreast?
Express your answer with the appropriate units.
ANSWER:
Correct
Part D
What is the speed of the car when they are abreast?
Express your answer with the appropriate units.
t
= 7.56 s
d
= 37.1 m
v
truck
=
15.9 m
s
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ANSWER:
Correct
Part E
Select the correct x
−
t
graph of each vehicle. Take x
= 0
at the initial location of the truck.
ANSWER:
v
car
=
25.7 m
s
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Problem 2.78 - Enhanced - with Feedback
During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be
launched from rest from the earth's surface and is to reach a maximum height of 930 m
above the earth's surface. The rocket's
engines give the rocket an upward acceleration so it moves with acceleration of 16.0 m/s
2
during the time T
that they fire. After
the engines shut off, the rocket is in free fall. Ignore air resistance. Assume that the acceleration due to gravity does not change
with the height of the rocket.
Part A
What must be the value of T
in order for the rocket to reach the required altitude?
Express your answer with the appropriate units.
ANSWER:
Correct
MCAT Passage Problem 2.90
The human circulatory system is closed - that is, the blood pumped out of the left ventricle of the heart into the arteries is
constrained to a series of continuous, branching vessels as it passes through the capillaries and then into the veins as it returns
to the heart. The blood in each of the heart
s four chambers comes briefly to rest before it is ejected by contraction of the heart
muscle.
The contraction of the left ventricle lasts 250 ms
and the speed of blood flow in the aorta (the large artery leaving the heart) is
0.80 m/s
at the end of the contraction.
Part A
What is the average acceleration of a red blood cell as it leaves the heart?
ANSWER:
T
= 6.6 s
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Problem 2.61 - Enhanced - with Feedback
A gazelle is running in a straight line (the x
-axis). The graph in shows
this animal's velocity as a function of time.
Part A
During the first 12.0 s
, find the total distance moved.
Express your answer in meters.
ANSWER:
Correct
Part B
During the first 12.0 s
, find the displacement of the gazelle.
Express your answer in meters.
310 m/s
2
31 m/s
2
3.2 m/s
2
0.32 m/s
2
x
path
=
92.0
m
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ANSWER:
Correct
Exercise 2.44 - Enhanced - with Feedback
An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way
down and passes a point a distance 50.0 m
below its starting point at a time 5.00 s
after it leaves the thrower's hand. Air
resistance may be ignored.
Part A
What is the initial speed of the egg?
Express your answer in meters per second.
ANSWER:
Correct
Part B
How high does it rise above its starting point?
Express your answer in meters.
ANSWER:
Correct
Part C
What is the magnitude of its velocity at the highest point?
Express your answer in meters per second.
ANSWER:
x
displacement
=
92.0
m
v
in
=
14.5
m/s
h
= 10.7
m
v
h . p .
=
0
m/s
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Part D
What is the magnitude of its acceleration at the highest point?
Express your answer in meters per second squared.
ANSWER:
Correct
Part E
What is the direction of its acceleration at the highest point?
ANSWER:
Correct
Exercise 2.14 - Enhanced - with Feedback
A race car starts from rest and travels east along a straight and level track. For the first 5.0 s
of the car's motion, the eastward
component of the car's velocity is given by v
x
(
t
) = (0.970 m/s
3
)
t
2
.
Part A
What is the acceleration of the car when v
x
v
= 14.3 m/s
?
Express your answer with the appropriate units.
ANSWER:
Correct
a
= 9.80
m/s
2
up
down
a
x
=
7.45 m
s
2
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Exercise 2.33 - Enhanced - with Solution
A juggler throws a bowling pin straight up with an initial speed of 9.70 m/s
.
For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Up-and-down motion in free fall
.
Part A
How much time elapses until the bowling pin returns to the juggler's hand?
Express your answer with the appropriate units.
ANSWER:
Correct
IDENTIFY: The pin has a constant downward acceleration of 9.80 m/s
2
and returns to its initial position.
SET UP: We can use the kinematics formulas for constant acceleration.
EXECUTE: The kinematics formulas give y
−
y
0
=
v
0
y
t
+
1
2
a
y
t
2
. We know that y
−
y
0
= 0
, so
t
= −
2
v
0
y
a
y
= −
2 ( 9.70 m / s )
− 9.80 m / s
2
= + 1.98 s
EVALUATE: It takes the pin half this time to reach its highest point and the remainder of the time to return.
Exercise 2.49
A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s
of its motion, the vertical
acceleration of the rocket is given by a
y
=
2.90 m/s
3
t
, where the +
y
-direction is upward.
Part A
What is the height of the rocket above the surface of the earth at t
t
= 10.0 s
?
Express your answer with the appropriate units.
ANSWER:
Correct
t
= 1.98 s
(
)
h
= 483 m
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Part B
What is the speed of the rocket when it is 295 m
above the surface of the earth?
Express your answer with the appropriate units.
ANSWER:
Correct
Key Example Variation Problem 2.7
Be sure to review Example 2.7 (Section 2.5) before attempting these problems.
VP 2.7.1
Part A
You throw a ball straight up from the edge of a cliff. It leaves your hand moving at 12.0 m/s
. Air resistance can be neglected.
Take the positive y
-direction to be upward, and choose y
= 0 to be the point where the ball leaves your hand. Find the ball's
position 0.300 s
after it leaves your hand.
Express your answer with the appropriate units.
ANSWER:
Correct
Part B
State whether the ball is above or below your hand 0.300 s
after it leaves your hand.
ANSWER:
Correct
Part C
υ
y
=
104 m
s
y
1
= 3.16 m
The ball is above your hand.
The ball is below your hand.
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Find the y
-component of the ball's velocity 0.300 s
after it leaves your hand.
Express your answer with the appropriate units.
ANSWER:
Correct
Part D
State whether the ball is moving upward or downward 0.300 s
after it leaves your hand.
ANSWER:
Correct
Part E
Find the ball's position 2.60 s
after it leaves your hand.
Express your answer with the appropriate units.
ANSWER:
Correct
Part F
State whether the ball is above or below your hand 2.60 s
after it leaves your hand.
ANSWER:
Correct
Part G
v
1
y
=
9.1 m
s
The ball is moving downward.
The ball is moving upward.
y
2
=
-1.9 m
The ball is above your hand.
The ball is below your hand.
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Find the y
-component of the ball's velocity 2.60 s
after it leaves your hand.
Express your answer with the appropriate units.
ANSWER:
Correct
Part H
State whether the ball is moving upward or downward 2.60 s
after it leaves your hand.
ANSWER:
Correct
VP 2.7.2
Part I
You throw a stone straight down from the top of a tall tower. It leaves your hand moving at 8.00 m/s
. Air resistance can be
neglected. Take the positive y
-direction to be upward, and choose y
= 0 to be the point where the stone leaves your hand.
Find the stone's position 1.50 s
after it leaves your hand.
Express your answer with the appropriate units.
ANSWER:
Correct
Part J
Find the y
-component of the stone's velocity 1.50 s
after it leaves your hand.
Express your answer with t0he appropriate units.
ANSWER:
v
2
y
=
-13.5 m
s
The ball is moving downward.
The ball is moving upward.
y
1
= -23.0 m
v
1
y
=
-22.7 m
s
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Part K
Find the y
-component of the stone's velocity when it is 8.00 m
below your hand.
Express your answer with the appropriate units.
ANSWER:
Correct
VP 2.7.3
Part L
You throw a football straight up. Air resistance can be neglected. When the football is 4.00 m
above where it left your hand,
it is moving upward at 0.500 m/s
. What was the speed of the football when it left your hand?
Express your answer with the appropriate units.
ANSWER:
Correct
Part M
How much time elapses from when the football leaves your hand until it is 4.00 m
above your hand?
Express your answer with the appropriate units.
ANSWER:
Correct
VP 2.7.4
Part N
You throw a tennis ball straight up. Air resistance can be neglected. The maximum height above your hand that the ball
reaches is H
. At what speed does the ball leave your hand?
Express your answer in terms of the acceleration due to gravity g
g
and the variable H
H
.
v
2
y
=
-14.9 m
s
v
0
=
8.87 m
s
t
= 0.854 s
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ANSWER:
Correct
Part O
What is the speed of the ball when it is a height H
/2
above your hand?
Express your answer numerically as a fraction of the speed v
0
v_0
at which it left your hand.
ANSWER:
Correct
Part P
At what height above your hand is the speed of the ball half as great as when it left your hand?
Express your answer in terms of H
H
.
ANSWER:
Correct
Key Example Variation Problem 2.5
Be sure to review Example 2.5 (Section 2.4) before attempting these problems.
VP 2.5.1
Part A
A sports car starts from rest at an intersection and accelerates toward the east on a straight road at 8.0 m/s
2
. Just as the
sports car starts to move, a bus traveling east at a constant 18 m/s
on the same straight road passes the sports car. When
the sports car catches up with and passes the bus, how much time has elapsed?
Express your answer with the appropriate units.
ANSWER:
v
0
=
√
2
gH
v
H
/ 2
=
0.71
v
0
h
=
3
H
4
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Part B
When the sports car catches up with and passes the bus, how far has the sports car traveled?
Express your answer with the appropriate units.
ANSWER:
Correct
VP 2.5.2
Part C
A car is traveling on a straight road at a constant 35 m/s
, which is faster than the speed limit. Just as the car passes a police
motorcycle that is stopped at the side of the road, the motorcycle accelerates forward in pursuit. The motorcycle passes the
car 14.5 s
after starting from rest. What is the acceleration of the motorcycle (assumed to be constant)?
Express your answer with the appropriate units.
ANSWER:
Correct
Part D
How far does the motorcycle travel before it passes the car?
Express your answer with the appropriate units.
ANSWER:
Correct
VP 2.5.3
Part E
t
= 4.5 s
d
= 81 m
a
= 4.8 m
s
2
d
= 510 m
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A police car is traveling north on a straight road at a constant 20.0 m/s
. An SUV traveling north at 30.0 m/s
passes the
police car. The driver of the SUV suspects he may be exceeding the speed limit, so just as he passes the police car he lets
the SUV slow down at a constant 1.60 m/s
2
. How much time elapses from when the SUV passes the police car to when the
police car passes the SUV?
Express your answer with the appropriate units.
ANSWER:
Correct
Part F
What distance does the SUV travel during this time?
Express your answer with the appropriate units.
ANSWER:
Correct
Part G
What is the speed of the SUV when the police car passes it?
Express your answer with the appropriate units.
ANSWER:
Correct
VP 2.5.4
Part H
At t
= 0 a truck starts from rest at x
= 0 and speeds up in the positive x
-direction on a straight road with acceleration a
T
. At
the same time, t
= 0, a car is at x
= 0 and traveling in the positive x
-direction with speed v
C
. The car has a constant negative
x
-acceleration: a
car −
x
= −
a
C
, where a
C
is a positive quantity. At what time does the truck pass the car?
Express your answer in terms of the variables v
C
v_C
, a
C
a_C
, and a
T
a_T
.
ANSWER:
t
= 12.5 s
d
= 250 m
v
=
10.0 m
s
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Part I
At what coordinate does the truck pass the car?
Express your answer in terms of the variables v
C
v_C
, a
C
a_C
, and a
T
a_T
.
ANSWER:
Correct
Key Example Variation Problem 2.8
Be sure to review Example 2.8 (Section 2.5) before attempting these problems.
VP 2.8.1
Part A
You throw a rock straight up from the edge of a cliff. It leaves your hand at time t
= 0 moving at 15.0 m/s
. Air resistance can
be neglected. Find both times at which the rock is 4.00 m
above where it left your hand.
Enter your answers in ascending order separated by a comma. Express your answer in seconds.
ANSWER:
Correct
Part B
Find the time when the rock is 4.00 m
below where it left your hand.
Express your answer with the appropriate units.
ANSWER:
t
=
2
v
C
a
T
+
a
C
x
=
2
v
C
2
a
T
a
T
+
a
C
2
(
)
t
= 0.30,2.77
s
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VP 2.8.2
Part C
You throw a basketball straight down from the roof of a building. The basketball leaves your hand at time t
= 0 moving at
7.00 m/s
. Air resistance can be neglected. Find the time when the ball is 5.00 m
below where it left your hand.
Express your answer with the appropriate units.
ANSWER:
Correct
VP 2.8.3
Part D
You throw an apple straight up. The apple leaves your hand at time t
= 0 moving at 5.50 m/s
. Air resistance can be
neglected. How many times (two, one, or none) does the apple pass through a point 1.30 m
above your hand? If the apple
does pass through this point, is the apple moving upward or downward at each of these times?
ANSWER:
Correct
Part E
At what times t
1
and t
2
does the apple pass through this point?
Enter your answers in ascending order separated by a comma. Express your answers in seconds.
t
= 3.31 s
t
= 0.523 s
The apple passes through this point twice, the first time it moves downward and the second time it moves
upward.
The apple passes through this point once and moves downward.
The apple does not pass through this point.
The apple passes through this point twice, the first time it moves upward and the second time it moves
downward.
The apple passes through this point once, and this is the stopping point of the apple's movement.
The apple passes through this point once and moves upward.
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ANSWER:
Correct
Part F
How many times (two, one, or none) does the apple pass through a point 1.80 m
above your hand? If the apple does pass
through this point, and is the apple moving upward or downward at each of these times?
ANSWER:
Correct
VP 2.8.4
Part G
You throw an orange straight up. The orange leaves your hand at time t
= 0 moving at speed v
0
. Air resistance can be
neglected. At what time(s) is the orange at a height v
2
0
/2
g
above the point where it left your hand? At these time(s) is the
orange moving upward, downward, or neither?
ANSWER:
t
1
, t
2
=
0.338,0.784
s
The apple passes through this point twice, the first time it moves upward and the second time it moves
downward.
The apple passes through this point twice, the first time it moves downward and the second time it moves
upward.
The apple passes through this point once and moves upward.
The apple does not pass through this point.
The apple passes through this point once, and this is the stopping point of the apple's movement.
The apple passes through this point once and moves downward.
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Part H
At what time t
does the orange reach this point?
Express your answer in terms of acceleration due to gravity g
g
and the variable v
0
v_0
.
ANSWER:
Correct
Part I
At what time(s) is the orange at a height 3
v
2
0
/8
g
above the point where it left your hand? At these time(s) is the orange
moving upward, downward, or neither?
ANSWER:
The orange does not pass through this point.
The orange passes through this point twice. The first time it moves upward and the second times it moves
downward.
The orange reaches this point once, and this is the stopping point of the orange's movement.
The orange passes through this point once and moves upward.
The orange passes through this point twice. The first time it moves downward and the second times it moves
upward.
The orange passes through this point once and moves downward.
t
=
v
0
g
The orange reaches this height once, and this is the stopping point of the orange's movement.
The orange does not pass through this point.
The orange passes through this point twice. The first time it moves downward and the second times it moves
upward.
The orange passes through this point once and moves upward.
The orange passes through this point once and moves downward.
The orange passes through this point twice. The first time it moves upward and the second times it moves
downward.
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Part J
At what times t
1
and t
2
does the orange pass through this point?
Enter your answers in ascending order separated by a comma. Express your answers in terms of acceleration due
to gravity g
g
and the variable v
0
v_0
.
ANSWER:
Correct
Problem 2.84
In your physics lab you release a small glider from rest at various points on a long, frictionless air track that is inclined at an
angle θ
above the horizontal. With an electronic photocell, you measure the time t
it takes the glider to slide a distance x
from the
release point to the bottom of the track. Your measurements are given in , which shows a second-order polynomial (quadratic) fit
to the plotted data. You are asked to find the glider
s acceleration,
which is assumed to be constant. There is some error in each
measurement, so instead of using a single set of x
and t
values, you
can be more accurate if you use graphical methods and obtain your
measured value of the acceleration from the graph.
Part A
How can you re-graph the data so that the data points fall close to a straight line? (
Hint:
You might want to plot x
or t
, or
both, raised to some power.)
ANSWER:
t
1
, t
2
=
v
0
2
g
,
3
v
0
2
g
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Part B
Select the correct graph with the equation for the straight line that is the best fit to the data points described in the previous
part.
ANSWER:
A plot of x
versus √
t
should be a straight line.
A plot of x
versus t
should be a straight line.
A plot of x
versus t
2
should be a straight line.
A plot of x
2
versus t
2
should be a straight line.
A plot of x
2
versus t
should be a straight line.
A plot of x
2
versus √
t
should be a straight line.
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Part C
Use the straight-line fit from the previous part to calculate the acceleration of the glider.
Express your answer to two significant figures and include the appropriate units.
ANSWER:
Correct
Part D
The glider is released at a distance 1.59 m
from the bottom of the track. Use the acceleration value you obtained in part C to
calculate the speed of the glider when it reaches the bottom of the track.
Express your answer to two significant figures and include the appropriate units.
ANSWER:
Correct
Bridging Problem: The Fall of a Superhero
The superhero Green Lantern steps from the top of a tall building. He falls freely from rest to the ground, falling half the total
distance to the ground during the last t
2
= 1.00 s
of his fall . What is
the height h
of the building?
a
= 7.6 m
s
2
v
=
4.9 m
s
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IDENTIFY
and SET UP
Part A
You are told that Green Lantern falls freely from rest. What does this imply about his acceleration?
Hint 1. Free fall
Free fall is the motion of an object under the sole influence of gravity. Therefore it moves with the acceleration due
to gravity
.
ANSWER:
Correct
The value of g
is also called the free-fall acceleration.
Part B
You are told that Green Lantern falls freely from rest. What does this imply about his initial velocity?
Hint 1. Initial velocity
What velocity does the object have when in rest?
ANSWER:
Correct
Part C
Take the direction of the positive y
-axis upward and its origin at the starting point.
The magnitude of acceleration is constant and is equal to g
.
The acceleration increases linearly with time.
The magnitude of acceleration is constant and is equal to zero.
The acceleration increases quadratically with time.
The initial velocity is equal to g
.
The initial velocity has a small non-zero value, so he can start moving.
The initial velocity depends on the height of the building.
The initial velocity is equal to zero.
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3/7/23, 11:31 PM
Ch 02 HW
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460266
57/58
You can divide Green Lantern’s fall into two parts: from the top of the building to the halfway point and from the halfway
point to the ground. You know that the second part of the fall lasts 1.00 s
. Decide what you would need to know about Green
Lantern’s motion at the halfway point in order to solve for the target variable h
. Then choose two equations, one for the first
part of the fall and one for the second part, that you’ll use together to find an expression for h
.
Select a set of two equations that apply.
Hint 1. Target quantities at the halfway point
At the point y
= −
h
/2
, we know the acceleration is a
y
= −
g
. We don't know the velocity v
at the halfway point or how
long it took to travel from the origin to the halfway point.
Hint 2. Number of equations and number of target variables
You have to find the value for h
. If you also introduce a new target variable in the system, you will need two equations
to solve for h
and the new target variable.
ANSWER:
Correct
You have identified a possible set of equations that involve the target variable h
.
EXECUTE
Part D
Use your two equations to solve for the height h
. Heights are always positive numbers, so your answer should be positive.
Express your answer with the appropriate units.
ANSWER:
Correct
Here we learned how to find the distance covered by a free falling object solving system of equations.
EVALUATE
v
1
=
gt
1
, where v
1
is the speed at the midpoint and t
1
is the time to fly the first half
v
1
=
√
gh
, where v
1
is the speed at the midpoint
h
=
g
(
t
1
+
t
2
)
2
2
, where t
1
is the time to fly the first half
v
1
+
gt
2
=
√
2
gh
, where v
1
is the speed at the midpoint
h
2
=
gt
2
1
2
, where t
1
is the time to fly the first half
h
= 57.1 m
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3/7/23, 11:31 PM
Ch 02 HW
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460266
58/58
Part E
To check your answer for h
, use one of the free-fall equations to find how long it takes Green Lantern to fall from the top of
the building to half the height.
Express your answer with the appropriate units.
Hint 1. Initial conditions
You are now considering a motion with constant acceleration, zero initial velocity, and a known distance covered,
equal to h
2
.
ANSWER:
Correct
Part F
To check your answer for h
, use one of the free-fall equations to find how long it takes Green Lantern to fall from the top of
the building to the ground.
Express your answer with the appropriate units.
Hint 1. Relation to Part E
You can use the same formula you used for Part E, but now the distance is equal to h
.
ANSWER:
Correct
Time t
total
turns out to be 1.00 s
greater than time t
1
, which means that your answer for h
is correct.
For the steps and strategies involved in solving this problem, you may view the Video Tutor Solution
.
Score Summary:
Your score on this assignment is 94.1%.
You received 24.47 out of a possible total of 26 points.
t
1
=
2.41 s
t
total
=
3.41 s
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