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Phy 231
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The problem:
A graph is constructed representing velocity vs. clock time for the interval between clock times t = 5 seconds and t = 13 seconds. The graph consists of a straight line from the point (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s).
• What is the clock time at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The midpoint of the clock time for the interval is found as follows:
Time midpoint = (13 s – 5 s) / 2 = 8 s / 2 = 4 s
The interval runs from clock time 5 s to clock time 13 s. Clock time 4 s isn't even in the interval, so it's clearly not the midpoint clock time.
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The midpoint of the velocity for the interval is found as follows:
Velocity midpoint = (40 cm/s – 16 cm/s) / 2 = 24 cm/s / 2 = 12 cm/s
If velocity changes from 16 cm/s to 40 cm/s, the 12 cm/s velocity isn't even in the interval, much less at the midpoint.
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
ds = (vAve)*dt
ds = change in position; vAve = average velocity; dt = change in time
vAve = (v0 + vf) / 2 = (16 cm/s + 40 cm/s) / 2 = 28 cm/s
dt = final time – initial time = 13 s – 5 s = 8 s
ds = (28 cm/s)*(8s) = 224 cm
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
During this interval the clock time changes 8 s because final time – initial time = 13 s – 5 s = 8 s.
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
During this interval the velocity changes 24 cm/s because final time – initial time = 40 cm/s – 16 cm/s = 24 cm/s.
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• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
average rate of change of velocity = average acceleration = a
a = dv / dt
dv = change in velocity; dt = change in time
a = (24 cm/s) / 8 s = 3 cm/s^2
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Rise of the graph = vertical coordinates = change in velocity = 40 cm/s – 16 cm/s = 24 cm/s
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• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Run of the graph = horizontal coordinates = change in clock time = 13 s – 5 s = 8 s.
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Slope = rise / run
Rise = 24 cm/s
Run = 8 s
Slope = (24 cm/s) / 8 s = 3 cm/s^2
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• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The slope of the graph tell for the acceleration of the object.
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• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
average rate of change of velocity = average acceleration = a
a = dv / dt
dv = change in velocity; dt = change in time
a = (24 cm/s) / 8 s = 3 cm/s^2
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15 min
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Very good, except for the questions about midpoint clock time and midpoint velocity, which you will likely find very easy to correct.
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