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Phy 121
Your 'cq_1_02.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_02.2_labelMessages **
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):
M= [(x1+ x2)/2] , [(y1 + y2)/2] = [(5+13)/2], [(16cm/s + 40cm/s)/2] = (9 sec, 28cm/s)
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Since the graph is velocity vs. clock time, 28cm/s is the velocity at the midpoint interval.
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How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Since vAve = 'ds/'dt, I would venture a guess that the object travels ('ds) = (40cm/s - 16cm/s) * (13s - 5s) = 24cm/s * 8s = 192 cm.
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Good reasoning, but you appear to be confusing the average velocity with the change in velocity.
If the velocity went from, say, 40 cm/s to 38 cm/s, by your reasoning you would multiply 2 cm/s by 8 s to get 16 m, much less than the 192 cm you calculated. Clearly this can't be right, because the velocity is falling much more slowly than before, that the average velocity and therefore the distance traveled will be greater than before.
The average velocity of this situaiton occurs at the midpoint of the interval, and you have already found it. The average velocity is 28 cm/s, not 24 cm/s. 24 cm/s is the change in the velocity.
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
13 sec - 5 sec = 8 seconds
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
40cm/s - 16cm/s = 24 cm/sec
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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):
aAve = 24cm/s / 8 sec = 3cm/s /second
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What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise = 'dv, so 24 units. 40 -16 =24
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Run = 'dt, so 8 units. 13 - 5 = 8
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
m = rise/run, so m=24/8 or m=3.
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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 tells me the object is tripling its vertical position for every one horizontal unit. The motion is increasing.
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Rise, run and slope all have units. You used the units correctly in the preceding, and will understand that the units of rise, run and slope are cm/s, s, and cm/s/s.
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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):
3cm/sec.
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3 cm/s could be an initial, final or average velocity, or a change in velocity. But it can't be a rate of change of velocity.
The units of the rate of change of velocity with respect to clock time will be units of velocity divided by units of time; in this case the units would be cm/s/s or cm/s^2.
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*#&!
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Very good work on most questions.
You do have a couple of errors, and you'll want to avoid those errors in the future.
You'll understand my notes. You're doing very good work, so I'm not going to ask for a revision, though if you aren't sure of the correct solution you're welcome to submit one.
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