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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.
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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 way I got the midpoint of the clock time is 13 sec + 5 sec = 18sec and then I divided it by 2 and my result was 9 sec. So the midpoint is 9 seconds.
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The way I got the midpoint of the velocity is taking 40cm/s + 16cm/s = 56cm/s and then divide it by 2 which is 28cm/s. The midpoint of the velocity is 28cm/s.
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The change of clock time is when you take 13sec - 5 sec = 8 sec. You take the ending time and subtract it by the beginning time to get the change of time.
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity changes 24cm/s because you take 40cm/s and subtract it by 16cm/s and got 24cm/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):
The average rate would be the change of velocity and divide it by change of clock time. Which would be 24cm/s/8sec = 3cm/s^2
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What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise is the change in position which is 40cm/s - 16cm/s = 24cm/s.
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The run is the change in time which is 13 sec - 5 sec = 8 seconds.
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The slope is the rise over run which is 24cm/s/8s = 3cm/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 tells you how far the displacement and the time interval which the velocity changes. Also its the acceleration which is the change in position over the change in time.
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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 the velocity * time interval is 3cm/s * 8s = 24cm
@& That average rate would be (change in velocity) / (change in clock time). You've already calculated that to be 8 cm/s^2.*@
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25 min
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&#Good responses. See my notes and let me know if you have questions. &#