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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):
The change in clock time is (13sec-5se=8sec), halfway through the interval, 4sec would have elapsed, adding 4 sec to the initial time, or 5sec+4sec=9sec.
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
in accordance with the logic above, the change in velocity would be half of the total change of 24, or 12cm/sec. The 12cm.sec change in velocity would be added to the original velocity of 16cm/sec to yield 28cm/sec.
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Good. Alternatively you could average the initial and final velocities.
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How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve=ds/dt
24cm/s= ds/8s
(24cm/s)*8s=ds
192cm=ds
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
13s-5s=8s
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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) = 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):
((40cm/s)-(16cm/s))/(13s-5s)=(24cm/sec)/(8s)=(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):
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):
13s-5s=8s
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
(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):
It shows how quickly the velocity is changing with respect to time.
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Excellent work. Check my one note.
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