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Phy 201
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).
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What is the clock time at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
13sec-5sec= 9seconds at the midpoint
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The slope is equal to 3. 3x9= 27cm/sec at the midpoint
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How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
I'm not exactly sure. (40cm/sec+ 16cm/sec + 27cm/sec)/2= 27.67cm/sec average based on these numbers. 27.67x8= 221.36cm
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
13-5= 8seconds
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
40-16= 24cm/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: ->->->->->->->->->->->-> scussion (start in the next line):
(24cm/sec)/(8sec)= 3cm/sec/sec
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What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
40-16=24
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
13-5= 8
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
24/8= 3 rise/run = slope
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@& The rise, run and slope all have units, which should be included.
The final calculation is 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: ->->->->->->->->->->->-> scussion (start in the next line):
The object is gradually accelerating during the 8 seconds. The graph is increasing from left to right.
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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: ->->->->->->->->->->->-> scussion (start in the next line):
(40-16)/(8)= 3cm/sec/sec ?
@& Right. Good.*@
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*#&!
&#This looks good. See my notes. Let me know if you have any questions. &#