phy201
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).
0. What is the clock time at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):_
9 seconds
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0. What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):_
28 cm/sec
I figured out the first 2 questions by averaging the clock time points and the velocity points repectively.
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0. How far do you think the object travels during this interval?
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average velocity is 28cm/s and the change in clock time is 8 sec so the distance the object travels would be 28 * 8= 224 cm
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0. By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):_
13s-5s=8 sec
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0. By how much does velocity change during this interval?
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40cm/s-16cm/s= 24 cm/s
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0. 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):_
24cm/s ( change in velocity)/ 8s( change in clock time) = 3 cm/sec( average rate of change of velocity)
This answer is incorrect. Be sure you compare with the link provided at the end of this document.
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0. What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):_
24
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0. What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):_
8
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0. What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):_
the slope would be 24/8=3
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0. What does the slope of the graph tell you about the motion of the object during this interval?
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the slope = the average rate of change in velocity
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0. What is the average rate of change of the object's velocity with respect to clock time during this interval?
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??? Isn’t this the same answer as above???
24cm/s ( change in velocity)/ 8s( change in clock time) = 3 cm/sec( average rate of change of velocity)
Yes but the units of your results are incorrect both here and there.
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20 MIN
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&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#