PHY 231
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 midpoint of this interval is 9, 28 ( 9 sec, 28 cm/s ). So, the clock time is 9 seconds.
13 - 5 = 8
40 - 16 = 24
8 / 2 = 4 and 24 / 2 = 12
5 + 4 = 9
16 + 12 = 28
( 9, 28 ) is midpoint between ( 5, 16 ) and ( 13, 40 ).
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity is 28 cm/s.
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
9 seconds - 5 seconds = 4 seconds
28 cm/s + 16 cm/s = 44 cm /s
44 cm/s / 2 = 22 cm/s
4 seconds * 22 cm / seconds = 88 cm
The object has traveled about 88 cm during this interval.
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I just realized that I believe I did a similar problem earlier, but I did incorrectly. Instead of adding and dividing the velocities by two, I subtracted them. I am not sure why it made since at the time; but, I knew better. I am confident this is the correct way to do it.
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The clock time changes 8 seconds during the original interval from ( 5, 16 ) to ( 13, 40 ), but only 4 seconds during the midpoint to endpoint intervals.
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity changes overall by 24 cm /s during the original interval from ( 5, 16 ) to ( 13, 40 ), but only 12 cm/ s during the midpoint to endpoint intervals.
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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 of change of velocity with respect to clock time is 3 cm/s for every one value of t.
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise of the graph between each individual point on the graph is 3 cm /s ( v ) for every one second ( t ), between both intervals. The overall rise between ( 5, 16 ) and ( 9, 28 ) is 12, while the overall rise between ( 5, 16 ) and ( 13, 40 ) is 24.
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These points and this interval are difficult to understand; To me, I am getting confused whether you are referencing ( 5, 16 ) and ( 9, 28 ) or ( 5, 16 ) and ( 13, 40 ). I think it is slightly ambiguous. Is there any way you could put a few antecedents in the preceding problems for future use??? I think it would be helpful to at least reference these two points in this question for these points ( or mention the midpoint and the first point if you don't want to give away answers ), especially for preceding questions.
I understand the suggestion.
However the interval in this problem begins with the event of the first point and ends with the event of the last, and this is a consistent convention.
Once we understand this convention, additional wording would be unnecessary, and would in fact clutter up our statements.
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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 one second ( t ) for every 3 cm/s ( v ). The overall run between ( 5, 16 ) and ( 9, 28 ) is 4, and the overall rule between ( 5, 16 ) and ( 13, 40 ) is 8.
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Between ( 5, 16 ) and ( 9, 28 ) & ( 5, 16 ) and ( 13, 40 ), the slope is:
Slope = ( y2 - y1 ) / ( x2 - x1 )
28 - 16 = 12
9 - 5 = 4
Slope = 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 you that the graph is positive and increasing. Thus, it tells you that during this interval that the objects velocity is increasing with 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):
During this interval of four seconds, the average rate of change of the object's velocity with respect to clock time is 3, or 3cm /s for every t.
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Copy and paste your work into the box below and submit as indicated:
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30 Minutes
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I thought some of the references were a little ambiguous...See #### Note
Good work. See my notes. You will understand.
Rises and runs have units, which then determine the units of slopes.
&#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. &#