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).
• What is the clock time at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
9 seconds
13-5 = 9 seconds
I am not sure about that answer?
That is the change in clock time, not the clock time at the middle
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
28.8 cm/s I crossed multiplied and solve for velocity
I am not positive about that answer?
The graph is a straight line, so the midpoint is velocity is halfway between 16 cm/s and 40 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):
I used the average rate equation: 16 cm / sec * 5 sec’s = 80 cm’s, & 40 cm / sec * 13 sec’s = 520 cm’s. Then I subtracted the two and got how far it travel which is approximately 440 cm’s.
If the clock had started, say, 4 seconds earlier, then it would have read 9 sec for the first point and 17 sec for the second. This would not affect the fact that the object traveled for a total of 8 seconds, so would not affect the distance it traveled. However it would affect your answer. Clock times are subject to when the clock was started, but that does not affect the time interval or the motion of the object on that interval.
The definition of average velocity is required here; combined with change in clock time you can get the correct displacement.
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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 changed from 5 seconds to 13 seconds so a difference of 8 seconds.
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
If it went from 16 cm / sec to 40 cm / sec, then it changed 24 cm / 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: ->->->->->->->->->->->-> (start in the next line):
The average rate of change of velocity with respect to clock time on this interval is: change in velocity divided by the change in clock time: 24 cm per sec / 8 seconds = 3 cm’s per second.
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
24 is the rise, because the difference between 40 and 16 is 24.
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• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
8 is the run, because the difference between 13 and 5 is 8.
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
3 is the slope, because slope = rise / run = 24 / 8 = 3.
Your units aren't completely correct here. Be sure to check the discussion in the link given at the end of this document.
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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 average rate of velocity is the same exact number for the slope of the line, so they related equally. It moves at a rate of 3 cem’s per second.
the units of that calculation are not 3 cm/s, and the interpretation differs from the one you give
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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):
The average rate of change of the object’s velocity with respect to clock time during this interval is the change in a divided by the change in b or the change in velocity divided by the change in clock time during this interval: 3 cm’s per second / 8 seconds = .375 cm’s per second is the average rate of change of the object’s velocity with respect to clock time.
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30 Minutes
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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. &#