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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.
** 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: ->->->->->->->->->->->-> scussion (start in the next line):
With velocity as the y-axis and clock time as the x-axis, we can just use the midpoint formula. (5 + 13)/2, (16 + 40)/2 = 9 sec, 28 cm/s
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
28 cm/s / 9 sec = 3cm/sec
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The velocity is never less than 16 cm/s and never greater than 40 cm/s.
So 3 cm/s could not be the average velocity.
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How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The difference in clock time is 8 seconds. The difference in velocity is 24cm/s. Assuming acceleration is constant we can divided the difference in velocity by the difference in clock time yielding 3cm/s. So 16 cm + 19 + 22 + 25 +28 +31 + 34 + 37 + 40 = 252cm
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As in your work on the preceding 'seed' question, you have correctly listed the velocities at start, and after each of the 8 seconds. However none of these represents the distance traveled during any of those seconds. (see also my note on the preceding problem)
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
8 seconds.
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
25 cm/s
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What is the velocity at the beginning of the 8-second interval?
What is the velocity at the end of this interval?
Therefore how much change is there in the velocity?
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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):
3cm/s
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The units are not correct, and you haven't indicated how you got this result.
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What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Slope formula: 40 - 16cm/s, 13 - 5 sec = 24cm/s, 8 sec. 24 is the rise.
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With units, the rise is 24 cm/s.
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
8 is the run.
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8 seconds, not just a unitless 8.
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
rise/run = 24/8 = 3
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The numbers are right but the units are missing in your calculation and in your result.
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You've got all the pieces but in some cases you need to put them together a little more completely.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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