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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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?
(5 + 13) / 2 = 9 sec
• What is the velocity at the midpoint of this interval?
(16 + 40)/ 2 = 28 cm/s
• How far do you think the object travels during this interval?
With an average velocity of 28cm/s, after 8 secs, 28 * 8 = 224cm
• By how much does the clock time change during this interval?
8 seconds
• By how much does velocity change during this interval?
24cm/s
• What is the rise of the graph between these points?
24
• What is the run of the graph between these points?
8
• What is the slope of the graph between these points?
24 / 8 = 3
• What does the slope of the graph tell you about the motion of the object during this interval?
The slope tells me that for every change in second the velocity increases by 3 cm/s
• What is the average rate of change of the object's velocity with respect to clock time during this interval?
3cm/s, should be equal to the slope
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10 mins
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The slope of the graph is (24 cm/s) / (8 s) = 3 cm/s^2, not 3 and not 3 cm/s.
The unit calculation is essential, and if the units don't work out the way they should it's a clear indication of an error either in units calculation or in procedure.
Otherwise very good.