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, 12 cm/s) to the point (13 sec, 36 cm/s).
What is the clock time at the midpoint of this interval?
9 sec
What is the velocity at the midpoint of this interval?
12 cm/s
How far do you think the object travels during this interval?
d=rt
r=3cm/s
t=8, so d=24 cm
By how much does the clock time change during this interval?
8 sec
By how much does velocity change during this interval?
24 cm/sec
What is the rise of the graph between these points?
24cm
What is the run of the graph between these points?
8 sec
What is the slope of the graph between these points?
3 cm/sec
What does the slope of the graph tell you about the motion of the object during this interval?
For every second the object travels 3 cm constantly.
What is the average rate of change of the object's velocity with respect to clock time during this interval?
3cm/sec
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Is this any better?
You aren't applying the definitions quite carefully enough, but you've got all the necessary ideas. Compare with the following and feel free to self-critique and/or ask detailed questions on anything you aren't sure of in the following:
The following is based on the two points (5 sec, 16 cm/s) and (13 sec, 40 cm/s). The previous choice of points had average velocity 24 cm/s, and change in velocity coincidentally equal to 24 cm/s. The change in velocity and average velocity are not the same, and the changed coordinates reveal this important distinction, whereas the original coordinates did not.
The clock time midway between 5 sec and 13 sec is (5 sec + 13 sec) / 2 = 9 sec.
The velocity midway between 16 cm/s and 40 cm/s is 28 cm/s.
The average velocity on the interval would be expected to lie between the initial and final velocities. Without additional information it is not certain that this is the case, but without additional information the most reasonable assumption would be that the average velocity is halfway between the initial and final velocities.
Basing a distance estimate on the assumption that the midpoint velocity of 28 cm/s is the average velocity, we would conclude that the object moves 28 cm/s * 8 s = 224 cm.
If it is known that acceleration is uniform, the average velocity is equal to the midpoint velocity and this estimate would be accurate. If acceleration is not uniform, then it probably isn't completely accurate.
The change in velocity from 16 cm/s to 40 cm/s is 24 cm/s, and this corresponds to the rise of the graph.
The change in clock time is 13 sec - 5 sec = 8 sec
The average rate of change of velocity with respect to clock time is
ave rate = (change in velocity) / (change in clock time) = 24 cm/s / (8 s) = 3 cm/s^2.