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).
標hat is the clock time at the midpoint of this interval?
oThe clock time at the midpoint of the interval would be at 9 seconds.
標hat is the velocity at the midpoint of this interval?
oThe velocity is about 25 cm/sec.
you don't say how this was calculated; however the velocity midway between 16 cm/s and 40 cm/s is 28 cm/s, not 25 cm/s.
菱ow far do you think the object travels during this interval?
oI think that it travels 24 cm/sec over the 8 seconds.
24 cm/s is not the average velocity on this interval; 8 seconds is the correct duration of the interval
What is the definition of average velocity?
What is the average velocity on this specific interval?
What is the change in clock time?
How are average velocity and change in clock time related by the definition of average velocity to the question of 'how far'?
稗y how much does the clock time change during this interval?
o8 seconds
稗y how much does velocity change during this interval?
oIt increases to 24 cm/sec
標hat is the rise of the graph between these points?
o8
標hat is the run of the graph between these points?
o24
Good, but you have reversed the rise and the run. The graph is of velocity vs. clock time, so the rise will represent the change in velocity, the run will represent the change in clock time.
標hat is the slope of the graph between these points?
o8/24 (or reduced to 1/3)
You didn't include units as part of this calculation.
If the units calculation doesn't result in the correct units, then you know you have either erred in the units calculation or in the procedure you used to solve the problem.
標hat does the slope of the graph tell you about the motion of the object during this interval?
oAs the clock time increases, the velocity of the object also increases.
The rise of this graph represents the change in velocity.
The run represents change in clock time.
What therefore does slope = rise / run represent?
標hat is the average rate of change of the object's velocity with respect to clock time during this interval?
oIt changes 3 cm/sec per second
This is correct. See above to be sure you understand how this final result is related to the v vs. t graph.
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20 minutes
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See my notes and be sure you can answer all the questions posed. If there's anything you aren't sure of, submit a revision consisting of a copy of this document, inserting your answers/revisions/questions/discussion at the appropriate places and marking your insertions with &&&&.
Here is a narrative solution by which you can check yourself. Try to answer the questions I've posed above without reference to this solution, then check it out to see how you did:
The lesser of the two known velocities is the 4 m/s initial velocity. The greater of the two is the 10 m/s final velocity. Lacking other information, the most reasonable estimate of the average velocity would be the velocity halfway between these two, which is easily found by averaging the two velocities to get "+sc2$+"(4 cm/s + 10 cm/s) / 2 = 7 cm/s.
'It is certainly possible that the ball changed velocity at a nonuniform rate. The ball might even have slowed down below 4 cm/s before speeding up to 10 cm/s, or it might have exceeded 10 cm/s at some point during the interval; it might have spent most of the interval moving at 4 cm/s before suddenly speeding up to 10 cm/s. In any of these cases, the velocity does not change at a uniform rate, and our 7 cm/s estimate of the average velocity could differ significantly from the average velocity.
Recall that the average velocity is the average rate of change of position with respect to clock time so that"+sc2$+"ave vel = (change in position) / (change in clock time).
The average rate of change of velocity with respect to clock time is (change in velocity) / (change in clock time). The change in velocity from 4 cm/s to 10 cm/s is 10 cm/s - 4 cm/s = 6 cm/s. If this occurs in 3 seconds, then we have"+sc2$+"average rate of change of velocity with respect to clock time = (change in velocity) / (change in clock time) = 6 cm/s / (2 s) = 3 cm/s^2.