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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: ->->->->->->->->->->->-> (start in the next line):
(13-5s)/2 = 4s
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The midpoint between two clock times is halfway between them.
4 s is not halfway between 5 s and 13 s.
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
40-16cm/s = 24cm/s / 8 seconds(time of whole interval) = 3 cm/s/s (acceleration). 16cm/s (initial velocity) + 3cm/s/s * 4s (midpoint) = 28 cm/s
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`` For a problem you actually reasoned out correctly, the following notes might seem picky. However solution of physics problems requires correct conceptual thinking, and carefully constructed expressions.40 - 16 cm/s is not 24 cm/s.
40 and 16 cm/s are not like terms and cannot be subtracted.
40 cm/s - 16 cm/s could be simplified, but 40 - 16 cm/s cannot.
40 cm/s - 16 cm/s = 24 cm/s.
40 cm/s - 16 cm/s is not equal to 24 cm/s / (8 s).
The = sign needs to be confined to statements of equality, not to a sequence of calculator steps or a train of thought.
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16cm/s (initial velocity) + 3cm/s/s * 4s (midpoint) = 28 cm/s
is a very good calculation, and correctly expressed.
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Since the graph is a straight line, it is also possible to say that the midpoint velocity is halfway between 16 cm/s and 40 cm/s. You might want to think about why your calculation gave the same result as would have been obtained by finding the mean of these two velocities.
Both ways of calculating the midpoint velocity provide important insights, and the reason both ways give the same result provides additional, and valuable, insight.
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How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Based on acceleration, it travels 3 cm/s faster each second, so second 0-1 is 16cm/s + 19cm/s (second 1-2) + 22cm/s (second 2-3) + 25 cm/s (second 3-4) = 82 cm
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In fact the displacement is equal to the product of the average velocity and the time interval.
In fact the object travels for 8 seconds at a speed greater than 16 cm/s, so it travels more than 8 s * 16 cm/s = 128 cm.
The object doesn't travel 16 cm in the first second, since its velocity is 16 cm/s only at the first instant of the first second. It ends up with a velocity of 19 cm/s at the end of the first second.
Note also that the motion doesn't stop at the midpoint clock time.
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Whole interval 13-5s = 8 seconds. Midpoint is 8s/2 = 4 s
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The midpoint is 4 s after the first instant, but this is not particularly useful in analyzing the motion in this case.
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Midpoint 25 cm/s - 16 cm/s = 9 cm/s
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The midpoint is not related to how much the velocity changes.
It's not clear where the 25 cm/s came from, but in any case the midpoint velocity is not relevant to this question. The change in velocity is the change from the first instant to the last.
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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):
9 cm/s / 4s = 2.25 cm/s
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This does not result from the definition of average rate of change. Neither 9 cm/s nor 4 s is a quantity that would be associated with that definition for this motion.
You should start with the definition and apply it step by step to answer this question.
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What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
2.25 * 4s = 9
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2.25 * 4 s is equal to 9 s, not to 9.
However this calculation is not related to the slope of the graph.
You should have a graph with two points and a straight line connecting them.
The rise and the run are found in the most straightforward way from the coordinates of those two points.
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
1 * 4 = 4
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The previous note applies here also.
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
9/s = 2.25
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To find the slope you need to find the correct rise and run.
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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):
It's speeding up more than two times the rate time is moving.
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The rate at which a quantity speeds up cannot be compare to how fast it is moving. The two quantities are unlike terms. They have different units and different meanings.
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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):
whole interval 40cm/s - 16 cm/s = 24 cm/s / 8 seconds = 3 cm/s/s
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You are misusing the = sign here, but your thinking and your result are correct.
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Midpoint 25 cm/s - 16 cm/s = 9cm/s / 4 seconds = 2.25 cm/s/s
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The average rate between the initial instant and the midpoint must be equal to the average rate for the entire interval, since the v vs. t graph is a straight line.
This would have been the case had you used the correct midpoint velocity for this calculation.
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You've done many good things here, so forgive all the notes. I would have inserted fewer notes if I didn't think you were ready to handle all of them.
You are thinking correctly about average rate, though you did make at least one error on an average rate.
I think you're overcomplicating the questions related to the graph.
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Be sure to include the entire document, including my notes. &#
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