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Phy 121
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):
The midpoint is 9 s. We know this because 13s - 5s = 8 and 8/2 = 4 so the midpoint 9 is 4 seconds away from both times given.
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
First we have to find y intercept using y= mx + b. So, y = (40-16)/ (13-5) * x +b
40 = (3) * 13 + b
1 = b
Then, to find the velocity for our midpoint 9,
Y = (3)* 9 + 1
Y = 28 cm/s
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
28 cm/s = (‘ds/ 9s)
Multiply both sides by 9s,
252 cm = ‘ds
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
13s - 5s = 8 seconds (same as run)
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
40 cm/s - 16 cm/s = 24 cm/s (same as rise)
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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):
(24 cm/s) / (8s) = average rate of change of velocity with respect to clock time.
(24 cm/s) * (1/8s) = 3 cm/s^2
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
24 cm/s (math shown earlier)
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• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
8s (math shown earlier)
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Slope = rise / run = change in velocity / change in clock time
Slope = (24cm/s) / 8s = 3 cm/s^2
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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):
The slope of the graph tells you the average rate of change in velocity with respect to clock time.
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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):
Avg rate of change of velocity = rise / run = change in velocity / change in clock time
Avg rate of change of velocity = (24cm/s) / 8s = 3 cm/s^2
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*#&!
&#Your work looks very good. Let me know if you have any questions. &#