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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 **
Samantha Rogers
PHY 201
Seed 2.2
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 of this interval is 9, 28 (9 sec, 28 cm/sec). So the clock time is 9 seconds.
13sec - 5sec = 8 sec
40cm/sec - 16cm/sec = 24cm/sec
8 / 2 = 4 and 24 / 2 = 12
5 + 4 = 9
12 + 16 = 28
(9, 28) is the midpoint between (5, 16) and (13, 40)
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity is 28cm/sec.
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):13 sec - 5 sec = 8 sec
28cm/sec * 8 sec = 224 cm.
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The clock changes by 8 seconds.
13 sec - 5 sec = 8 seconds.
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
40cm/sec - 16cm/sec = a change of 24cm/sec.
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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):
Change of velocity is 24cm/sec
aAve = (24cm/sec) / 8sec
= 3 cm/sec^2
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
rise of the graph is 40cm/sec - 16cm/sec = 24cm/sec
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• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
run of the graph is 13sec - 5sec = 8 sec.
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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
= (24 cm/sec) / (8sec) = 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 represents the average rate of change, which is the acceleration.
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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):
aAve = `dv / `dt
aAve = (24cm/sec) / (8sec)
aAve = 3cm/sec^2
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*#&!*#&!
&#This looks very good. Let me know if you have any questions. &#