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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.
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Asst 2 Question 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):
9 s
9 s is the point halfway between 5 s and 13 s. (13 s - 5 s = 8 s/. 8 s / 2 = 4 s. 13 s - 4 s = 9 s; 5 s + 4 s = 9 s.)
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
28 cm/s
40 cm/s - 16 cm/s = 24 cm/s. 24 cm/s / 2 = 12 cm/s. 40 cm/s - 12 cm/s = 28 cm/s; 16 cm/s + 12 cm/s = 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):
About 24 cm
vAve = ( 40 cm/s - 16 cm /s ) / (13 s - 5 s)
= 24 cm/s / 8 cm/s
= 3 cm/s/s
3 * 8 = 24 cm
@& From the given information the velocity is never less than 16 cm/s, so the object would have to move more than 24 cm in 3 seconds.
The quantity you identify as vAve is not the rate of change of position with respect to clock time, since the number you subtract in you numerator are not positions. So that quantity is not the average velocity.
You did calculate a valid average rate of change, 3 cm/s/s. What is it the rate of change of?*@
If the average velocity is 3 cm/s, and the interval lasts 8 seconds, the object should travel about 24 cm.
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
8 s
13 s - 5 s = 8 s
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
24 cm/s
40 cm/s - 16 cm /s = 24 cm/s
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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):
‘dv/’dt = 24 cm/s / 8 s = 3 cm/s/s
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
‘dv correlates to the rise of the graph. So, the rise = 24 cm/s
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• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
‘dt correlates to the run of the graph. So run = 8 s
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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/s / 8 s = 3 cm/s/s
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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 tells us that the acceleration is constant on this straight-line graph because the average acceleration between any two points is the same as between any two points (3).
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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):
3 cm/s/s.
@& There is one small glitch in your otherwise very clear and well-expressed thinking about these rates.
A revision should clarify everything for you.
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