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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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cq_1_022
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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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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 sec
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
28 cm/ 9 sec=3.1 cm/sec or
3 cm/sec
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
40 cm
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Velocity at midpoint is 40+16=56/2
=28 cm/s
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The initial velocity is 16 cm/s and the final is 40 cm/s. The velocity will always be at least 16 cm/s and can be as great as 40 cm/s.
The velocity will never be 3 cm/s.
There is no quantity equal to 28 cm. However there is a quantity equal to 28 cm/s, and as it turns out that is the midpoint velocity, halfway between the initial and the final.
You wouldn't divide the midpoint or average velocity by the time interval. That would have no meaning.
However there certainly are circumstances where you would multiply it by the time interval.
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
8 seconds
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
the vAve went from being 3.2 to being 3.07 by the end of the interval it changed by -.17
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velocity changed from 16 to 40, changed 24 cm/s
v0=16c m/s
vf= 40 cm/s
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The average velocity for the interval doesn't change. It's 28 cm/s.
Not clear how you got 3.2 or 3.07.
What was the initial velocity on the interval?
What was the final velocity?
By how much did the velocity therefore change?
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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):
-.17cm/s/8 sec= -.02125 cm/s/s
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aAve= dv/dt
24 cm/s/ 8 sec= 3 cm/s/s
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What is the definition of the average rate of change of velocity with respect to clock time?
What does this definition tell you that you need to do with the quantities you've been given?
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
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):
run=8 seconds
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
m=3 cm/s/s
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Slope= 3 cm/s/s which is also the acceleration
To get slope. M= rise/run= 24 cm/s/8 s=3 cm/s/s
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3 is the correct number.
The slope also has units, which should be included at every step of your calculation.
Can you indicate the calculation you used to get the slope, using units throughout?
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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 steeper the slope the greater the velocity. The less steep it is the lower the velocity
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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):
-.17cm/s/8 sec= -.02125 cm/s/s
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Very good.
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