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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?
The clock time would be (5s + 13s)/2 = 9 sec
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Velocity would be (16 cm/s + 40 cm/s)/2 = 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):
average velocity * time
28 cm/s * 4 sec = 112 cm
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@& The interval lasted 8 seconds, not 4 seconds. Otherwise good.*@
By how much does the clock time change during this
interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The clock time changes by +4 seconds. 9 sec - 5 sec = 4
sec
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Velocity changes by 12 cm/s. 28 cm/s - 16 cm/s = 12
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):
12 cm/s / 4 sec = 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):
12 cm/s
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
4 sec
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
+3
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@& The rise and run have units and meanings, which gives the slope units and meaning.*@
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 positive slope tells you the velocity of the object
is increasing.
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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):
12 cm/s / 4 sec = 3 cm/s^2
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*#&!
&#Good responses. See my notes and let me know if you have questions. &#