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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 **
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 lies halfway between the two clock times and halfway between the two velocities. (5 sec + 13 sec) / 2 = 9 sec.
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•What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The midpoint of the velocity between 16 cm/s and 40 cm/s occurs midway between the two velocities, at (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):
The object moves at average velocity 28 cm/s for 8 seconds. So it travels 28 cm/s * 8 s = 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 change in clock time from t = 5 sec to t = 13 sec
`dt = 13 sec - 5 sec = 8 sec.
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•By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The change in velocity during this interval is 24cm/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):
ave rate = change in velocity / change in clock time = ( 24 cm/s ) / (8 s) = 3 (cm/s) / s = 3 (cm/s) * (1/s) = 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):
16 cm/s to 40 cm/s
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):
The run is 8sec
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•What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
( 24 cm/s ) / (8 s) = 3 (cm/s) / s = 3 (cm/s) * (1/s) = 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 is the same as the average rate of change of velocity with respect to clock time. The slope represents this average rate of change.
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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):
The slope is equal to the average rate of change of an object’s velocity which is 3 cm/s
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&#Good responses. Let me know if you have questions. &#