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Phy 121
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):
The clock time at the midpoint would be 9 seconds. This is obtained by adding the start and end times and dividing by 2. 5 sec + 13 sec = 18 seconds / 2 = 9 seconds.
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity at the midpoint would be 28 cm/sec. This can be found by adding the beginning and ending velocities and dividing that answer by 2. 16 cm/sec + 40 cm/sec = 56 cm/sec / 2 = 28 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):
I would estimate that the object travels 224 cm during this interval. Since this is a straight line, the velocity at the midpoint would be the average velocity for the interval. We would just take this average velocity and multiply it by the change in clock time, 8 seconds, to get the distance the ball traveled.
28 cm/sec * 8 seconds = 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 time changes by 8 seconds during this interval. This can be found by finding the difference in the end and start clock times. 13 seconds - 5 seconds = 8 seconds.
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity changes by 24 cm/sec onver the interval. This is determined by subtracting the starting velocity from the ending velocity.
40 cm/sec - 16 cm/sec = 24 cm/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):
The average rate of change of velocity with respect to clock time of this interval is 3 cm/sec/sec. This is found by taking the change in average velocity from start to end and dividing it by the change in clock time.
24 cm/sec / 8 seconds = 3 cm/sec/sec.
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What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise in the graph between these points is 24 units. This is found by subtracting the y values for each point. 40 - 16 = 24.
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&#Very good responses. Let me know if you have questions. &#