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Phy 121
Your 'cq_1_04.1' 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_04.1_labelMessages **
The problem:
A ball is moving at 10 cm/s when clock time is 4 seconds, and at 40 cm/s when clock time is 9 seconds.
• Sketch a v vs. t graph and represent these two events by the points (4 sec, 10 cm/s) and (9 s, 40 cm/s).
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
the rise is 30 cm/sec
the run is 8 sec
the slope is 3.75 cm/sec^2
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The run is from 4 sec to 9 sec, so the run is 5 sec.
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• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
the area of the graph beneath this segment corresponds to the product of the average velocity and the time duration. The average velocity in this case is the initial velocity at 10cm/s plus the final velocity at 40 cm / s divided by 2 to give vAve = 25 cm /sec. The product of this and the time duration of 8 sec is `ds=vAve(25cm/sec)(8sec) = 200 cm. On a graph of velocity vs time the area beneath the segment representing the interval on a graph is the product of the average velocity and the time duration.
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Again the run is 5 sec.
Otherwise excellent.
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Very good, but be sure you understand that the width or run is 5 sec, not 8 sec.
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