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PHY 231
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.
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cq_1_041
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PHY 231
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.
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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):
I graphed (4 sec, 10 cm/s) and (9 s, 40 cm/s) with time along the x axis and position along the y axis.
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
I also sketched the straight line between points. (Increasing to the right)
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
Rise = 40cm-10 cm= 30 cm
Run = 9s-4s = 5s
Slope = 30cm/5s = 6 cm/s
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@& 30 cm is not a quantity associated with this situation.
30 cm/s is.
The vertical coordinates are in cm/s, not cm. So the rise is in cm/s.
This changes the units of the slope.
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Completely understand the correct units, I should have reviewed my answers.
Rise = 40cm/s -10 cm/s = 30 cm/s
Run = 9s-4s = 5s
Slope = 30cm/s /5s = 6 cm/s^2
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• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
.5(40 cm + 10 cm) * 5 s
125 cm
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@& Very good. Just continue to be careful about units.*@