cq_1_041

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.

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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).

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Sketch a straight line segment between these points.

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What are the rise, run and slope of this segment?

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Rise = 30, Run = 5, so the slope = 6

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What is the area of the graph beneath this segment?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

The area is found by multiplying its average height by its width. Its avg. height is 25, found by averaging the 10 cm/s and 40 cm/s, and its width is 5 s. When multiplied I found the area = 125.

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10 minutes

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You didn't answer the first couple of questions, but I'm confident you understand this.