Phy201
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.
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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:
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Sketch a straight line segment between these points.
answer/question/discussion:
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What are the rise, run and slope of this segment?
answer/question/discussion: Rise = 30cm
Run = 5sec
Slope = 30cm/5sec = 6cm/sec
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What is the area of the graph beneath this segment?
answer/question/discussion: I assume you are implying to make a right triangle.
The area beneath this segment is 75cm
The region is the trapezoid defined by the segment. The trapezoid extends down to the horizontal axis. Its average altitude is 25 cm/s (midway between the altitudes 10 cm/s and 40 cm/s) and its width is 5 s (from 4 sec to 9 sec is 5 sec), so that its area is 25 cm/s * 5 s = 125 cm.
The 75 cm you got is the area of the right triangle defined by the two points; however the 10 cm/s * 5 s rectangle immediately beneath this triangle must be included.
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10 min
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&#Good responses. See my notes and let me know if you have questions. &#