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:
Answer: I have graphed the two points.
• Sketch a straight line segment between these points.
answer/question/discussion:
Answer: I have drawn a line segment.
• What are the rise, run and slope of this segment?
answer/question/discussion:
Answer: The rise of the line segment is (40cm/s-30cm/s)=10cm/s, which is the change in velocity. The run of the line segment is (9sec-4sec)=5sec, which is the change in clock time. The slope is 10cm/s / 5sec = 2 cm/sec/sec.
• What is the area of the graph beneath this segment?
answer/question/discussion:
The area under the graph is the change in position of the ball. The area is 75cm^2. So the change in position of the ball is 75 cm.
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10 minutes
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I know this is a very small detail but the area under the line segment is 75cm^2 but the change in position is 75cm. Why the difference in units?
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