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phy 201
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 sketched a graph with velocity on the y-axis and time on the x-axis and plotted the two given points.
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): Ok.
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): rise of segment = 30 cm/s
run of segment = 5 seconds
Slope = rise / run = 30 cm/s / 5 s = 6
@& Good, but the result has units, which I believe you know how to find. The correct units are 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): The area beneath this segment is part of the graph.
I am not sure I understand what you are asking.
I believe that the area beneath the graph of a curve deals with integration.
@& The area would involve integration for most v vs. t curves, and the area can be equated with the integral of the velocity function between the two clock times. This is a good thing to know, though you aren't required to know it in your course.
In the present case the v vs. t graph is linear, so the region below the graph is a trapezoid. You can find its area using basic geometry.
Check the link given below. I expect that you'll understand that information well, but if not be sure to ask questions.*@
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15 minutes
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&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#