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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):
My graph has velocity (cm/s) along the y axis and time (s) along the x axis. The velocity axis goes from 0 to 40 marked by increments of 5 cm/s, and the time axis is from 0 to 10 marked by increments of 1 s. The first point is towards the left and bottom section of the graph, and the second point is in a more positive direction, towards the top and right side of the graph.
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I sketched a straight line between the points, it is in a positive direction almost cutting the graph in half.
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise between the two given points is 30 cm/s. The run between the given points is 5 s. The slope is 30 cm/s / 5 s = 6 cm/s/s
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• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Area = [ (10 cm/s + 40 cm/s) / 2 ] (5 s)
= (50 cm/s / 2) (5 s)
= 25 cm/s * 5 s
= 125 cm
The area of the graph beneath this segment is 125 cm.
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&#Very good responses. Let me know if you have questions. &#