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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.
** CQ_1_04.1_labelMessages **
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):
Clock time is on the x axis of my graph and velocity is on the y axis; I have plotted these two events using (4, 10) and (9, 40).
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I drew a straight line between these points.
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Rise = change in y = 40 cm/s - 10 cm/s = 30 cm/s
Run = change in x = 9s - 4s = 5s
Slope = rise / run = 30cm/s / 5s = 6 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 of my graph can be found by extending a vertical line from (9, 40) down to the x axis and also extending a vertical line from (4, 10) down to the x axis. I shaded the region between these two new vertical lines and my shaded region creates a trapezoid. Area of a trapezoid = 0.5 * base * (h1 + h2). By filling in this equation, my area = 0.5 * 5s * (40 cm/s + 10 cm/s) = 125 cm. Overall this means that my object was displaced 125 m during the time interval of 5 seconds.
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15 minutes
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&#This looks very good. Let me know if you have any questions. &#