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course phy 231
June 30, 11:06
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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):
The graph of v vs. t would have velocity on the y-axis and clock time on the x-axis. At 4 sec you would plot a value of 10 cm/s and at 9 sec a value of 40 m/s.
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Sketch a straight line segment between these points.
answer/question/discussion (start in the next line):
After sketching a straight line segment between the points (9, 40) and (4, 10), the slope of the line would represent the acceleration of the ball which would be (40 m/s - 10 m/s)/(9 sec - 4 sec) = rise/run = (30 m/s)/(5 sec) = 6 m/s^2.
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What are the rise, run and slope of this segment?
answer/question/discussion (start in the next line):
rise = (40 m/s - 10 m/s)
run = (9 sec - 4 sec)
slope = rise/run = 30 m/s/5 sec = 6 m/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 the line is the distance the ball traveled with the given parameters. `ds = (40 m/s + 10 m/s)/2 * 5 sec = 125 meters. Or, area underneath = (10m/s * 5 sec) + .5 * 5 sec * 30m/sec = 50 meters + 75 meters = 125 meters.
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