course phy 201
2/11/2010 7:48 pm
Seed 4.1The 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):
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Sketch a straight line segment between these points.
answer/question/discussion (start in the next line):
On a graph of v vs t, v would be on the vertical axis and t would be on the horizontal axis. This graph has a straight line and as time increases the velocity increases.
So the graph is increasing at an increasing rate.
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What are the rise, run and slope of this segment?
answer/question/discussion (start in the next line):
Rise is the change in position on the vertical axis the run is the change in position on the horizontal position and the slope= rise/run.
Rise = 10cm/s= 40 cm/s = -30cm/s
run = 4 sec - 9 sec= -5 sec
slope = -30cm/s / -5 sec = 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):
Area is the length * width
-30cm/s * -5 sec = 150cm
Your slope is correct, and overall your work is good.
However you used the rise of your graph to calculate the area. You should have used the 'average altitude' of the trapezoid, not the rise, to calculate the area.
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