Phy 231
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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):
With the given two points, the graph just looks linear and it is increasing at a constant rate.
The rise is 40 cm/s – 10 cm/s = 30 cm/s, The run is 9s – 4 s = 5 s. The slope can then be found by 30 cm/s / 5 s = 6 cm/s/s. The graph has a slope of 6 cm/s/s.
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The line gives a better view at how the line is linear and increasing at a constant rate.
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise is 40 cm/s – 10 cm/s = 30 cm/s, The run is 9s – 4 s = 5 s. The slope can then be found by 30 cm/s / 5 s = 6 cm/s/s. The graph has a slope of 6 cm/s/s.
Good.
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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 the graph beneath the segment is the same as the total distance traveled. I found that it was 180 cm. 40 cm/s / 2 = 20 cm/s and then the total time is 9 sec, so 20 cm/s * 9 s = 180 cm.
Good explanation, but your result isn't correct.
The time interval lasts only 5 seconds.
The initial velocity isn't 0 so the average velocity isn't half the final velocity.
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15 mins
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Solution
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