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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.
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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).
• Sketch a straight line segment between these points.
• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
we know then that `dv = 40 cm/s - 10 cm/s = 30 cm/s=`dv
we know that `dt = 9sec -4sec = 5sec = `dt
so from formulas we know average acceleration = `dv/`dt = 6 cm/sec^2
we want to see if the graph will give us the same data. Given the points (4sec 10 cm/s) and (9 s, 40 cm/sec) we know the rise is 40-10 = 30 cm/sec which gives `dv
we know that run = 9-4 = 5 seconds = `dt.
So the slope = 30cm/sec / 5 sec = 6 cm/sec^2 which is acceleration.
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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 this segment forms a trapezoid whose area would be equal to `ds. To find the area of a trapezoid you average the 2 altitudes and then multiply by the base. For this graph we have a straight line which represent the average velocity of the interval. The width of the trapezoid is the run.
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Good description on the calculation of area, but you should also have calculated it. The area is 25 cm/s * 5 s =125 cm, as I suspect you already know.
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