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Phy 121
Your 'cq_1_09.1' 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_09.1_labelMessages **
A ball accelerates uniformly as it rolls 20 cm down a ramp, starting from rest, in 2 seconds.
What are its average velocity, final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds = (v0 + vf) / 2 * `dt
20 = (0 + vf) / 2 * 2
20 = 0 + vf
Vf = 20 cm/s
vAve = 20 cm / 2 s = 10 cm/s
a = 20 cm/s / 2 s = 10 cm/s^2
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If the time interval is in error so that it is 3% longer than the actual time interval, then what are the actual values of the final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
‘dt = 1.94 s
Vf = 20.6 cm/s
A = 20.6 cm/s / 1.94 s = 10.6 cm/s^2
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
3% and 6%
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If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
they are different
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
the final velocity is double the acceleration, so the percent error will be greater.
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10 minutes
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That isn't the reason for the doubling of the percent error.
The reason is that if the time interval is off by 3%, then since the process of calculating the acceleration involves dividing twice by the time interval, that 3% error occurs in two divisions, resulting in a 6% error.
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