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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: ->->->->->->->->->->->-> scussion:
`dv = 20cm/2s = 10cm/s
vf = 10cm/s and v0 = 0cm/s
vAve = (10cm/s + 0cm/s)/2 = 10cm/s/2 = 5cm/s
a= `dv/`dt = (10cm/s)/2s = 5cm/s
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If a ball starts from rest and accelerates at 5 cm/s^2 for 2 seconds it reaches a final velocity of 10 cm/s, so its average velocity is 5 cm/s. In 2 seconds it would therefore move 10 cm.
So your conclusions aren't consistent with the given informatoin.
You've confused a couple of definitions. Try reasoning this out again, starting from the definitions of average velocity and acceleration. If you start with those definitions you're unlikely to mix anything up.
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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: ->->->->->->->->->->->-> scussion:
`dt actually equals 2s *1.03 = 2.06s
`dv = 20cm/2.06s = 9.71cm/s
vf = 9.71cm/s and v0 = 0cm/s
a= `dv/`dt = (9.71cm/s)/2s = 4.86cm/s
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In addition to mixing up some of the quantities (per previous note) you used the original 2-second time interval in this last step, when you should have used the interval that was 3% longer.
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
For vf : %error = error/quantity*100% = (10cm/s -9.71cm/s)/10cm/s *100% = 2.9%
For a: : %error = error/quantity*100% = (5cm/^2s -4.86cm/s^2)/5cm/s *100% = 2.8%
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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: ->->->->->->->->->->->-> scussion:
I think they are close enough to be considered equal. 2.9% vs. 2.8 %
I assume the difference is due to rounding along the way.
They are the same (effectively) because each was deteremined by dividing some quantity by the `dt. Therefore they are both equally affected by the adjustment to the `dt. Therefore the % error should be equal.
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> scussion:
I don't believe they are significantly different
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I believe you got careless about identifying your quantities and mixed some of them up.
You'll want to pretty much start over and be sure to use the definitions carefully.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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