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PHY 231
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.
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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: ->->->->->->->->->->->-> :
vAve = 20 cm / 2 s
= 10 cm/s
vf = 20 cm/s
a = 20 cm/s / 2 s
= 10 cm/s/s
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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: ->->->->->->->->->->->-> :
vAve= 20 cm / 2.06s
=10.31 cm/s
vf = 20.62 cm/s
a = 20.62 cm/s / 1.94 s
=10.63 cm/s/s
Compare values and find actual values using the above as the error rate:
VF
20 cm/s - .62 cm/s
Acceleration
10 cm/s/s - .63 cm/s/s
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• What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
3% for vf and 6% for Acceleration
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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: ->->->->->->->->->->->-> :
NA
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• If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
The vAve is off by 3% because it is based on a relationship with time which has a 3% error rate. The acceleration has twice the error rate because it is based off both a relationship with time (3% error) and vf (3% error).
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@& Very good.
Note furthermore that 1.03^3 is vey close to 1.06. The former is the factor by which the time interval would be less, given a 3% error on the 'low side'. Componding two 3% errors corresponds to a factor of 1.03 compounded with another factor of 1.03. These factors multiply, so the combined factor is 1.06. This corresponds to a 6% error.
Similarly a factor of .97 when squared is very close to .94, which indicates that a 3% error on the other side of 1 would also result in a 6% error in the final result.
The square of a number close to 1 is in general about twice as far from 1, which is easily proven within either an algebra or a calculus context.
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@& No need for a revision, but check also the discussion at the link.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
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