cq_1_091

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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 **

cq_1_091

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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.

&#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.

&&&& Gosh this IS a mess! I was staining my deck all weekend long. I'm gonna blame the fumes!

@&

It happens to us all once in awhile.

I've got a deck to finish power-washing and stain before the dry weather is over. So you might see me in the same state soon.

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vAve = `ds/`dt = 20cm/2s = 10cm/s

vAve = (vf+v0)/2 Multiply each side by 2

2vAve= vf+v0 subtract v0 from each side

2vAve -v0 = vf Reverse

vf=2vAve -v0

vf = 2 * 10cm/s -0cm/s = 20cm/s

aAve = (vf - v0)/`dt

aAve = (20cm/s -0cm/s)/2s = 10cm/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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> scussion:

`dt actually equals 2s *1.03 = 2.06s

vAve= 20cm/2.06s = 9.71cm/s

vf = 9.71cm/s and v0 = 0cm/s

a= `dv/`dt = (9.71cm/s)/2s = 4.86cm/s

@&

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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dt actually equals 2s *1.03 = 2.06s

vAve= 20cm/2.06s = 9.71cm/s

vf = 2*vAve = 19.4cm/s (approx)

a= `dv/`dt = (19.4cm/s)/2.06s = 9.42cm/s^2

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What is the percent error in each?

answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> scussion:

For vf : %error = error/quantity*100% = (20cm/s -9.71cm/s)/10cm/s *100% = 2.9%

For a: : %error = error/quantity*100% = (10cm/^2s -4.86cm/s^2)/10cm/s *100% = 2.8%

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For vf : %error = error/quantity*100% = (20cm/s -19.4cm/s)/20cm/s *100% = 3%

For a: : %error = error/quantity*100% = (10cm/s^2 -9.42cm/s^2)/10cm/s^2 *100% = 5.8%

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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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The percent error is not the same

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Right. For acceleration your percent difference in acceleration to about 6%.

In calculating acceleration the time interval was used twice, each time contributing 3% to the error.

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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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*#&!

&#This looks good. See my notes. Let me know if you have any questions. &#