Phy231
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.
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What are its average velocity, final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
v0=0cm/s 'dt=2s 'ds=20cm vAve='ds/'dt=20cm/2s=10cm/s vAve=(vf+v0)/2 10cm/s=(vf+0cm/s)/2 20cm/s=(vf+v0) vf=20cm/s
a='dv/'dt a=(20cm/s-0cm/s)/2s a=20cm/s/2s a=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:
2s-3%=1.94s 'ds=(vf+v0)/2*'dt 'ds/'dt=(vf+v0)/2 ('ds/'dt)*2=Vf+v0 vf=('ds/'dt)*2-v0 vf=(20cm/1.94s)*2-v0 vf=10.309cm/s*2-0cm/s vf=20.619cm/s
a='dv/'dt a=(20.619cm/s-0cm/s)/1.94s a=10.628cm/s^2
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
Percentage error in vf is 20cm/s/20.619cm/s=.97 1.00-.97=.03 so 3%
Percentage error in a is 10cm/s^2/10.628cm/s^2=.941 1.000-.941=.059 so 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: ->->->->->->->->->->->-> scussion:
They are not the same.
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> scussion:
They are different because the 'dt is used only once to find vf, but used twice to find a. Because of this then the a error should be twice as much.
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15min
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3/16/10 9pm
&#This looks very good. Let me know if you have any questions. &#