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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: ->->->->->->->->->->->-> :
Given: uniform accel, ‘ds=20cm, v0=0, ‘dt=2s
Find vAve, vf and a:
Find a:
‘ds=v0 ‘dt + .5 a ‘dt^2
a=(‘ds - v0 ‘dt)/(.5 ‘dt^2)
a=(20cm - (0cm/s)(2s))/(.5 * (2s)^2)=10cm/s^2
find vf:
vf=v0 + a ‘dt
vf=0cm/s + 10cm/s^2 * 2s=20cm/s
find vAve:
vAve=(vf + v0)/2
vAve=(20cm/s + 0cm/s)/2=10m/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: ->->->->->->->->->->->-> :
‘dt + 3%=2s + 2s *.03=2.06s
a becomes:
a=(20cm - (0cm/s)(2.06s))/(.5 * (2.06s)^2)=9.43cm/s^2
vf becomes:
vf=0cm/s + 9.43cm/s^2 * 2.06s=19.43cm/s
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• What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
vf error = (20-19.43)/20=2.85 percent
a error = (10-9.43)/10=5.7 percent
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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: ->->->->->->->->->->->-> :
Percent error not same.
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• If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
There is more of an effect on accel because it ws the first property to calculate, then the new a was plugged in to find vf.
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The key is that to find average velocity you divide displacement by `dt; this quantity is used to find final velocity and hence change in velocity; then we divide once more by `dt.
Two divisions by `dt, which is off by 3%, combine to give us an error of about 6%.
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