cq_1_091

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Phy 201

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: ->->->->->->->->->->->-> :

The average velocity is vAve= ‘ds/ ‘dt= 20cm/2s= 10cm/s

The final velocity is: vAve= (vf+v0)/2 = 10cm/s=(vf+0)/2= 20cm/s=vf

The acceleration is: ‘dv/’dt= (20cm/s-0cm/s)/2=10cm/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: ->->->->->->->->->->->-> :

3 percent longer would make the seconds 2.06 s.

Therefore the vAve= 20cm/2.06=9.7cm/s

The vf would = 9.7cm/s=(vf+0)/2.06=19.98cm/s

Your vf should be 19.4 cm/s.

vAve = (vf + v0) / 2, not (vf + v0) / `dt.

The acceleration would be= (19.98-0)/2=9.99cm/s/s

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• What is the percent error in each?

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

The percent error would be 3 % for all of them.

That's not the case. You should actually calculate your percent errors.

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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: ->->->->->->->->->->->-> :

This is because both of these variables are found by knowing the ‘dt and if the ‘dt was off by 3 seconds then the other figures will be as well.

Good reasoning but you've overlooked something important.

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• If the percent errors are different explain why it must be so.

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

They are not.

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10 minutes

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Solution

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