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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.
** 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: ->->->->->->->->->->->-> :
vAve=(20m/s)/2s= 10m/s
vf= 0+ 10 * 2= 20m/s
a= (20-0)/2= 10m/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: ->->->->->->->->->->->-> :
2 * 1.03= 2.06s
a= (20-0)/2.06= 9.71m/s/s
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If the time interval is longer it will also affect your result for the final velocity, which will no longer be 20 cm/s.
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vf= 0+ 9.71 * 2.06= 20m/s
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
a= 3% error
vf= 0%
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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: ->->->->->->->->->->->-> :
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
because the final velocity is not changed by the time it takes to reach that velocity. Only the acceleration and average velocity would be different.
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5 mins
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self-critique #$&*
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&#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.
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Solution
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You overlooked the fact that the change in the time interval affects not only the time interval but the resulting calculations of average and final velocity, so there is a twofold effect on the calculation of acceleration.
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