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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: ->->->->->->->->->->->-> :
20cm/2s = 10cm/s (vAve). Because 10cm = (0m/s + vf)/2, vf must equal 20cm/s. a = (20-0cm/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: ->->->->->->->->->->->-> : If the 'dt above is 3% longer than it should be, that means I must subtract 3% for 2 seconds, which equals 1.94s. So, vAve in this case
equals 10.31cm/s, vf = 20.62cm/s, and a = 10.63cm/s^2.
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> : 20/20.62 = 97% accuracy and 3% error (for vf). 10/10.62 = 94% accuracy and 6% error (acceleration).
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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: ->->->->->->->->->->->-> : I don't have the same. Are you talking about vAve or vf? vAve and vf are the same in my calculations but acceleration is different.
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> : vAve is calcuated by 20cm/1.94, assuming 'ds is accurate. This gives a different number than (vf-v0)/1.94s, which is 20.62m/s / 1.94s, so
the percentage points would be a little off from each other because the numbers being divided are a little different considering the adjustments in time ('ds stays the
same in velocity, but final velocity and time are changed with the new 3% adjustement in time).
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The calculations do show that but the question is why.
One of the two quantities is calculated using a single division by `dt. The process used to calculate the other from the given information requires two divisions by `dt.
Specifically how does this result in a 3% error for one quantity and a 6% error for the other?
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15-20 min
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4:03 pm EST 6/11/14
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Good work in calculating the 3% and 6% uncertainties.
You haven't completely answered the second question. Don't spend more than 10 minutes on it, but see if you can nail down the specifics of how that happened.
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