Your 'cq_1_9.1' 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?
o vAve = `ds/`dt = 20 cm/ 2 s = 10 cm/s
o vf = 2 vAve – v0 = 2 (10 cm/s) – 0 = 20 cm/s
o aAve = `dv/`dt = 20 cm/s / 2 s = 10 cm/s/s
`dv = vf – v0 = 20 cm/s – 0 cm/s = 20 cm/s
• 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?
o Vf = 19.42 cm/s
o aAve = 9.43 cm/s/s
• What is the percent error in each?
o Vf= 3%
20cm/s – 19.418 cm/s = 0.6/19.418 * 100 = 3%
o aAve = 6%
10 cm/s/s – 9.426 cm/s/s = 0.6/9.426 * 100 = 6%
• If the percent error is the same for both velocity and acceleration, explain why this must be so.
o The percent errors are different.
• If the percent errors are different explain why it must be so.
The percent errors are different because to find final velocity there was an uncertainty of only the one number in the equation; that is the average velocity. However, when solving for acceleration there were two numbers of uncertainty. The change in velocity changed as well as the time. Both numbers in the equation changed therefore causing a higher percent error than final velocity
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20 min
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Very good work. Essentially, the 3% uncertainty in time interval caused an uncertainty in average and final velocities, and in the change in velocity.
Then when you divided change in velocity by time interval the 3% uncertainty was present in both numerator and denominator, which approximately doubled the uncertainty to near 6%.