PHY201
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?
answer/question/discussion:
Vavg = ‘ds/ ‘dt = 20cm/2s = 10cm/s
Vf = 2*vavg - v0 = 2*10cm/s - 0 = 20cm/s
‘dv = vf-v0 = 20cm/s - 0 = 20cm/s
A = ‘dv/ ‘dt = 20cm/s / 2s = 10cm/s/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?
answer/question/discussion:
2 * .97 = 1.94s
Vavg = 20cm / 1.94s = 10.3cm/s
Vf = 2 * 10.3cm/s = 20.6cm/s
‘dv = 20.6 – 0 = 20.6cm/s
A = ‘dv / ‘dt = 20.6cm/s / 1.94s = 10.6cm/s/s
• What is the percent error in each?
answer/question/discussion:
vf % error = ((20.6 – 20)/20.6) *100 = 2.9%
a % error = ((10.6 – 10) /10.6) *100 = 5.7%
• If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion:
They are not the same.
• If the percent errors are different explain why it must be so.
answer/question/discussion:
The percent error for the acceleration is about double the percent error for the final velocity because the error in the final velocity adds to the error in the acceleration. It adds to the error in the acceleration because the error present in the calculation of the final velocity carries through when using the final velocity to calculate the acceleration. So the error is essentially double for the acceleration due to its being present in both calculations.
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1.5 hours
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Very good. The key is that the time interval is used twice, once in calculating velocities then again in calculating rate of change of velocity; this compounds the error.