PHY 201
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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Seed 9.1
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:
v0 = 0 cm/s
'ds = 20 cm
'dt = 2
'ds = (vF + v0)/2 * 'dt
2'ds/'dt = vF + v0
vF = 2'ds/'dt - v0
= 2(20)/2
vF = 20 cm/s
vAve = 20+0 /2 = 10 cm/s
a = 'dv/'dt
a = 20/2
a = 10 cm/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:
.97 * 2 = 1.94 seconds
vF = 40/1.94
vF = 20.6 cm/s
a = 20.6/1.94
a = 10.6 cm/s/s
What is the percent error in each?
answer/question/discussion:
For velocity it changed by .6 cm/s
.6/20 = .03 = 3%
For acceleration it changed by .6 cm/s
.6/10 = .058 = 6%
If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion:
If the percent errors are different explain why it must be so.
answer/question/discussion:
The percent error is basically double for acceleration than what it is for vF. If you think about it, when solving for the new value of vF we divided by 'dt which which was 3% less than last time. Therefore the new vF will increase by 3%.
When we solved for the new acceleration, we not only divided by the 3% less time, but we also divided it into the vF that increased by 3%. This is why its error was double.
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30 minutes
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Excellent work.