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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: ->->->->->->->->->->->-> :
Average velocity is 10 cm/s , final velocity is 20 cm/s , acceleration is 10 cm/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: ->->->->->->->->->->->-> :
Final is 19.417 , Acceleration is 9.426
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
For the final 3% , For the acceleration 6%
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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: ->->->->->->->->->->->-> :
No
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
It must be because the avg velocity was used to get acceleration and then acceleration was gotten by two modified values
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That's pretty much correct.
Basically, to get average velocity you divide displacement by time interval. You base your final velocity on that result, then divide by the time interval once more.
With each division by the time interval, with its 3% uncertainty, you get a 3% uncertainty. When you combine the 3% uncertainties you get a 6% uncertainty.
In terms of calculus, which isn't required for your course but of which you have a solid knowledge, we can reason this out in terms of differentials. I'm including this in case you find it of interest:
a = 2 s / t^2
where I'm using s and t instead of `ds and `dt, so the derivatives will be clearer.
Regarding a = 2 s / t^2 as a function of t, the differential is
da = da / dt * dt = -4 s / t^3 dt = 2 * (-2 s / t^2) * dt / t = a * (2 dt/t)
so that
da / a = 2 dt / t.
dt is regarded as the uncertainty in t, so dt / t is the uncertainty in t, as a proportion of t.
Similarly da / a is the uncertainty in a , as a proportion of a.
The uncertainty in a as a proportion of a is double the uncertaity in t as a proportion of t.
The percent difference is just that proportion multiplied by 100.
So the percent uncertainty in a is double the percent uncertainty in t.
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&#Very good data and responses. Let me know if you have questions. &#