course phy 201
i am resubmitting cq 1 9.1 with answers to your comments
cq_1_91phy 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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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:
Ds = 20cm , v0 = 0 , dt = 2sec
Vave = ds/dt Vf = 2 * vAve
Vave = 20/2 Vf = 2 * 10m/s
Vave = 10m/s Vf = 20m/s
Vf = v0 + at
A = vf - v0 / t
A = 20 - 0 / 2
A = 10m/s ^2
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 * .03 = .06sec
Ds = 2 + .06 = 2.06sec
Vave = 9.71m/s
Vf = 19.42
A = 9.43 m/s^2
What is the percent error in each?
answer/question/discussion: 3 % for each
Vave = .03/10 * 100 = .3
Vf = .03 / 20 * 100 = .15
&&& vf = .03 * 20 = .6
vf error = 20 - .6 = 19.4 m/s &&&
The error in vf is 20 cm/s - 19.42 cm/s = .58 cm/s, which is about 3% of the original 20 cm/s.
&&& A = .03 * 10m/s = .3
You've already determine that
a = 9.43 m/s^2 .
This is compared to the original value a = 10 m/s^2.
The difference is (9.43 m/s^2 - 10 m/s^2) = .57 m/s^2.
As a percent of the original 10 m/s^2, this is
.57 m/s^2 / (10 m/s^2) = .057, or 5.7%. This is about 6%.
A = 20 - 0 / 2.06
A = 9.7 m/s^2 instead of using the 3% error in the time could I have just figured the percent error of the A itself? A = 10 * .03 = .3 ; A = 10 - .3 = 9.7m/s^2 &&&
What is the discrepancy in your acceleration, and what is this as a percent of the originally calculated acceleration?
&&& I used the original 3 % error in time since that is directly related to finding acceleration&&&
If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: because the two are directly related and are found by using the two same variables. Any change in vf or v0 will directly effect vave and a
If the percent errors are different explain why it must be so.
answer/question/discussion:
I used the same error of 3% for all variables but since the amounts are different then their percent errors will be different as well
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25min
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You appear to be assuming that there is a 3% error in the acceleration. This is not supported by your correct acceleration calculations, which indicate a 6% error.
See my notes and let me know if you don't understand completely. If you do, there's no need to resubmit, but do see my notes in the link below:
&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#