Phy 121
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.
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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: ->->->->->->->->->->->-> : final v=10m/s. avg v=5m/s. a=5m/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: ->->->->->->->->->->->-> :
9.7m/s and 4.7m/s^2
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> : 3%
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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: ->->->->->->->->->->->-> : Because the uncertainty was that percentage for time and time is directly related to acceleration and velocity, and thus affect their uncertainty.
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
They arent.
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10mins.
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You have made a few errors, though in general you understand the scheme and are getting a number of solutions correct.
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
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.
** **
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: ->->->->->->->->->->->-> :
I am given the displacement 20 cm, the initial velocity 0 cm/s, and the time interval 2 seconds, then I can solve for the rest of the variables. I am given the time interval and the displacement, so I can solve for the average velocity by: 20 cm / 2 s = 10 cm/s. Then if I know the average velocity and was given the initial velocity, then the final velocity is twice that of the average velocity: 10 m/s * 2 = 20 cm/s. Then, the change in velocity is 20 cm/s as well. Finally, if I know the change in velocity and the change in time, then I can calculate the acceleration: 20 cm/s / 2 s = 10 cm/s^2. Therefore the average velocity is: 10 m/s, the final velocity is 20 cm/s, and the 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: ->->->->->->->->->->->-> :
The time is at an error, which it is supost to be 3 % longer than the 2 seconds, so: 2 * .03 = .06 + 2 = 2.06 seconds. Then, if the time interval is 2.06 s then the final velocity is: 20 cm/s / 2.06 s = 9.71 cm/s * 2 = 19.42 cm/s / 2.06 s = 9.43 cm/s^2. So, the final velocity is 19.42 cm/s and the acceleration is 9.43 cm/s^2.
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
The percent error in the average velocity is the number without the error divided by the number with the error: 10 cm/s / 9.71 cm/s = 1.03 is the percent error of the average velocity. Then, the percent error for the acceleration is: 10 cm/s^2 / 9.43 cm/s^2 = 1.06.
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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: ->->->->->->->->->->->-> :
The percent error is not exactly the same, but approximately the same. The percent error is the same because if the error was on the time, the time affects both the average velocity and the acceleration. So, if will change both of them at the same rate.
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
Mine are a little off, but still very much alike. I am not sure how precise you want the percent errors to be.
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30 minutes
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June 24 around 3:30 pm