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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: ->->->->->->->->->->->-> :
The acceleration is 3.16 cm/s^2. The vf is 6.32 cm/s. The average velocity is 3.16cm/s also.
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Now that I rework the problem, I get completely different answers. Seeing as the displacement is 20 cm and it's accelerating at a constant rate. I used the formula ds = (vf+v0) / 2 * dt . So if I fill in with what I have, it's 20 = (vf + 0)/2 * 2. Working this out, I got that vf = 20. So the Dv/Dt = a. That is 10 cm/s^2 which should seem right because If it accelerates 10 cm per second per second, it would takes 2 seconds to go 20 cm.
@& If it accelerates at 10 cm/s/s then in 2 sec its velocity would change by 20 cm/s.
If the initial velocity happens to be zero, as is the case here, then by coincidence the object also moves 20 cm during that time interval. However that result does not follow from the acceleration alone, but from the initial velocity, the time interval and the acceleration.*@
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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 acceleration would then be 3.26 cm/s^2 and the vf would be 6.52 cm/s.
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The 3% time difference should make the acceleration 10.31 cm/s^2. The vf would then be 20.6 cm/s. I got this by taking the time and multiplying it by .03. Then I went back and put in 20cm/s / 2.06 since the time is now 3% longer. For the final velocity, I added 3% to it.
@& Good, but you overlooked the fact that the final velocity is now 20.6 cm/s, not 20 cm/s.*@
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
The percent error for acceleration and average velocity is 3.06%. The percent error for final velocity is 3.06% as well.
&&&& If we changed the time by 3% , then the percent error should be 3% in both.
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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: ->->->->->->->->->->->-> :
I think this is because the average velocity and acceleration are the same. For vf, I think the reason for the percent error being the same is because the answer for vf derives from the avg velocity and acceleration.
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
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15 minutes
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This is my revised version.
@& Good, but you did overlook one thing. Be very sure you understand. The uncertainty in the acceleration is not 3% but 6%.*@
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new 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. &#