PHY 231
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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 ball's average velocity is 10 cm/s; its final velocity is 20 cm/s. It's acceleration is 10 cm/s^2.
Calculations Below:
vAve = 'ds / 'dt
vAve = 20 cm/ 2 s
vAve = 10 cm/s
vAve = ( v0 + vf ) / ( 2 )
10 cm/s = ( 0 cm/s + vf ) / ( 2 )
20 cm/s = 0 cm/s + vf
20 cm/s = vf
vf^2 = ( v0^2 )+ 2 * a * `ds
400 cm^2/s^2 = ( 0 cm^2/s^2 ) + ( 2 * a * 20 cm )
400 cm^2/s^2 = 2 * a * 20 cm
200 cm^2/s^2 = a * 20 cm
10 cm/s^2 = a
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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: ->->->->->->->->->->->-> :
If the time interval is three percent longer, the overall time elapsed is 2.06 seconds.
The final velocity drops by .6 cm/s to 19.4 cm/s. The acceleration drops by approximately .6 cm/s^2 to 9.409 cm/s^2.
Calculations:
2 seconds ( 3/100 ) = .06 ( three percent of 2 )
2 + .06 = 2.06 seconds
vAve = 'ds / 'dt
vAve = 20 cm/ 2.06 s
vAve = 9.708737864 cm/s or about 9.7 cm/s
vAve = ( v0 + vf ) / ( 2 )
9.7 cm/s = ( 0 cm/s + vf ) / ( 2 )
19.4 cm/s = 0 cm/s + vf
19.4 cm/s = vf
vf^2 = ( v0^2 )+ 2 * a * `ds
376.36 cm^2/s^2 = ( 0 cm^2/s^2 ) + ( 2 * a * 20 cm )
376.36 cm^2/s^2 = 2 * a * 20 cm
188.18 cm^2/s^2 = a * 20 cm
9.409 cm/s^2 = a
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• What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
The percent error in the final velocity is approximately three percent, while the percent error in the acceleration is approximately six percent.
Calculations:
% Error in Final Velocity:
100 - (( Actual / Theoretical ) * 100 ) =
100 - (( 19.4 cm/s / 20 cm/s ) * 100 ) =
100 - (( .97 ) * 100 ) =
100 - ( .97 * 100 ) =
100 - 97 = 3%
% Error in Acceleration:
100 - (( Actual / Theoretical ) * 100 ) =
100 - (( 9.409 cm/s^2 / 10 cm/s^2 ) * 100 ) =
100 - (( .9409 ) * 100 ) =
100 - ( .9409 * 100 ) =
100 - 94.09 = 5.91%
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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: ->->->->->->->->->->->-> :
Although the differences in-between the actual and theoretical final velocity and acceleration values are approximately the same at .6 respectively for both, neither has the same percent error.
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• If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
The reason the percent error for the acceleration is almost twice as great as the final velocity is because of the law of proportions. Even though they both differ by about .6 less their real values, resulting more or less from the inherent error in estimation/significant figures/rounding, the absolute difference of .6 accounts for a portion/value half as great in the acceleration versus the final velocity, doubling the percent error in the acceleration.
#$&*
________________________________________
Copy and paste your work into the box below and submit as indicated:
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25 Minutes
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Good explanation, but there is an alternative explanation more related to the process. Check out the link. I believe yOu will understand; no need for a revision unless I'm wrong about that
&#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. &#