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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.
** 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 20 cm/2 seconds or 10 cm/s
Because it started from rest, the final velocity would be (0 + vf)/2 = 10 cm/s or 20
cm/s
The acceleration is (20 cm/s - 0 cm/s)/2 s or 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 average velocity would be 20 cm/1.94 s = 10.31 cm/s
This would make the final velocity (0 + vf)/2 = 10.31 cm/s or 20.62 cm/s
That would make the acceleration (20.62 cm/s - 0 cm/s)/1.94 s = 10.63 cm/s^2
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
Velocity:
From 10 cm/s to 10.31 cm/s, the % error is 3%
(Obtained by dividing 10 by 10.31, subtracting that from 1)
Final velocity:
From 20 cm/s to 20.62 cm/s, the % error is 3%
(Obtained by dividing 20 by 20.62, subtracting that from 1)
Acceleration:
From 10 cm/s^2 to 10.63 cm/s^2, the % error is 6%
(Obtained by dividing 10 by 10.63, subtracting that from 1)
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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 did not find that the percent error for both velocity and acceleration were the same.
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
In order to get the first acceleration answer, I divided the change in velocity by the
time, which was 20 cm/s divided by 2 s.
For the second answer, The change in velocity had gone up to 20.62 cm/s while the change
in time had gone down to 1.94 s. This was a 3% change up for velocity and a 3% change
down for time, which resulted in an error of 6%.
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*#&!
@&
You have described how you got the result, and everything is correct.
You haven't explained why this happens.
The key is that you have divided twice by the time interval, once in calculating average velocity, and again in calculating the acceleration. Each division changed the result by about 3%.
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&#Good responses. See my notes and let me know if you have questions. &#