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Phy 231
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: ->->->->->->->->->->->-> :
Use the formula: 'ds = (vf + v0)/2 *'dt
You can use this formula sice the ball is being uniformly accelerated.
20 = (vf + 0)/2 * 2
Divide both sides by 2
10 = (vf + 0)/2
Multiply both sides by 2. We get vf = 20 cm
Now we can get the average velocity: vave = (vf + v0)/2
vave = (20 + 0)/2
Vave = 10
a = 'dv /'dt
a = (20) / 2
a = 10 m/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: ->->->->->->->->->->->-> :
To find the actual time interval: .03 *2 = .06. The actual time interval is 2.06 seconds
The final velocity is: 20 = (vf + 0)/2 *2.06
Divide both sides by 2.06
9.71 = (vf)/2
Multiply both sides by 2
vf = 19.4
a = 'dv / 'dt
a = 20/2.06
a = 9.7 cm/s^2
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
The percent error in velocuty is 20 - 19.4 = 6% error
The percent error in acceleration is (10 -9.7) = 3% error
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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: ->->->->->->->->->->->-> :
They were not the same.
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
The percent error for acceleration is greater because you first find velocity, and then divide by the error in velocity to find acceleration. Therefore, the acceleration would have a greater error in the answer.
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&#Good responses. Let me know if you have questions. &#