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Phy 121
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cq_1_091
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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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Copy the problem below into a text editor or word processor.
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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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
‘ds = v0 * ‘dt + .5 * a * ‘dt^2
20cm = 0cm/s * 2s + .5 * a * (2s)^2
20cm = .5 * a * 4s^2
20cm = 2s^2 * a
a = 10cm/s^2
The acceleration of the given ball is 10cm/s^2.
vf = v0 +a *’dt
vf = 0cm/s + 10cm/s * 2s
vf = 0 cm/s + 20cm/s
vf = 20cm/s
The final velocity of the given ball is 20cm/s.
vAve = (v0 + vf) / 2
vAve = (0cm/s + 20cm/s) / 2
vAve = 20cm/s / 2
vAve = 10cm/s
The average velocity of the given ball is 10cm/s.
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Good, but note that while it's necessary to be able to use them as well as you do, you don't need all those equations to answer this question, and there is insight to be gained from direct reasoning.
The average velocity is 20 cm/ (2 s) = 10 cm/s.
Initial velocity is 0, acceleration is uniform, so final velocity has to be 20 cm/s.
Change in velocity is 20 cm/s, which occurs in 2 seconds, so acceleration is 20 cm/s / (2 s) = 10 cm/s^2.
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Very good.
Note:
Direct reasoning, which for this case is more efficient than the equations:
20 cm / (2 s) = 10 cm/s, ave vel.
Init vel. is zero so assuming unif. accel. final vel = 2 * ave vel. = 20 cm/s
Change in vel from 0 cm/s to 20 cm/s is 20 cm/s.
accel is change in vel / time interval = 20 cm/s / (2 s) = 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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
‘ds = v0 * ‘dt + .5 *a * ‘dt^2
20cm = 0cm/s * 1.94s + .5 * a * (1.94s)^2
20cm = .5 * a * 3.76s^2
20cm = 1.88s^2 * a
10.64cm/s = a
The actual acceleration is 10.64cm/s^2.
vf = v0 + a * ‘dt
vf = 0cm/s + 10.64cm/s^2 * 1.94s
vf = 0cm/s + 20.64cm/s
vf = 20.64cm/s
The actual final velocity is 20.64cm/s.
vAve = (v0 + vf) /2
vAve = (0cm/s + 20.64cm/s) / 2
vAve = 20.64cm/s / 2
vAve = 10.32cm/s
The actual average velocity is 10.32cm/s.
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• What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
The percent error in the acceleration is 6%.
The percent error of final velocity is 3%.
The percent error of the average velocity is 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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
The percent error is not the same for both the velocity and acceleration.
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• If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
This is because the there is a larger difference between the acceleration than that of the velocity.
&&&&The acceleration has a larger percent error because the acceleration is the velocity per unit time, whereas velocity is just the distance over time.
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The key is that in calculating acceleration we divide once by the time interval to get velocity information, then divide change in velocity to get the acceleration. Each division adds approximately a 3% error.
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*#&!
&#This looks good. See my notes. Let me know if you have any questions. &#