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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: ->->->->->->->->->->->-> :
‘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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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 and that of the velocity.
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This is circular. You're basically saying that it's bigger because it's bigger.
What is the root reason that it's so?
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*#&!*#&!
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You're doing really well, but see if you can come up with a good explanation for why the percent errors differ. (see my note)
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