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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: ->->->->->->->->->->->-> :
vAve = 'ds / 'dt = 20 cm / 2 s = 10 cm/s
vf = ?
vAve = (vf + v0) /2
10 cm/s = vf / 2
vf = 20 cm/s
aAve = 'dv / 'dt = 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: ->->->->->->->->->->->-> :
'dt = 2.00 s +-3%
'dt = 2.00 s - (2*.03) = 1.94 s
vAve = 'ds /'dt = 20 cm / 1.94 s = 10.3 cm/s
vf = 20.6 cm/s
aAve = 10.6 cm/s^2
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve error = (10 cm/s - 10.3 cm/s) / 10.3 cm/s = 2.91%
t error = (2 s - 1.94 s) / 1.94 s = 3.09 %
vf error = (20 cm/s - 20.6 cm/s) / 20.6 cm/s = 2.91%
aAve error = (10 cm/s^2 - 10.6 cm/s^2) / 10.6 cm/s^2 = 5.66 %
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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: ->->->->->->->->->->->-> :
If the percent error is in the distance travelled
'ds +-3%
vAve = 'ds / 'dt = (10 cm / 2s - 10.3 cm / 2s) / (10.3 cm / 2 s) = 2.91%
aAve = 'dv / 'dt
percent error would be in the 'dv
v ave = vf + v0 / 2
5.1455 = vf / 2
vf = 10.291 cm/s
aAve error = ('dv / 'dt - 'dverror / 'dt) / ('dv error / 'dt) = 2.82% almost the same.
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
If the percent error is in 'dt then the percent error in VAve and AAve will be different.
dt = +-3%
vAve = (('ds / 'dt) - ('ds / 'dt+-3%)) / ('ds / 'dt +-3%) = 3 %
AAve = (('dv / 'dt) - ('dv / 'dt+-3%)) / ('ds / 'dt +-3%) = 1.6 %
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1/2 hour
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&#Very good responses. Let me know if you have questions. &#