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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?
vAve = ‘ds/’dt, vAve= 10cm/s; vf = (2’ds/’dt),vf = 20cm/s; a=(vf^2)/2’ds, a = 10cm/s^2
answer/question/discussion:
• 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?
‘dt / 1.03 = 1.94s; vf = (2’ds/’dt), vf = 20.6 cm/s; a = (vf^2)/2’ds, a = 10.6cm/s^2
answer/question/discussion:
• What is the percent error in each?
Vf has a 3% error and acc has a 6% error. 20.6cm/s/20cm/s = 1.03 which is the percentage multiplier or divider. 10.6cm/s^2/10cm/s^2 = 1.06 which is the percentage multiplier or divider.
answer/question/discussion:
• If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion:
• If the percent errors are different explain why it must be so.
The percent errors differ between final velocity and acceleration because of the fact that velocity changes over time and acceleration is uniform.
answer/question/discussion:
The decreased time interval results in a 3% change in average velocity, which is then divided by the decreased time interval, enhancing the effect of the change.
This is why a 3% error in time interval results in a 6% error in the calculated acceleration.
In terms of the equations, v0 is 0 so the third equation of motion gives you `ds = 1/2 a `dt^2, so that a = 2 `ds / `dt^2.
If `dt changes by factor 1.03, then the calculated value of a will change by 1 / 1.03^2 = 1 / 1.06 = .94.
Note that 1.03^2 is about 1.06; the square of a number close to 1 is about twice as far from 1.
1 / 1.06 = .94. The reciprocal of a number close to 1 is about equally close to 1, but 'on the other side'. That is, 1.06 is .06 units from 1, and .94 is .06 units
from 1. Similaraly 1 / .94 = 1.06.
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.5 hours
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Would like to have feedback. Thanks, Joshua
Perhaps I could use some clarification on the last question.
&#Your work looks good. See my notes. Let me know if you have any questions. &#