Your 'cq_1_9.1' 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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Seed quest 9.1
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 = 20cm
v0 = 0cm/s
dt = 2 s
vave = ds/dt
vave = 20cm/2s
vave = 10cm/s
vave = (vf + v0)/2
10cm/s = (vf + 0cm/s)/2
20cm/s = vf + 0
vf = cm/s
aave = dv/dt
aave = (20cm/s - 0cm/s)/2s
aave = 20cm/s / 2 s
aave = 10 cm/s/s
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:
Here is where I have problems with percentages…how do I determine the actual time interval? I always start with what I know (100)
100 = x + .03x
100 = 1.03x
100/1.03 = x
97 = x
2 = x + .03x
2 = 1.03x
2/1.03 = x
The actual time is 1.94sec
You ended up with a time interval 3% shorter than the original. That won't make a lot of difference in your final percent, but it's worth thinking through what a 3% difference means.
Your given time interval is 2 seconds.
3% of 2 seconds is .03 * 2 s = .06 s.
So 3% longer than 2 s is 2.06 s.
This is in fact the same as 1.03 * 2 s = 2.06 s.
Vave = ds/dt
Vave = 20cm/1.94s = 10.3cm/s
Vave = (vf +v0)/2
10.3 * 2 = vf + 0
20.6cm/s = vf
Aave = dv/dt
Aave = 20.6 / 2 = 10.3 cm/ss
you would divide by the new time interval; for your calculations this would be 20.6 cm/s / (1.94 s) = 21.2 cm/s.
Note that 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 1.2 cm/s greater than the original result of 20 cm/s. The proportional error is therefore 1.2 cm/s / (20 cm/s) = .06, or 6%.
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.
What is the percent error in each?
answer/question/discussion:
Vave = 10cm/s implies +-1
1/10 = .1 = 10% error
Vf = 20cm/s implies +- 1
1/20 = .2 = 20% error
Aave = 10cm/s/s implies +-1
1/10 = .1 = 10% error
Vave = 10.3cm/s implies +- .1
.1/10.3 = .0097 = .009 = .9 % error
Vf = 20.6cm/s implies +- .1
.1/20.6 = .00485 = .005 = .5% erro
Aave = 10.3cm/s/s implies +-.1
.1/10.3 = .0097 = .01 = 1% error
If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion:
Because acceleration is change in velocity divided by change in time. If there is a certain percent error in one, it should be the same in the other.
Also because v0 = 0 then vAve and aAve are the same?
If the percent errors are different explain why it must be so.
answer/question/discussion:
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30 MINS
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Be sure you see my notes here. You aren't calculating percent errors correctly, and there is an error in your second calculation of acceleration.
&#Let me know if you have questions. &#