phy 201
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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 = 20cm , v0 = 0 , dt = 2sec
Vave = ds/dt Vf = 2 * vAve
Vave = 20/2 Vf = 2 * 10m/s
Vave = 10m/s Vf = 20m/s
Vf = v0 + at
A = vf - v0 / t
A = 20 - 0 / 2
A = 10m/s ^2
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: 2 * .03 = .06sec
Ds = 2 + .06 = 2.06sec
Vave = 9.71m/s
Vf = 19.42
A = 9.43 m/s^2
What is the percent error in each?
answer/question/discussion: 3 % for each
Vave = .03/10 * 100 = .3
Vf = .03 / 20 * 100 = .15
The error in vf is 20 cm/s - 19.42 cm/s = .58 cm/s, which is about 3% of the original 20 cm/s.
A = .03 / 10* 100 = . 3
What is the discrepancy in your acceleration, and what is this as a percent of the originally calculated acceleration?
If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: because the two are directly related and are found by using the two same variables. Any change in vf or v0 will directly effect vave and a
If the percent errors are different explain why it must be so.
answer/question/discussion:
I used the same error of 3% for all variables but since the amounts are different then their percent errors will be different as well
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25min
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You did almost everything correctly, but you didn't calculate the percent errors correctly.
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