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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 = 20cm / 2s = 10cm/s
vf = 2vAve - v0 = 2(10) - 0 = 20cm/s
a = `dv/`dt = vf - v0/2 = 20 - 0/2 = 10cm/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?
actual time: 3 = 2 - t/t * 100
.03 = 2 - t/t
.03t = 2-t
.03t + t = 2
t(.03+1) = 2
t = 1.94175s
vAve = `ds/`dt = 20cm/1.94175s = 10.3cm/s
vf = 2vAve - v0 = 2(10.3) - 0 = 20.6cm/s
a = 20.6 - 0/1.94175 = 10.6cm/s^2
What is the percent error in each?
vAVe = abs.val. * [10 - 10.3/10.3] * 100 = 2.91% = 3%
vf = [20 - 20.6/20.6] * 100 = 2.91% = 3%
a = 5.66 = 6%
If the percent error is the same for both velocity and acceleration, explain why this must be so.
The percent error is not the same for both velocity & acceleration.
If the percent errors are different explain why it must be so.
The percent errors are different because the error in the time interval is magnified when carried through equations for velocity to the acceleration determination.
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20min.
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Please respond relative to goal of C.
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