Phy 231
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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:
v0 = 0 cm/s
'ds = 20 cm
'dt = 2 s
vAve = 'ds / 'dt = 20 cm / 2 s = 10 cm/s
vf = 2 * vAve = 2 * 10 cm/s = 20 cm/s (this is only true since v0 = 0 cm/s.
a = (vf - v0) / 'dt = 20 m/s / 2 s = 10 m/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:
'dt = 2 s * 1.03 = 2.06 s
vAve = 'ds / 'dt = 20 cm / 2.06 s = 9.71 cm/s
vf = 2 * 9.71 cm/s = 19.42 cm/s
a = (vf - v0) / 'dt = 19.42 cm/s / 2.06 s = 9.43 cm/s/s
What is the percent error in each?
answer/question/discussion:
The difference between the two vAve values is 10 cm/s - 9.71 cm/s = 0.29 cm/s, which is a difference of 0.0298, or approx 3%.
The difference between the two a values is 10 m/s/s - 9.43 m/s/s = 0.57 m/s/s, which is a difference of 0.0604, or approx 6%.
If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: I calculated different percent errors
If the percent errors are different explain why it must be so.
answer/question/discussion: This seems logical because velocity and acceleration are proportional, not equal.
A 3% increase the same as a change by factor 1.03. For example, if we multiply the 2 second time interval by 1.03, we get 2.06 seconds, the same result we obtained in solving the present problem.
If we divide a number by 1.03 the result is very nearly the same as if we had multiplied it by .97. This is because 1 / 1.03 = .97, correct to two sigfificant figures. The equality is not exact, because to 4 significant figures 1 / 1.03 = .97087; however it is very close. So a 3% increase in the denominator of a division corresponds to a 3% decrease in the quotient.
Now when we divide the displacement by the new time interval, which is 1.03 times the original time interval, we see why the result is very nearly 3% less than the original result. Thus the average velocity is 3% less than the original average velocity.
When we double the resulting average velocity to get the final velocity, it is also 3% less than the original. Our change in velocity is thus 3% less than the original.
When we divide this reduced change in velocity by the new time interval, we see that the result is reduced by (very nearly) another 3%. Thus when we calculate our acceleration, the result is very nearly 6% less than the original.
We can also see this by considering the formula for this calculation. When v0 = 0, the third equation of motion is easily solved for the acceleration. We obtain a = 2 `ds / (`dt)^2.
When `dt is increased by 3%, the denominator becomes (1.03)^2 times as great as in the original calculation. (1.03)^2 = 1.0609, so that 1.03^2 is close to 1.06. Thus 1 / (1.03)^2 = 1 / (1.06). 1 / 1.06 is close to .94; that is, a 6% increase in a quantity is associated with (nearly) a 6% decrease in its reciprocal.
So when `dt is increased by 3%, (`dt)^2 is increased by about 6% and 1 / (`dt)^2 is decreased by about 6%. Thus our result a = 2 `ds / (`dt)^2 is reduced by about 6%.
In general, if epsilon is a number much smaller that 1 (e.g., in this problem epsilon would be our 3%, or .03, which is much less than 1) the following apply:
1 / (1 + epsilon) = 1 - epsilon and
1 / (1 - epsilon) = 1 + epsilon.
(1 + epsilon)^n = 1 + n * epsilon, as long as n isn't too big.
In particular this means that to the extent that epsilon is small, the following hold:
sqrt(1 + epsilon) = 1 + epsilon / 2 and
(1 + epsilon)^2 = 1 + 2 epsilon.
In words:
The reciprocal of a number close to 1 is just about as close to 1, but 'on the other side' of 1 (i.e., if the number is smaller than 1 its reciprocal is bigger than 1 by about the same amount, and vice versa).
The square root of a number close to 1 is about twice as close to 1 and 'on the same side' of 1.
The square of a number close to 1 is about twice as far from 1 and 'on the same side' of 1.
These results are easily derived using basic algebra.
They are also derived in a broader context using the calculus of differentials (e.g., d(x^2) = 2 x dx; for x near 1 the change in x^2 is therefore about 2 times the change in x; d(sqrt(x)) = 1 / (2 sqrt(x)) dx, so when x is near 1 the change in sqrt(x) is about 1/2 the change in x; d(1/x) = -1 / x^2 dx so when x is near 1 the change in 1/x is about equal to the change in x, but opposite in sign).
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15 mins
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