cq1_09_1 solution and discussion

The following is a solution to the given problem.

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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?

Since initial velocity is 0, the final velocity is double the average velocity, or 20 cm/s.

The change in velocity is therefore 20 cm/s, and the average rate of change of velocity with respect to clock time is `dv / `dt = (20 cm/s) / (2 s) = 10 cm/s^2.

(see note at end for discussion of equation-based solution to this question)

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?

If the 2-sec time interval is 3% longer then since .03 * 2 = .06, the time interval is 2.06 seconds. This yields an average velocity of 20 cm/s / (2.06 s) = 9.7 cm/s and a final velocity of 19.4 cm/s.

19.4 cm/s differs from the original 20 cm/s by .6 cm/s, which is (.6 cm/s) / (20 cm/s) = .03 or 3% of the 20 cm/s result.

The resulting acceleration would be `dv / `dt = 19.4 cm/s / (2.06 s) = 9.4 cm/s.
This differs from the original 10 cm/s^2 acceleration by .6 cm/s^2, which is (.6 cm/s^2) / (10 cm/s^2) = .06, or 6 percent, of the original acceleration.

Thus a 3% change in time interval results in a 3% change in the calculated final velocity, and a 6% change in the calculated acceleration.

What is the percent error in each?

As demonstrated in the preceding, the percent error in velocity is 3% and the percent error in acceleration is 6%.

The calculation of velocity involves division by velocity. Division by a quantity with error 3% results in an error of approximately 3%.

To calculate acceleration we first calculate change in velocity, which involves a 3% error. Then we divide by the time interval, which results in another 3% error for a total of 6%.

The reasoning behind this is not difficult to understand. If we divide `ds by 1.03 * `dt, we get

`ds / (1.03 `dt) = `ds / `dt * (1 / 1.03) = `ds /`dt * .97, i.e., we get .97 * vAve , or 97% of the average velocity based on the ‘central’ value of `dt. Thus our 3% error.

Since we started from rest and accel is uniform, vf = 2 * vAve and our vf, and hence our `dv, will be .97 * vf, still off by 3%.

To find acceleration we divide our `dv, with its 3% error, by 1.03 * `dt. In the same manner as before, this introduces an additional error of approximately 3%, so that the error in finding acceleration is about 6%.

If epsilon is a small quantity, then to the extent that epsilon is small we have the following:

·        1 / (1 + epsilon) = 1 – epsilon

·        1 / (1 - epsilon) = 1 + epsilon

·        (1 +- epsilon)^2 = 1 +- 2 * epsilon

·        (1 +- epsilon)^n = 1 – n * epsilon

·        1 / (1 +- epsilon)^2 = 1 -+ 2 * epsilon

·        1 / (1 +- epsilon)^n = 1 -+ n * epsilon

These results are easily verified for epsilon values of .01, .02, .03, etc.. Significant discrepancies from these approximations become apparent when epsilon isn’t particularly small—around, say epsilon = .10.

Equation-based solution to first question.

If initial velocity is 0 and average velocity is 10 cm/s, then final velocity is 20 cm/s. You don't need formulas to reason this out, but it's very desirable to also be able to use the formulas.

You quoted the equation `ds = (vf + v0) / 2 + 2 `dt. This equation is not valid:

Among other things, the units of `ds are meters (units of length) while the unit of (vf + v0) / 2 is m/s (length units / time units), and the unit of 2 `dt is sec (time units). So you couldn't add (vf + v0) / 2 to 2 `dt; since the units are different these would be unlike terms.

If you multiplied (vf + v0) / 2 by 2 `dt you would get units m/s * s = m (length units) and your two sides would have the same units, but the formula would still be incorrect, since `ds = (vf + v0) / 2 * `dt (first equation of motion.

If you are going to use equations, use the four equations of uniformly accelerated motion as your basis.

An equation that applies to this situation, in which you know v0, `ds and `dt, is the first equation `ds = (vf + v0) / 2 * `dt. I think this is the equation you intended.

To solve this equation for vf:

Starting with

`ds = (vf + v0) / 2 * `dt
we multiply both sides by 2 / `dt to get
`ds * 2 / `dt = (vf + v0) / 2 * `dt * (2 / `dt), which simplifies to
(vf + v0) = 2 `ds / `dt. Then subtract v0 from both sides to get
vf = 2 `ds / `dt - v0.

Substitute the given information and you will get vf = 20 cm/s.

The following is a solution to the given problem.

Please compare the given solution with your solution and submit a self-critique of any error(s) in your solutions, and/or and additional questions or comments you might have.

Simply copy your posted document into a text editor and insert revisions, questions, and/or self-critiques, marking your insertions with ####. Submit using the Submit Work Form.

Since initial velocity is 0, the final velocity is double the average velocity, or 20 cm/s.

The change in velocity is therefore 20 cm/s, and the average rate of change of velocity with respect to clock time is `dv / `dt = (20 cm/s) / (2 s) = 10 cm/s^2.

If the 2-sec time interval is 3% longer then since .03 * 2 = .06, the time interval is 2.06 seconds. This yields an average velocity of 20 cm/s / (2.06 s) = 9.7 cm/s and a final velocity of 19.4 cm/s.

19.4 cm/s differs from the original 20 cm/s by .6 cm/s, which is (.6 cm/s) / (20 cm/s) = .03 or 3% of the 20 cm/s result.

The resulting acceleration would be `dv / `dt = 19.4 cm/s / (2.06 s) = 9.4 cm/s. This differs from the original 10 cm/s^2 acceleration by .6 cm/s^2, which is (.6 cm/s^2) / (10 cm/s^2) = .06, or 6 percent, of the original acceleration.

Thus a 3% change in time interval results in a 3% change in the calculated final velocity, and a 6% change in the calculated acceleration.

As demonstrated in the preceding, the percent error in velocity is 3% and the percent error in acceleration is 6%.

The calculation of velocity involves division by velocity. Division by a quantity with error 3% results in an error of approximately 3%.

To calculate acceleration we first calculate change in velocity, which involves a 3% error. Then we divide by the time interval, which results in another 3% error for a total of 6%.

The reasoning behind this is not difficult to understand. If we divide `ds by 1.03 * `dt, we get

`ds / (1.03 `dt) = `ds / `dt * (1 / 1.03) = `ds /`dt * .97, i.e., we get .97 * vAve , or 97% of the average velocity based on the ‘central’ value of `dt. Thus our 3% error.

Since we started from rest and accel is uniform, vf = 2 * vAve and our vf, and hence our `dv, will be .97 * vf, still off by 3%.

To find acceleration we divide our `dv, with its 3% error, by 1.03 * `dt. In the same manner as before, this introduces an additional error of approximately 3%, so that the error in finding acceleration is about 6%.

If epsilon is a small quantity, then to the extent that epsilon is small we have the following:

· 1 / (1 + epsilon) = 1 – epsilon

· 1 / (1 - epsilon) = 1 + epsilon

· (1 +- epsilon)^2 = 1 +- 2 * epsilon

· (1 +- epsilon)^n = 1 – n * epsilon

· 1 / (1 +- epsilon)^2 = 1 -+ 2 * epsilon

· 1 / (1 +- epsilon)^n = 1 -+ n * epsilon

These results are easily verified for epsilon values of .01, .02, .03, etc.. Significant discrepancies from these approximations become apparent when epsilon isn’t particularly small—around, say epsilon = .10.

The resulting acceleration would be `dv / `dt = 19.4 cm/s / (2.06 s) = 9.4 cm/s.
This differs from the original 10 cm/s^2 acceleration by .6 cm/s^2, which is (.6 cm/s^2) / (10 cm/s^2) = .06, or 6 percent, of the original acceleration.