Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I had a motion detector record the position of a dynamics cart and automatically plot Position vs Time and Velocity vs Time plots in Logger Pro on the computer. If the instrument uncertainty in position x was given by $\pm0.05$ (or more generally any value $\pm e$) what will be the propagated instrument uncertainty on the second derivative of position, $\ddot{x}$-i.e. acceleration?

Thanks in advance

share|cite|improve this question
Uncertainty is not simply propagated. With uncertainty of $\pm 0.05$ on value, uncertainty of derivatives could be anything (just look at $e \cos(m t)$ for arbitrary large $m$). To compute velocity or acceleration, you have to approximate the whole curve. – Jean-Claude Arbaut Mar 25 '13 at 14:49

I think there must be some more assumptions to be made before your question can be answered. But here's my best shot:

If you calculate acceleration at $t_i$ as

$\frac{x_{i+1}-2x_i+x_{i-1}}{\Delta t^2}$

then each $x_n$ contributes $\Delta x$ to the uncertainty, which is then $4\Delta x/\Delta t^2$. But wait, there's more. Your recorded times are not going to be exact either. So the uncertainty in $\Delta t$, call it $\delta t$, will also contribute to an uncertainty in acceleration. But it doesn't sound like you are concerned with that right now, so I will stop here, but it's something to keep in mind.

share|cite|improve this answer
But $\Delta x/\Delta t^2$ is big. – Jean-Claude Arbaut Mar 25 '13 at 15:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.