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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

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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
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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.

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