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Q. A ball falls freely from rest with an acceleration g. The variation with time t of its displacement s is given by s = 1/2 gt^2. The percentage uncertainty in the value of t is ±3% and that in the value of g is ±2%. Calculate the percentage uncertainty in the value of s.

A. Δs/s = Δg/g + 2 Δt/t = 0.02 + (2 × 0.03) = 0.08 = ±8%

Question of the day

Can you explain why in the answer the fractional uncertainty for time (Δt/t) is multiplied by 2 before being added to the fractional uncertainty for gravity (Δg/g)?

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  • $\begingroup$ Suppose that you just have a quantity $q = t^2$ Then differentiating this expression gives $\Delta Q = 2 t \Delta t \Rightarrow \frac{\Delta q}{q} = 2 \frac {\Delta t}{t}$. $\endgroup$ – Farcher May 12 '18 at 7:52
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$gt^2=g*t*t$, so the fractional uncertainty is $0.02+0.03+0.03=0.02+(2\times0.03)$. Alternatively, as what Farcher has said, you can differentiate, and the $2$ will be brought down from the power to become the coefficient.

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