Physics PDF solutions | Engineering Mathematics and Sciences

A marathon runner completes a 42.188-km course in 2 h, 30 min, and 12 s. There is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the elapsed time.

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SOLUTION:

Part A

The percent uncertainty in the distance is

\begin{align*} \text{\%\:uncertainty}_{\text{distance}} & =\frac{25\:\text{m}}{42.188\:\text{km}}\times \frac{1\:\text{km}}{1000\:\text{m}}\times 100\% \\ & =0.0593\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Part B

The uncertainty in time is

\begin{align*} \text{\%\:uncertainty}_{\text{time}} & =\frac{1\:\text{s}}{9012\:\text{s}}\times 100\% \\ & =0.0111\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Part C

The average speed is

\begin{align*} \text{average speed} & =\frac{42.188\:\text{km}}{9012\:\text{s}}\times \frac{1000\:\text{m}}{1\:\text{km}} \\ & = 4.681\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Part D

The percent uncertainty in the speed is the combination of uncertainties of distance and time.

\begin{align*} \text{\%\:uncertainty}_{\text{speed}} & =\text{\%\:uncertainty}_{\text{distance}}+\text{\%\:uncertainty}_{\text{time}} \\ & =0.0593\%+0.0111\% \\ & =0.0704\% \\ \end{align*}

Therefore, the uncertainty in the speed is

\begin{align*} \delta _{speed} & =\frac{0.0704\%}{100\%}\times 4.681\:\text{m/s} \\ & = 0.003\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}