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PROBLEM:
(a) A car speedometer has a 5.0% uncertainty. What is the range of possible speeds when it reads 90 km/h?
(b) Convert this range to miles per hour. (1 km=0.6214 mi)
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SOLUTION:
Part A
The uncertainty in the velocity of the car is computed as
\begin{align*}
\delta _v & =\frac{5.0\:\%}{100\:\%}\times 90.0\:\text{km/hr} \\
& = 4.5\:\text{km/hr}
\end{align*}Therefore, the range of the possible speeds is
\begin{align*}
\text{Range} & =90.0\:\pm 4.5\:\text{km/hr} \\
\text{Range} & :85.8\:\text{km/hr}\:-\:94.5\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}So the range of the possible speeds is 85.5 km/hr to 94.5 km/hr.
Part B
Convert the range to mi/h
For 85.5 km/hr
\begin{align*}
85.5\:\text{km/hr}=\left(85\:\frac{\text{km}}{\text{hr}}\right)\left(\frac{0.6214\:\text{mi}}{1\:\text{km}}\right)=53.13\:\text{mi/hr}
\end{align*}For 94.5 km/hr
94.5\:\text{km/hr}=\left(94.5\:\frac{\text{km}}{\text{hr}}\right)\left(\frac{0.6214\:\text{mi}}{1\:\text{km}}\right)=58.72\:\text{mi/hr}Therefore, the range can be represented as 53.13 mi/hr to 58.72 mi/hr.
\text{Range} : 53.13 \ \text{mi/hr} - 58.72 \ \text{mi/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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