College Physics Problem 1.24

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.

(a) Calculate the percent uncertainty in the distance.

(b) Calculate the uncertainty in the elapsed time.

(c) What is the average speed in meters per second?

(d) What is the uncertainty in the average speed?


Part A

The percent uncertainty in the distance is

\displaystyle \%\:uncertainty_{distance}=\frac{25\:m}{42.188\:km}\times \frac{1\:km}{1000\:m}\times 100\%

\displaystyle \%\:uncertainty_{distance}=0.0593\%          ☚

Part B

The uncertainty in time is

\displaystyle \%\:uncertainty_{time}=\frac{1\:s}{9012\:s}\times 100\%

\displaystyle \%\:uncertainty_{time}=0.0111\%          ☚

Part C

The average speed is

\displaystyle average\:speed=\frac{42.188\:km}{9012\:s}\times \frac{1000\:m}{1\:km}

\displaystyle average\:speed=4.681\:m/s          ☚

Part D

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

\displaystyle \%\:uncertainty_{speed}=\%\:uncertainty_{distance}+\%\:uncertainty_{time}

\displaystyle \%\:uncertainty_{speed}=0.0593\%+0.0111\%

\displaystyle \%\:uncertainty_{speed}=0.0704\%

Therefore, the uncertainty in the speed is

\displaystyle \delta _{speed}=\frac{0.0704\%}{100\%}\times 4.681\:m/s

\displaystyle \delta _{speed}=0.003\:m/s          ☚