College Physics Problem 1.21

A person measures his or her heart rate by counting the number of beats in 30 s. If 40±1  beats are counted in 30±0.5 s, what is the heart rate and its uncertainty in beats per minute?


In order to compute for the heart rate in beats per minute, we need to solve for the base. The base is

\displaystyle \frac{40\:beats}{30\:sec\:}\times \frac{60\:sec}{1\:min}=80\:beats/min

Then we compute for the percent uncertainty by combining the uncertainties of the number of beats and time. That is

\displaystyle \%\:uncertainty=\frac{1\:beat}{40\:beats}\times 100\%+\frac{0.5\:s}{30.0\:s}\times 100\%

\displaystyle \%\:uncertainty=2.5\%+1.7\%

\displaystyle \%\:uncertainty=4.2\%

Based on this percent uncertainty, we compute for the tolerance

\displaystyle \delta _A=\frac{\%\:uncertainty}{100\:\%}\times A

\displaystyle \delta _A=3.3\:beats/min

Therefore, the heart rate is

\displaystyle 80\pm 3\:beats/min          ☚