# College Physics Problem 1.24

## Solution:

### 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 $\displaystyle \delta _{speed}=\frac{0.0704\%}{100\%}\times 4.681\:m/s$ $\displaystyle \delta _{speed}=0.003\:m/s$          ☚