Chapter 12.6, Problem 44PPSE

Solution

To find: The time taken by the boat to return to home port.

The time taken by the boat to return to home port is 6.75 hours i.e., 6 hours 45 minutes.

Given information:

Given, a fishing boat leaves its home port and travels 150 mi directly east. It then changes course and travels 40 mi due north. The average speed of the return trip is 23 mi/h.

Calculation:

The boat is 150 mi east and 40 mi north of the home port.

The vector representing the position of the boat from the home port is $〈150,40〉$

The direct distance between the boat and the home port is given by the magnitude of the vector.

The magnitude of the vector $〈150,40〉$ is

  $\begin{array}{l}d=\sqrt{{\left(150\right)}^2+{\left(40\right)}^2}\\ d=\sqrt{22500+1600}\\ d=\sqrt{24100}\\ d\approx 155.24\end{array}$

The time taken at a speed of 23 mi/h is given by:

  $\begin{array}{l}t=\frac{d}{s}\\ t=\frac{155.24}{23}\\ t=6.75\text{ hours}\end{array}$

Conclusion:

The time taken by the boat to return to home port is 6.75 hours i.e., 6 hours 45 minutes.