## College Physics (4th Edition)

(a) The boat's speed relative to the bank is $1.80~mi/h$ (b) It takes 0.80 hours to cross the river. (c) The boat will finish 0.80 miles upstream. (d) The boat should have headed at an angle of $32.2^{\circ}$ upstream.
(a) Let the upstream direction be the +x-direction and let the direction across the river be the +y-direction. We can find find the boat's velocity relative to the riverbank in component form: $v = [(3.00~sin~60.0^{\circ} - 1.60)~\hat{i} + 3.00~cos~60.0^{\circ}\hat{j}~]~mi/h$ $v = (1.00~\hat{i} + 1.50~\hat{j}~)~mi/h$ We can find the boat's speed relative to the bank: $\sqrt{(1.00~mi/h)^2+(1.50~mi/h)^2} = 1.80~mi/h$ The boat's speed relative to the bank is $1.80~mi/h$ (b) We can use the y-component of the velocity to find the time it takes to cross the river: $t = \frac{1.20~mi}{1.50~mi/h} = 0.80~h$ It takes 0.80 hours to cross the river. (c) We can use the x-component of the velocity to find the distance upstream the boat will finish: $\Delta x = (1.00~mi/h)(0.80~h) = 0.80~mi$ The boat will finish 0.80 miles upstream. (d) Let $\theta$ be the angle upstream at which the boat should head. In order to go straight across the river, the upstream component of the boat's velocity must be equal in magnitude to the speed of the current. We can find $\theta$: $(3.00~mi/h)~sin~\theta = 1.60~mi/h$ $sin~\theta = \frac{1.60~mi/h}{3.00~mi/h}$ $\theta = sin^{-1}(\frac{1.60~mi/h}{3.00~mi/h})$ $\theta = 32.2^{\circ}$ The boat should have headed at an angle of $32.2^{\circ}$ upstream.