Answer
(a) $22.8i+7.4j$
(b) $7.4mi/h$, $22.8mi/h$
Work Step by Step
(a) Given the boat heading 24mi/h N $72^{\circ}$ E, the angle $\theta=90-72=18^{\circ}$ from the shore, the velocity of the boat relative to the water as a vector in component form is $\vec {v_b} =24cos18^{\circ} i + 24sin18^{\circ} j\approx22.8i+7.4j$
(b) Given the water is flowing directly south, we can assume the velocity of the water as $\vec {v_w} =-v_wj$, the true velocity of the boat is then $\vec{v_t} =\vec {v_b} +\vec {v_w} =22.8i+(7.4-v_w)j$. We know that the true direction of the boat is directly east, so its vertical component should be zero, which gives the speed of the water $v_w=7.4mi/h$. The true speed of the boat can be found as $|\vec {v_t} |=\sqrt {(22.8)^2+(0)^2}=22.8mi/h$
