A boat is travelling on a bearing of $α=60^\circ$ at a constant speed $u=3.0$ $m/s$. A man is swimming at a constant speed $v$ in order to reach the boat from a point a distance $l=100\ m$ due east of the boat.
Find, as a bearing, the direction in which the swimmer should head in order to reach the boat with the minimum speed.
What then is the minimum value of $v$ which will enable the swimmer to reach the boat.
What is the time taken for the swimmer to reach the boat with this speed?
I guess the picture looks like this. Let $t$ be the time the man reaches the boat. So, from the figure, $$tu\sin\theta=tv\sin\varphi$$ and $$tu\cos\theta-tv\cos\varphi=l.$$ That is, $$v=\sqrt{v^2\sin^2\varphi+v^2\cos^2\varphi}=\sqrt{u^2\sin^2\theta+\left(u\cos\theta-\frac{l}{t}\right)^2}\geq u\sin\theta.$$ The minimum value $v_\min=u\sin\theta$ is achievable by swimming northward (i.e., with $\varphi=90^\circ$). The time $t^*$ it takes will be $\frac{l}{u\cos\theta}$.
With $\theta=60^\circ$, $u=3.0\ \text{m}/\text{s}$, and $l=100\ \text{m}$, we have $v_\min=u\sin\theta=2.6\ \text{m}/\text{s}$. The time it takes is $t^*=\frac{l}{u\cos\theta}=67\ \text{s}$.