A boat moves at 4 m/s in still water. It crosses a river 80 m wide that flows at 2 m/s…

Step 1. Shortest time. Point straight across: $t = \dfrac{80\ \text{m}}{4\ \text{m/s}} = 20\ \text{s}$.

Step 2. Drift. In that time the river carries the boat $2\ \text{m/s} \times 20\ \text{s} = 40\ \text{m}$ downstream.

Step 3. Landing opposite. $\sin\phi = \dfrac{2\ \text{m/s}}{4\ \text{m/s}} = 0.5$, so $\phi = 30^\circ$ upstream from the line straight across.

Step 4. That crossing. The part across is $\sqrt{(4\ \text{m/s})^2 - (2\ \text{m/s})^2} = 3.46\ \text{m/s}$, so $t = \dfrac{80\ \text{m}}{3.46\ \text{m/s}} = 23.1\ \text{s}$.