A man crosses a river in a boat. If he cross the river in minimum time he takes $10 min$ with a drift $120 m$. If he crosses the river taking shortest path, he takes $12.5 min$. Find the velocity of the boat with respect to water?

$$\begin{gathered} Let v_r = velocity of river \\ v{br}_{ } = velocity of river in still water and \\ \omega = width of river \\ Given, t{min}_{ } = 10 min or \omega /v_{br}=10 -----\left(i\right) \end{gathered}$$

$$\begin{gathered} Drift in this case will be, x=v_{r}t \\ 120=10vr -----\left(ii\right) \\ Shortest ath is taken when vb is along AB. In this case, \end{gathered}$$

$$\begin{gathered} v_b=\sqrt{v_{tr}^2-v_r^2} \\ Now, 12.5=12.5=\frac{a}{v_b}=\frac{6}{\sqrt{v_{br}^2-v_r^2}}----\left(iii\right) \\ Solving these three equations, we get v_{br}=20 m/min. \end{gathered}$$