A man crosses a river in a boat. If he crosses the river in minimum...

Let& ${v}_{r}$ = velocity of river

${v}_{br}$ = velocity of river in still water and

$w$ = width of river

Given, {t}_\min = 10min

or $\dfrac{w}{v\text{br}} = 10$ ...(i)

Drift, in this case, will be,

$x = {v}_{r}t$

$∴120 = 10{v}_{r}$ ...(ii)





The shortest path is taken when v_b is along $AB$. In this case,

${v}_{b} = \sqrt{{v}_{br}^{2} - {v}_r^{2}}$

Now, $12\text{.}5 = \dfrac{w}{v\text{b}} = \dfrac{w}{{v}_{br}^{2} - {v}_r^{2}}$ ...(iii)

Solving these three equations, we get

${v}_{br} = 20m\min^{- 1}$,

${v}_{r} = 12m\min^{- 1}$

and $w = 200m$