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
(a) width of the river
(b) velocity of the boat with respect to water
(c) speed of the current

Options

A. `v_(br)=2m/min, v_(r )=12 m/min` and `w=200 m`.
B. `v_(br)=20m/min, v_(r )=125 m/min` and `w=200 m`.
C. `v_(br)=20m/min, v_(r )=12 m/min` and `w=200 m`
D. `v_(br)=200m/min, v_(r )=12 m/min` and `w=200 m`.

Answer

B

Explanation

Let `v_(r )` =velocity of river
`v_(br)` = velocity of river in still water and
w = width of river
Given, `t_(min) = 10 min` or `(w)/(v_(br))=10` …(i)
Drift in this case will be,
`x = v_(r )t`
`therefore 120 = 10v_(r )` …(ii)

Shortest path is taken when `v_(b)` is along AB. In this case,
`v_(b)=sqrt(v_(br)^(2)-v_(r )^(2))`
Now, `12.5=(w)/(v_(b))=(w)/(sqrt(v_(br)^(2)-v_(r )^(2)))` ...(iii)
Solving these three equations, we get
`v_(br)=20m/min, v_(r )=12 m/min` and `w=200 m`.