Let the velocity of the river be 'v' along C to B
and the boat is rowed along A to C so that the resultant velocity of the boat rows it along A to B
For the boat to take the shortest distance, it should reach the point directly opposite to the width of the river that is B
For that, the boat should be rowed at an angle against the flow of river along A to C so that the boat covers the 1km distance with the resultant velocity
**********B***x km***C**************
**********!************/**************...
**********!***********/***************...
**********!**********/****************...
**********!*********/*****************...
**********!********/******************...
**1 km**!*******/**y km*************
**********!******/********************...
**********!*****/*********************...
**********!****/**********************...
**********!***/***********************...
**********!**/************************...
**********!*/*************************...
**********A***************************
A to B:
Distance = 1 km
Velocity of the boat is the resultant of boat's velocity in still water and river's velocity
A to C:
Distance = y km
Speed of the boat = 5km/hr
C to B:
Distance = x km (Drift of the boat due to river's velocity)
Speed of the river = v km/hr
Time in both cases = 15/60 hr = 1/4 = y/5 = x/v
So, y=5/4 and x=v/4
In right triangle ABC,
AB^2 + BC^2 = AC^2
x^2 + 1^2 = y^2
(v/4)^2 + 1 = (5/4)^2
v = 3 km/hr = velocity of river along C to B
OR THE QUESTION CAN BE SIMPLY SOLVED AS FOLLOWS:
In still water,
Distance=1km
Speed , s=5km/hr
So, Time taken = 1/5 hr
In the flowing river,
Distance = 1km as the boat is supposed to take the shortest route (directly opposite)
Time = 15 min = 15/60hr =1/4 hr
Resultant speed of boat in river , u = 1/(1/4) = 4 km/hr
Let the speed of river be v
Using pythagoras theorem in the velocity triangle
u^2=s^2-v^2
4^2=5^2-v^2
v=3km/hr
