About which has a speed of 5km/h in still water crosses a river of width 1km along the shortest possible path in 15minutes.the velocity of the river water(in km/h) is-

							
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************** 
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**1 km**!*******/**y km************* 
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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