A boat which has a speed of 5 km `hr^(-1)` in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. Calculate the velocity of the river water.

Explanation

The situation is shown in Fig. of previous example
OA=A, represents velocity of the river water.
OB=B, represents velocity of the boat in still water which is 5 km `hr^(-1)`
The boat crosses the 1 km wide river along the shortest possible path OC in 15 minutes.
Hence OC =R represents resultant velocity of the boat which is `( 1 km )/( 15 m t) = 4 k m h r^(-1)`
Here, OB = AC = 5 km `hr^(-1) ,OC =4 km hr^(-1) , OA=?`
In `Delta OAC, OA = sqrt((AC)^2 -(OC)^2) =sqrt( (5)^2 -(4)^2) = 3`
Here the velocity of the river water is 3 km `hr^(-1)`