A boat which has a speed of `5km hr^(-1)` in still water crosses a river of with 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 `5km hr^(-1)`
The boat crosses the 1km wide river along the shortest possible path OC in 15 minutes.
Hence OC=R represents resultant velocity of the boat which is `(1km)/(15mt)=4km hr^(-1)`
Here , `OB=AC =5 km hr^(-1), OC=4 Km hr^(-1),OA=?`
In`DeltaOAC, OA=sqrt((AC)^(2)-(OC)^(2))=sqrt((5)^(2)-(4)^(2))=3`
Here the velocity of the river water is `3km hr^(-1)`