47

Q 47. A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to water is 3 km/h. 

(a) If he heads in a direction making an angle q with the flow, find the time he takes to cross the river.

(b) Find the shortest possible time to cross the river.

Answer: (a) The component of velocity perpendicular to direction of flow of river = 3sinθ  km/h =3000sinθ/60 m/min = 50 sin θ  m/min. 

Time taken to cross the river = 500 m/ 50sin θ  min = 10/sin θ  min. 

(b) For the shortest possible time the denominator "sin θ" should be maximum. And maximum possible value of "sin θ" is 1 for θ=90°. So shortest possible time = 10/1  min =10 minutes.