Two boats `A` and `b` move away form a buoy anchored at the middle of a river along mutually perpendicular staight line. The boat `A` moves along the stream and the boat `B` across the river. After moving off an equal distance of `500` metre form the buoy both the voats returned to their original position. Find the ratio of the time taken by boat `A` to that taken by boat `B` if the velocity of ecah boat with respect to water is `20m//s` and the stream velocity si `10m//s`.

Explanation

For boat `A`
Tima taken to move a distance `500m` along the stream `= 500//(20+10)`
Time taken to move a distance `500m` Against the stream `- 500//(20-10)`
`:.` Time taken by boat `A` to reach the original position
`=(350)/(30) +(500)/(10) = (2000)/(30)` second.
For boat `B`. The situation is shown in (fig.)

The boat `B` moves across the river for `500m` and comes back to the original postion i.e.,
`V_(B)sin theta-10 =10`, i.e. `20 sin theta - 10 = 0`,
`:. sin theta = 1//2` or `theta = 30^(@)`
Now velocity of boat `B` across the river `= 20`
`cos 30^(@) = 20 (sqrt(3)//2) = 10sqrt(3)m//s`.
Time taken to move a distance `500m` across
the river `=(500)/(10sqrt(3))` second
Time taken to move a distance `500m` to come
back `= (500)/(10sqrt(3))` second
Total time taken `= (500)/(10sqrt(3)) +(500)/(10sqrt(3)) = (1000)/(10sqrt(3)) = (100)/(sqrt(3))`
`:.` Required ration `= (2000//30)/(100//sqrt(3))`
`= (2000sqrt(3))/(3000) = (2)/(sqrt(3)) = 1.155`