Question

A swimmer swims in still water at a speed `=5 km//hr`. He enters a `200m` wide river, having river flow speed `=4km//hr` at point `A` and proceeds to swim at an angle of `127^(@) (sin 37^(@) = 0.6)` with the river flow direction. Another point `B` is located directly across `A` on the other side. the swimmer lands on the other bank at a point `C`, form which he walks the distance `CB` with a speed `= 3km//hr`. The total time in which he reaches from `A` to `B` is
A. `5`minutes
B. `4`minutes
C. `3`minutes
D. None

Answer 1

Correct Answer - B
`V_(RY) = V_(my) +V_(wy)`
`= 5sin 127^(@)`
`= 5 cos 37^(@)`
`= 9xx 0.8`
`4 km//H`
`= (4xx5)/(10) m//s`
time taken to cross river `t_(AB) = (200 xx 18)/(4xx5)sec`
`t = 180sec = 3min`
`V_(R(X)) = V_(mx) +V_(mx) rArr V_(R(X)) = 5cos 127^(@) +4`
`V_(R(X)) = 5(-sin 37^(@))+4 rArr V_(R(X)) =- 3+4`
`rArr V_(R(X)) = 1km//h = (5)/(18) m//s`
Horizontal drag `BC`
`=V_(Rx) xx t = (5)/(18) (m)/(s) xx 180 sec = 50m`
Time taken to reach form `B` to `C`
`t_(BC) = (dis tan ce)/(speed) rArr t_(BC) = (50)/(3xx(5)/(18)m//s)`
`= 60 sec = 1min`
total time `= 4min`.