The speed of a motor launch with respect to water in a stream is `8 ms^(-1)` while water current's speed is `3 m s^(-1)`. When the launch began travelling upstream, a float was dropped from it. After travelling a distance of `4.8 km` upstream, the launch turned back and caught up with the float. What is the total time which elapsed during the process?

Explanation

Here, speed of motor launch, ` v_m = 8 ms^(-1)` speed of stream of water`, `v_w = 4 ms^(-1)`
For upstream motionof motor launch , the relative velocity of motor launch w.r.t. water
` = v_m- v_w = 8 - 4 = 4 ms^(-1)`
Time taken by launch to travel a distance ` 4.8. km` upstream
` t_1 = ( 4.8 km) /( 4 ms^(-1)` = ( 4800 m) /( 4 ms^(-1)` = 1200s`.
For downstream motion of motor launch Distance movedby float with downstream in ` 1200 s`. ltBrgt `= 4 xx 1200 = 4800 m`
Before turning about, the distance between launch and float ` = 48 00 = 4800 = 9 6 00 m`
This distance is to be coverd by launce with its own velocity ( =8 ms^(-1)`) because the stream velocity is shared bothn by launch as well as by float. Tiem taken toreach the float,
` t_2 = ( 9600)/ 8 = 1200 s`
Total time taken, to reach the float is
` t= t_1 + t_2 = 1200 + 1200 = 2400 s= 400 min`.