The speed of a motor launch with respect to still 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 gloat. What is the total time which elapsed during the process?

Explanation

`v=8 ms^(1), u=3ms^(-1)`
.
`t_(AB)=t_(AC+t_(CB)`
`(x)/(u)(4.8)/(v-u) +(4.8+x)/(v-u)` ltbtgt `(x)/(3)=(4.8)/(8-3)+(4.8+x)/(8+3) rArr x=28.8//5 km`
Time take: `(x)/(u) =(28.8xx1000)/(5xx3) =1920s=32 min`.