Question

Let us consider a boat which moves with a velocity $v_{bw}$ = 5 km/h relative to water. At time t=0, the boat passes through a piece of code floating in water while moving downstream. If it turns back at time t=$t_1$ when and where does the boat meet the cork again? Assume $t_1$, = 30 min.

Answer 1

Time of travelling of boat from A to B ($t_1$) and then B to C ($t_1^{\prime }$) = time of moving the cork from A to C.
Velocity of boat from A to B
${\overrightarrow{v}}_{b,w}+{\overrightarrow{v}}_w$ = ( 5 + u ) km/h
And velocity of boat from B to C
${\overrightarrow{v}}_{b,w}+{\overrightarrow{v}}_w$ = ( 5 - u ) km/h
Distance moved by boat in time $t_1$,
AB = (5 + 4)$t_1$,
And distance moved by boat in time $t_1^{\prime }$ = BC = (5 - u)$t_1^{\prime }$,
Distance moved by cork during this time
AC = u$(t_1+t_1^{\prime })$
But AB = AC + BC
(5 + u)$t_1$ = u$(t_1+t_1^{\prime })+(5-u)t_1^{\prime }$
5$t_1=5t_1^{\prime }$
$t_1^{\prime }=t_1$ = 30 min
Hence, the cork meets the boat again after 1 h.