Question

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

• A. 30 min
• B. 1 hr
• C. 45 min
• D. 2 hr

Answer 1

The correct option is B

1 hr

Time of traveling of boat from A to B$\left(t_1\right)$ then B to C$\left(t_1\right)$ = 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/hr

And velocity of boat from B to C $\overrightarrow{v_{b,w}}+\overrightarrow{v_w}=(5-u)$km/hr
Distance moved by boat in time $t_1,AB=(5+u)t_1$
And distance moved by boat in time $t_1=BC=(5-u)t_1^{\prime }$
Distance moved by cork during this time, AC=$u(t_1+t_1^{\prime })$

But AB=AC+BC
$(5+u)t_1=u(t_1+t_1^{\prime })+(5-u)t_1^{\prime }$
$5t_1=5t_1^{\prime }\Rightarrow t_1^{\prime }=t_1=30$ min
Hence cork meets again the boat after 1 hr.

Ok, now lemme show you the power of relative motion! Another approach is to solve it from the frame of the cork, from which the velocity of the boat will be +5 km/ hr for the first $t_1$= 30 min. So, it will have moved 5 X 30/60 km = 2.5 km. Now, according to the question, the boat takes an U-turn after this time, so it will have a velocity of -5 km/hr wrt the cork, and it has to cover a distance of 150 km. So, how much time will it take, yes, easy right? Just the same time, $t_1$= 30 min again (= $\frac{2.5X60}{5}min$). So, the total is 60 min = 1 hr!