Question

A ferry boat transfers passengers between boat jetty $A$ and $B$, where $B$ is across the river downstream as shown. The boat goes from $A$ to $B$ and then comes back after waiting for $12\min$. While going from $A$ to $B$, the boat moves with minimum velocity relative to water and while coming back it moves with double the minimum velocity relative to water as for going from $A$ to $B$. If river stream has a speed of $5km/h$ and distance between $A$ and $B$ is $1km$, then the total time (in minute) for the whole trip i.e. going from $A$ to $B$, waiting and then coming back to $A$, will be

Answer 1

Answer: 48

For ${\vec{v}}_{BR}$ to be minimum
$\Rightarrow v_R\sin 60^{\circ }=v_{BR}$
$\Rightarrow v_{BR}=\frac{5\sqrt{3}}{2}\frac{km}{h}$
and $t_1=\frac{AB\sin 60^{\circ }}{V_{BR}\sin 30^{\circ }}=\frac{2}{5}h=24\min$
Waiting time, $t_2=12\min$
When boat starts return, then
$v_{BR}=2\times \left(\frac{5\sqrt{3}}{2}\right)\frac{km}{h}$
$=5\sqrt{3}\frac{km}{h}$

Now,
$\Rightarrow v_{BG}=5km/h$
$\Rightarrow t_3\times v_{BG}=AB$
$\Rightarrow t_3=\frac{1}{5}h=12\min$
Total time, $t_1+t_2+t_3$
$=24+12+12=48\min$