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 $5\, km / h$ and distance between $A$ and $B$ is $1 \,km$, 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

For $\vec{v}_{B R}$ to be minimum
$\Rightarrow v_{R} \sin 60^{\circ}=v_{B R} $
$\Rightarrow v_{B R}=\frac{5 \sqrt{3}}{2} \frac{ km }{ h }$
and $t_{1}=\frac{A B \sin 60^{\circ}}{V_{B R} \sin 30^{\circ}}=\frac{2}{5} h =24\, \min$
Waiting time, $t_{2}=12\, \min$
When boat starts return, then
$v_{B R}=2 \times\left(\frac{5 \sqrt{3}}{2}\right) \frac{ km }{ h }$
$=5 \sqrt{3} \frac{ km }{ h }$

Now,
$\Rightarrow v_{B G}=5 \,km / h $
$\Rightarrow t_{3} \times v_{B G}=A B$
$\Rightarrow t_{3}=\frac{1}{5} h =12 \,\min$
Total time, $t_{1}+t_{2}+t_{3}$
$=24+12+12=48\, \min$