Question

Two boats, $A$ and $B$, move away from a buoy anchored at the middle of a river along the mutually perpendicular straight lines: the boat $A$ along the river, and the boat $B$ across the river. Having moved off an equal distance from the buoy the boats returned. The ratio of times of motion of boats $\frac{{\tau }_A}{{\tau }_B}$ is $\frac{x}{10}$ if the velocity of each boat with respect to water is $\eta =1.2$ times greater than the stream velocity. Find $x$ rounded off to the nearest integer.

Answer 1

Let $l$ be the distance covered by the boat $A$ along the river as well as by the boat $B$ across the river. Let $v_0$ be the stream velocity and $v^{\prime }$ the velocity of each boat with respect water. Therefore time taken by the boat $A$ in its journey
$t_A=\frac{l}{v^{\prime }+v_0}+\frac{l}{v^{\prime }-v_0}$
and for the boat $B$ $t_B=\frac{l}{\sqrt{{v^{\prime }}^2-v_0^2}}+\frac{l}{\sqrt{{v^{\prime }}^2-v_0^2}}=\frac{2l}{\sqrt{{v^{\prime }}^2-v_0^2}}$
Hence, $\frac{t_A}{t_B}=\frac{v^{\prime }}{\sqrt{{v^{\prime }}^2-v_0^2}}=\frac{\eta }{\sqrt{{\eta }^2-1}}(where\eta =\frac{v^{\prime }}{v})$
On substitution $\frac{t_A}{t_B}=1.8$