# two boats A and B move away from a buoy anchored at the middle of a river along mutually perpendicular straight lines. The boat A moves along the stream and the boat B across the river. After moving off an equal distance from the buot, both the boats returned to their original position. Calculate the ratio of the times taken by boat A to that taken by boat B if the velocity of each boat with respect to still water is 2 times the stream velocity

Aman Bansal
592 Points
11 years ago

et $u$ be the stream velocity, then velocity of each boat(with respect to water) is $v = 1.2 u.$

Moving away from buoy : If A moves away a distance $d$ in time ${t_A}$ and $B$ in ${t_B}$, then

${T_B} = \frac{d}{{{v_B}}} = \frac{d}{{v\cos \theta }} = \frac{d}{{v.\frac{{\sqrt {{v^2} - {u^2}} }}{v}}}$

$\;\,\, = \frac{d}{{\sqrt {{v^2} - {u^2}} }}$$\left[ {{\bf{since}}{\rm{ }}\;v\sin \theta = u\,\;\;\;{\rm{i}}{\rm{.e}}{\rm{.}}\;\;\sin \theta = u{\rm{/}}v} \right]$
and$\;\;\;\;\;{t_A} = \frac{d}{{v + u}}$

Coming back to buoy: If $A$ takes a time of ${t_A}'$ seconds and $B$ takes ${t_B}'$ seconds in coming back to buoy, then+

$t_B^' = \frac{d}{{\sqrt {{v^2} - {u^2}} }}$
and$\;\;\;\;\;t_A^' = \frac{d}{{v - u}}$

Therefore, total time of motion of B is

Hence$\;\;\;\;\;\frac{{{T_A}}}{{{T_B}}} = \frac{{1.2}}{{\sqrt {{{(1.2)}^2} - 1} }} = 1.8$

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Thanks

Aman Bansal

${T_B} = {t_B} + {t_B}' = 2{t_B} = \frac{{2d}}{{\sqrt {{v^2} - {u^2}} }}$

and that for A is

${T_A} = {t_A} + {t_A}' = \frac{d}{{v + u}} + \frac{d}{{v - u}}$
$\;\,\, = \frac{{2vd}}{{{v^2} - {u^2}}}$