question_answer

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 boat returned. What is the ratio of times of motion of boats \[\frac{{{\tau }_{A}}}{{{\tau }_{B}}},\] if the velocity of each boat with respect to water is 1.2 times greater than the stream velocity?  

A)  2.3      

B)        1.8    

C)  0.5                  

D) 0.2

Correct Answer: B

Solution :

[b] Suppose the stream velocity is \[{{\text{v}}_{\text{s}}}\text{=v}\], then the velocity of each boat with respect to water is \[{{\text{v}}_{b}}=1.2\text{ v}\]. Let each boat travel a distance\[\ell \].

Then for boat A, time of motion \[{{\tau }_{A}}=\frac{\ell }{{{v}_{b}}+{{v}_{s}}}+\frac{\ell }{{{v}_{b}}-{{v}_{s}}}\]

\[=\left[ \frac{\ell }{1.2\,v+v}+\frac{\ell }{1.2\,v-v} \right]=\frac{60\text{ }\ell }{11\,v}\]                ?..(i).

For the boat B, time of motion \[{{\tau }_{B}}=\frac{\ell }{\sqrt{{{v}_{b}}^{2}-{{v}_{s}}^{2}}}+\frac{\ell }{\sqrt{{{v}_{b}}^{2}-{{v}_{s}}^{2}}}=\frac{2\ell }{\sqrt{{{v}_{b}}^{2}-{{v}_{s}}^{2}}}\]         ?(ii)

\[=\frac{2\ell }{\sqrt{{{\left( 1.2\,v \right)}^{2}}-{{v}^{2}}}}=\frac{3.01\text{ }\ell }{v}\]

The ratio \[\frac{{{\tau }_{A}}}{{{\tau }_{B}}}=\frac{\left( 60\,\ell /11\,v \right)}{\left( 3.01\,\,\ell /v \right)}\approx 1.8\]