question_answer

Two boys are standing at the ends A and B of a ground where \[AB=a\]. The boy at B starts running in a direction perpendicular to AB with velocity \[{{v}_{1}}\]. The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is

A) \[a/\sqrt{{{\text{v}}^{\text{2}}}\text{+v}_{\text{1}}^{\text{2}}}\]            

B) \[a/(v+{{v}_{1}})\]

C) \[a/(v-{{v}_{1}})\]           

D) \[\sqrt{{{a}^{2}}/\text{(}{{\text{v}}^{\text{2}}}\text{-v}_{\text{1}}^{\text{2}}\text{)}}\]

Correct Answer: D

Solution :

          

Velocity of A relative to B is given by
\[\text{v}\,{{\overrightarrow{_{\text{A}}}}_{\text{/B}}}=\text{v}\,\overrightarrow{_{\text{A}}}-\text{v}\,\overrightarrow{_{B}}=\overrightarrow{\text{v}}={{\overrightarrow{\text{v}}}_{1}}\]	...(1)
By taking x-components of equation (1), we get
\[0=\text{v}\,\text{sin}\theta \text{-}{{\text{v}}_{1}}\Rightarrow \sin \theta =\frac{{{\text{v}}_{1}}}{\text{v}}\]	...(2)
By taking Y-components of equation (1), we get
\[{{\text{v}}_{y}}=\text{v cos}\theta \]	...(3)
Time taken by boy at A to catch the boy at B is given by
\[t=\frac{Relative\text{ }displacement\text{ }along\text{ }Y\text{ }-\text{ }axis}{Relative\text{ }velocity\text{ }along\text{ }Y\text{ }-\text{ }axis}\]
\[=\frac{a}{\text{v cos}\theta }=\frac{a}{\text{v}\text{.}\sqrt{1-{{\sin }^{2}}\theta }}=\frac{a}{\text{v}\text{.}\sqrt{1-{{\left( \frac{{{\text{v}}_{1}}}{\text{v}} \right)}^{2}}}}\]
[From equation (1)]
\[=\frac{a}{\text{v}\text{.}\sqrt{\frac{{{\text{v}}^{2}}-\text{v}_{1}^{2}}{{{\text{v}}^{2}}}}}=\frac{a}{\sqrt{{{\text{v}}^{2}}-\text{v}_{1}^{2}}}=\sqrt{\frac{{{a}^{2}}}{{{\text{v}}^{2}}-\text{v}_{1}^{2}}}\]