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)] | |
\[=\frac{a}{\text{v}\text{.}\sqrt{\frac{{{\text{v}}^{2}}-\text{v}_{1}^{2}}{{{\text{v}}^{2}}}}}=\frac{a}{\sqrt{{{\text{v}}^{2}}-\text{v}_{1}^{2}}}=\sqrt{\frac{{{a}^{2}}}{{{\text{v}}^{2}}-\text{v}_{1}^{2}}}\] |
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