Two boys are standing at the ends $\mathrm{A}$ and $\mathrm{B}$ of a ground where $\mathrm{AB}=a$. The boy…

Two boys are standing at the ends $\mathrm{A}$ and $\mathrm{B}$ of a ground where $\mathrm{AB}=a$. The boy at $\mathrm{B}$ starts running in a direction perpendicular to $\mathrm{AB}$ with velocity $v_1$. The boy at A starts running simultaneously with velocity $v$ and catches the other in time $t$, where $t$ is:
  1. $\frac{a}{\sqrt{v^2+v_1^2}}$
  2. $\frac{a}{v+v_1}$
  3. $\frac{a}{v-v_1}$
  4. $\sqrt{\frac{a^2}{v^2-v_1^2}}$

Solution

If boy \(A\) catches boy \(B\) in time \(t\),
\(\begin{aligned}
& \Rightarrow(\mathrm{vt})^2=\left(\mathrm{v}_1 \mathrm{t}\right)^2+\mathrm{a}^2 \\
& \Rightarrow \mathrm{t}^2=\frac{\mathrm{a}^2}{\mathrm{v}^2-\mathrm{v}_1^2} \\
& \Rightarrow \mathrm{t}=\frac{\mathrm{a}}{\sqrt{\mathrm{v}^2-\mathrm{v}_1^2}}
\end{aligned}\)

Asked in: NEET 2005

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