Search any question & find its solution
Question:
Answered & Verified by Expert
Two boys are standing at the ends $A$ and $B$ of a ground, where $A B=a$. The boy at $B$ starts running in a direction perpendicular to $A B$ 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
Options:
Solution:
1385 Upvotes
Verified Answer
The correct answer is:
$\sqrt{\frac{a^2}{v^2-v_1^2}}$
Distance covered by boy $A$ in time $t$
$A C=v t...(i)$
Distance covered by boy $B$ in time $t$
$B C=v_1 t...(ii)$
Using Pythagorus theorem
$\begin{aligned}& & A C^2 & =A B^2+B C^2 \\& \text { or } & (v t)^2 & =a^2+\left(v_1 t\right)^2 \\& \text { or } & v^2 t^2-v_1^2 t^2 & =a^2 \\& \text { or } & t^2\left(v^2-v_1^2\right) & =a^2\end{aligned}$
$\therefore$ $t=\sqrt{\frac{a^2}{\left(v^2-v_1^2\right)}}$
$A C=v t...(i)$
Distance covered by boy $B$ in time $t$
$B C=v_1 t...(ii)$
Using Pythagorus theorem
$\begin{aligned}& & A C^2 & =A B^2+B C^2 \\& \text { or } & (v t)^2 & =a^2+\left(v_1 t\right)^2 \\& \text { or } & v^2 t^2-v_1^2 t^2 & =a^2 \\& \text { or } & t^2\left(v^2-v_1^2\right) & =a^2\end{aligned}$
$\therefore$ $t=\sqrt{\frac{a^2}{\left(v^2-v_1^2\right)}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.