Question

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 along a straight line with velocity $v$ and catches the other boy in a time $t$, where t is

• A. $\frac{a}{v+v_1}$
• B. $\frac{a}{v-v_1}$
• C. $\frac{a}{\sqrt{v^2+v_1^2}}$
• D. $\frac{a}{\sqrt{v^2-v_1^2}}$

Answer 1

The correct option is D $\frac{a}{\sqrt{v^2-v_1^2}}$

From the diagram, using pythagoras theorem

$a^2+(v_{1}t{)}^2=v^2{t}^2$
$\Rightarrow t^2[v^2-v_1^2]=a^2$

$\Rightarrow t=\frac{a}{\sqrt{v^2-v_1^2}}$