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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
$\dfrac{a}{\sqrt{v^2 + v_1^2}}$
B
$\sqrt{\dfrac{a^2}{v^2 - v_1^2}}$
C
$\dfrac{a}{(v - v_1)}$
D
$\dfrac{a}{(v + v_1)}$
Detailed Solution
Let the two boys meet at point C after time $t$. The boy from B runs along BC, perpendicular to AB, and the boy from A runs straight along AC.
BC $= v_1 t$ and AC $= vt$.
Triangle ABC is right-angled at B, so $AB^2 = AC^2 - BC^2$
$a^2 = v^2t^2 - v_1^2t^2 = t^2(v^2 - v_1^2)$
$t^2 = \dfrac{a^2}{v^2 - v_1^2}$
$t = \sqrt{\dfrac{a^2}{v^2 - v_1^2}}$
BC $= v_1 t$ and AC $= vt$.
Triangle ABC is right-angled at B, so $AB^2 = AC^2 - BC^2$
$a^2 = v^2t^2 - v_1^2t^2 = t^2(v^2 - v_1^2)$
$t^2 = \dfrac{a^2}{v^2 - v_1^2}$
$t = \sqrt{\dfrac{a^2}{v^2 - v_1^2}}$