MHTCET 16 Apr 2026 Shift 1 Paper

Concept:The motion forms a right-angled triangle, so use Pythagoras theorem to relate the distances travelled by both boys.Explanation:Let point $A$ be the origin and line $AB$ be horizontal.The initial distance between the boys is $AB = a$.Boy at $B$ runs perpendicular to $AB$ with speed $V_1$, so his distance from point $B$ after time $t$ is $V_1 t$.Thus, at the catching point, the horizontal separation from $A$ is still $a$, and the vertical separation is $V_1 t$.Boy at $A$ runs with speed $V$ and catches the other boy in time $t$, covering the straight-line distance $V t$.By Pythagoras theorem:$$(V t)^2 = a^2 + (V_1 t)^2$$Simplify:$$V^2 t^2 = a^2 + V_1^2 t^2$$Rearranging:$$t^2 (V^2 - V_1^2) = a^2$$$$t^2 = \frac{a^2}{V^2 - V_1^2}$$Taking square root:$$t = \left[\frac{a^2}{V^2 - V_1^2}\right]^{\frac{1}{2}}$$This is physically valid when $V > V_1$.Answer:The correct option is Option A: $\displaystyle \left[\frac{a^2}{V^2 - V_1^2}\right]^{\frac{1}{2}}$.