MHTCET 25 Apr 2025 Shift 2 Paper

Concept:The girl at A follows the shortest possible path to intercept the girl at B, which is the straight-line distance between their positions at time $t$.Explanation:Let the distance covered by the girl starting from B in time $t$ be $x$.Since she runs perpendicular to $AB$ with velocity $V_1$,$$x = V_1 t$$Let the distance covered by the girl starting from A in the same time be $y$.Since she runs with velocity $V_2$,$$y = V_2 t$$The starting distance between them is $AB = b$.The path of the girl from B is perpendicular to $AB$, so at time $t$ the two positions form a right triangle.As A takes the shortest path, $y$ is the hypotenuse of this right triangle:$$y = \sqrt{b^2 + x^2}$$Squaring both sides:$$y^2 = b^2 + x^2$$Substitute $y = V_2 t$ and $x = V_1 t$:$$V_2^2 t^2 = b^2 + V_1^2 t^2$$Rearrange and solve for $t$:$$t^2 (V_2^2 - V_1^2) = b^2$$$$t = \frac{b}{\sqrt{V_2^2 - V_1^2}}$$This matches option D.Answer:$$t = \frac{b}{\sqrt{V_2^2 - V_1^2}}$$Option D is correct.