Q140 / 150 · MHT CET 2026 (16 April Shift 1) · Physics
Two boys are standing at points A and B on ground where distance AB = a. The boy at point B starts running perpendicular to line AB with velocity '$V_1$'. The boy at point A starts running simultaneously with velocity 'V' and catches the other boy in time 't'. The value of 't' is
- Correct answerA$\left[\dfrac{a^2}{(V^2 - V_1^2)}\right]^{\frac{1}{2}}$
- B$\left[\dfrac{a^2}{(V_1^2 - V^2)}\right]^{\frac{1}{2}}$
- C$\left[\dfrac{a^2}{(V^2 - V_1^2)}\right]$
- D$\left[\dfrac{a^2}{(V_1^2 - V^2)}\right]$
Solution
Let the meeting point be C.
In the right-angled triangle ABC, the initial distance AB is $a$.
Since the boy at B runs perpendicular to AB, the angle at B is $90^{\circ}$.
The distance travelled by the boy from B is $BC = V_1 t$.
The distance travelled by the boy from A is $AC = V t$.
Using Pythagoras theorem in $\triangle ABC$:
$(AC)^2 = (AB)^2 + (BC)^2$
$(V t)^2 = a^2 + (V_1 t)^2$
$V^2 t^2 - V_1^2 t^2 = a^2$
$t^2 (V^2 - V_1^2) = a^2$
$t = \left[\dfrac{a^2}{V^2 - V_1^2}\right]^{\dfrac{1}{2}}$
Answer: $\left[\dfrac{a^2}{(V^2 - V_1^2)}\right]^{\frac{1}{2}}$
In the right-angled triangle ABC, the initial distance AB is $a$.
Since the boy at B runs perpendicular to AB, the angle at B is $90^{\circ}$.
The distance travelled by the boy from B is $BC = V_1 t$.
The distance travelled by the boy from A is $AC = V t$.
Using Pythagoras theorem in $\triangle ABC$:
$(AC)^2 = (AB)^2 + (BC)^2$
$(V t)^2 = a^2 + (V_1 t)^2$
$V^2 t^2 - V_1^2 t^2 = a^2$
$t^2 (V^2 - V_1^2) = a^2$
$t = \left[\dfrac{a^2}{V^2 - V_1^2}\right]^{\dfrac{1}{2}}$
Answer: $\left[\dfrac{a^2}{(V^2 - V_1^2)}\right]^{\frac{1}{2}}$
Hence, the correct answer is option: 'A'.