Question

two boys a and b ate standing at ends of A & B of ground where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity V1. The boy at A starts running with with velocity 'V' and catches the other in time 't', where 't' is ............

Answer 1

$t = \frac{a}{ \sqrt{ {v}^{2} + {v1}^{2} } }$
shhi hi ??

Answer 2

Answer:

t = $\frac{a^{2} }{(V^{2}-V_{1}^{2} )}$

Explanation:

Distance covered by boy A in time t ⇒ AC = Vt

Distance covered by boy B in time t ⇒ BC = V₁t

Using Pythagoras Theorem :

⇒ AC² = AB² + BC²

⇒ (Vt)² = a² + (V₁t)²

⇒ a² = V²t² - V₁²t²

⇒ a² = t²(V²-V₁²)

⇒ t² = a²/ (V²-V₁²)

∴ t = $\frac{a^{2} }{(V^{2}-V_{1}^{2} )}$