Question

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

• A. \([\frac{a^2}{(V^2-V_1^2)}]^{\frac{1}{2}}\)
• B. \([\frac{a^2}{(V_1^2-V^2)}]^{\frac{1}{2}}\)
• C. \([\frac{a^2}{(V^2-V_1^2)}]\)
• D. \([\frac{a^2}{(V_1^2-V^2)}]\)

Hint:

In time t, A covers Vt along a hypotenuse of a right triangle with sides a and V1 t.

Answer 1

The Correct Option is A

Step 1: Relative velocity method:
Along AB, A must cancel B's sideways motion. Let A move at angle $\theta$ to AB: $V\sin\theta = V_1$.

Step 2: Closing speed:
A's component along AB is $V\cos\theta = \sqrt{V^2 - V_1^2}$ and must cover distance $a$.

Step 3: Time:
$t = \dfrac{a}{\sqrt{V^2 - V_1^2}}$, option (A).

Final Answer:
The catch time is a over root of V^2 - V1^2. \[ \boxed{\text{(A) }\left[\dfrac{a^2}{V^2-V_1^2}\right]^{1/2}} \]