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: Understanding the Concept:
B runs perpendicular to AB, so in time \(t\) he moves \(V_1t\) from B. A catches him at that point, so A's path is the hypotenuse of a right triangle with sides \(a\) and \(V_1t\).

Step 2: Write the condition:
A's distance is \(Vt\). By Pythagoras:
\[ (Vt)^2 = a^2 + (V_1t)^2 \]

Step 3: Solve for t:
\[ t^2(V^2 - V_1^2) = a^2 \Rightarrow t = \left[\frac{a^2}{V^2 - V_1^2}\right]^{1/2} \]

Step 4: Check the options:
Option (A) matches, and it needs \(V > V_1\) so that A can catch B. Option (B) has \(V_1^2 - V^2\), which would be negative when \(V > V_1\). Options (C) and (D) are missing the square root, so they would be \(t^2\) and not \(t\).

Final Answer:
Pythagoras gives t = a / sqrt(V^2 - V1^2). \[ \boxed{\text{(A) }\left[\dfrac{a^2}{V^2-V_1^2}\right]^{1/2}} \]