Question

Two boys are standing at points A and B on ground, where distance \(\mathrm{AB}=\mathrm{x}\). The boy at B stars running perpendicular to \(A B\) with velocity \(\mathrm{v}_1\). The boy at A starts running simultaneously with velocity v and meets the other boy in time \(t\). The value of \(t\) is

• A. \([\frac{x}{v-v_{1}}]^{1/2}\)
• B. \([\frac{x}{v_{1}-v}]^{1/2}\)
• C. \([\frac{x^{2}}{v^{2}-v_{1}^{2}}]^{1/2}\)
• D. \([\frac{x^{2}}{v_{1}^{2}-v^{2}}]^{1/2}\)

Hint:

In time t, the boy at B moves v1 t perpendicular to AB, while A covers a distance vt along the hypotenuse.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
They meet when A, running at speed \(v\), reaches the position of B after time \(t\). The motion forms a right triangle.

Step 2: Distances in time t
B moves \(v_1t\) perpendicular to AB. A moves \(vt\) along the line joining A to the meeting point.

Step 3: Pythagoras
\((vt)^{2}=x^{2}+(v_1t)^{2}\)
\(t^{2}(v^{2}-v_1^{2})=x^{2}\).

Step 4: Solve for t
\(t=\left[\dfrac{x^{2}}{v^{2}-v_1^{2}}\right]^{1/2}\).

Final Answer:
The time of meeting is \(\left[\dfrac{x^2}{v^2-v_1^2}\right]^{1/2}\). \[ \boxed{\text{(C) }\left[\dfrac{x^{2}}{v^{2}-v_{1}^{2}}\right]^{1/2}} \]