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(A) Let the boy at $B$ move along the $y$-axis and the boy at $A$ move along the $x$-axis. After time $t$, the position of the boy from $B$ is $(0, V_

Vedclass · Accepted Answer

$[\frac{a^2}{(V^2 - V_1^2)}]^{1/2}$

Vedclass · Answer

(A) Let the boy at $B$ move along the $y$-axis and the boy at $A$ move along the $x$-axis. After time $t$, the position of the boy from $B$ is $(0, V_1 t)$. The position of the boy from $A$ is $(a, Vt)$ if we consider the relative displacement. Alternatively, using the relative velocity approach: The boy at $A$ must have a velocity component $V_1$ along the $y$-direction to match the boy at $B$, and a velocity component $V_x$ along the $x$-direction to cover the distance $a$. By Pythagorean theorem, $V^2 = V_x^2 + V_1^2$, so $V_x = \sqrt{V^2 - V_1^2}$. The time taken to cover the distance $a$ is $t = \frac{a}{V_x} = \frac{a}{\sqrt{V^2 - V_1^2}}$. Squaring both sides, $t = [\frac{a^2}{(V^2 - V_1^2)}]^{1/2}$.

Two boys are standing at points $A$ and $B$ on the 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

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