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(A) Let the boy at $A$ be at the origin $(0, 0)$ and the boy at $B$ be at $(x, 0)$.The boy at $B$ moves along the $y$-axis with velocity $v_1$. His po

Vedclass · Accepted Answer

$\frac{x}{\sqrt{v^2 - v_1^2}}$

Vedclass · Answer

(A) Let the boy at $A$ be at the origin $(0, 0)$ and the boy at $B$ be at $(x, 0)$.The boy at $B$ moves along the $y$-axis with velocity $v_1$. His position at time $t$ is $(x, v_1 t)$.The boy at $A$ moves with velocity $v$ to meet the boy at $B$ at time $t$. His position at time $t$ is $(v_x t, v_y t)$ such that $\sqrt{(v_x t)^2 + (v_y t)^2} = vt$.Since they meet at $(x, v_1 t)$,we have $v_x t = x$ and $v_y t = v_1 t$.Using the velocity magnitude condition: $(v_x t)^2 + (v_y t)^2 = (vt)^2$.Substituting the values: $x^2 + (v_1 t)^2 = v^2 t^2$.Rearranging for $t^2$: $v^2 t^2 - v_1^2 t^2 = x^2$.$t^2 (v^2 - v_1^2) = x^2$.$t = \frac{x}{\sqrt{v^2 - v_1^2}}$.

Two boys are standing at points $A$ and $B$ on the ground,where distance $AB = x$. The boy at $B$ starts running perpendicular to $AB$ with velocity $v_1$. The boy at $A$ starts running simultaneously with velocity $v$ and meets the other boy in time $t$. The value of $t$ is:

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