Two boys are standing at the ends $A$ and $B$ of a ground where $AB = a$. The boy at $B$ starts running in a direction perpendicular to $AB$ with velocity ${v_1}.$ The boy at $A$ starts running simultaneously with velocity $v$ and catches the other boy in a time $t$, where $t$ is
Two boys are standing at the ends $A$ and $B$ of a ground where $AB = a$. The boy at $B$ starts running in a direction perpendicular to $AB$ with velocity ${v_1}.$ The boy at $A$ starts running simultaneously with velocity $v$ and catches the other boy in a time $t$, where $t$ is
$a/\sqrt {{v^2} + v_1^2} $
$\sqrt {{a^2}/({v^2} - v_1^2)} $
$a/(v - {v_1})$
$a/(v + {v_1})$
Two boys are standing at the ends $A$ and $B$ of a ground where $AB = a$. The boy at $B$ starts running in a direction perpendicular to $AB$ with velocity ${v_1}.$ The boy at $A$ starts running simultaneously with velocity $v$ and catches the other boy in a time $t$, where $t$ is
Let two boys meet at point $C$ after time $ 't'$ from the starting. Then $AC = v\,t$, $BC = {v_1}t$
${(AC)^2} = {(AB)^2} + {(BC)^2}$ $⇒$ ${v^2}{t^2} = {a^2} + v_1^2{t^2}$
By solving we get $t = \sqrt {\frac{{{a^2}}}{{{v^2} - v_1^2}}} $
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