Question

There are two men $A$ and $B$ starts with same velocity $6m/s$ at same time from a junction of two roads inclined at ${45}^{\circ }$ to each other. If they are travelling on different roads, then the absolute value of the rate(in $m/s$) at which they are separating is $3\sqrt{a-b}$ then $\frac{b^2}{a}$ is

Answer 1

Let $R$ and $S$ be the positions of men $A$ and $B$ at any time $t.$
Since, velocities are same
$\Rightarrow OR=OS=x$
$\Rightarrow \frac{dx}{dt}=6m/s$
Now in $△ORS,$
$\cos {45}^{\circ }=\frac{x^2+x^2-y^2}{2x^2}\Rightarrow \frac{1}{\sqrt{2}}=1-\frac{y^2}{2x^2}\Rightarrow \frac{y^2}{2x^2}=1-\frac{1}{\sqrt{2}}\Rightarrow y=\pm x\sqrt{2-\sqrt{2}}$
$\Rightarrow \frac{dy}{dt}=\pm \sqrt{2-\sqrt{2}}\times \frac{dx}{dt}\Rightarrow \frac{dy}{dt}=\pm 6\times \sqrt{2-\sqrt{2}}m/s\Rightarrow |\frac{dy}{dt}|=6\sqrt{2-\sqrt{2}}m/s\Rightarrow 3\sqrt{a-b}=3\sqrt{8-4\sqrt{2}}$

Hence $\frac{b^2}{a}=\frac{16\times 2}{8}=4$