Two men $A$ and $B$ start with velocities $v$ at the same time from the junction of two roads inclined at $45^{\circ}$ to each other. If they travel by different roads,find the rate at which they are being separated.

(A) Let two men start from the point $C$ with velocity $v$ each at the same time. Also,$\angle BCA = 45^{\circ}$.
Since $A$ and $B$ are moving with the same velocity $v$,they will cover the same distance in the same time.
Therefore,$\Delta ABC$ is an isosceles triangle with $AC = BC$. Let $AC = BC = x$ and the distance between them at any instant $t$ be $y = AB$.
Draw $CD \perp AB$. In $\Delta ACD$ and $\Delta DCB$:
$\angle CAD = \angle CBD$ (since $AC = BC$)
$\angle CDA = \angle CDB = 90^{\circ}$
Therefore,$\angle ACD = \angle DCB = \frac{1}{2} \times \angle ACB = \frac{1}{2} \times 45^{\circ} = 22.5^{\circ} = \frac{\pi}{8}$.
In $\Delta ACD$,$\sin(\frac{\pi}{8}) = \frac{AD}{AC} = \frac{y/2}{x}$.
Thus,$y = 2x \sin(\frac{\pi}{8})$.
Differentiating with respect to $t$:
$\frac{dy}{dt} = 2 \sin(\frac{\pi}{8}) \frac{dx}{dt} = 2v \sin(\frac{\pi}{8})$.
Using $\sin(\frac{\pi}{8}) = \frac{\sqrt{2-\sqrt{2}}}{2}$,we get:
$\frac{dy}{dt} = 2v \cdot \frac{\sqrt{2-\sqrt{2}}}{2} = v \sqrt{2-\sqrt{2}}$.