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, then find the rate at which they are being separated.

$ \sqrt{2-\sqrt{2}}v$

The correct answer is Option (2) → $ \sqrt{2-\sqrt{2}}v$ ##

Let two men start from the point $C$ with velocity $v$ each at the same time.

Also, $\quad \angle BCA = 45^\circ$

Since, $A$ and $B$ are moving with same velocity $v$, so they will cover same distance in same time.

Therefore, $\Delta ABC$ is an isosceles triangle with $AC = BC$.

Now, draw $CD \perp AB$.

Let at any instant $t$, the distance between them is $AB$.

Let $\quad AC = BC = x \quad \text{and} \quad AB = y$

In $\Delta ACD$ and $\Delta DCB$,

$\angle CAD = \angle CBD \quad [∵AC = BC]$

$\angle CDA = \angle CDB = 90^\circ$

$∴\quad \angle ACD = \angle DCB$

or $\quad \angle ACD = \frac{1}{2} \times \angle ACB$

$\Rightarrow \quad \angle ACD = \frac{1}{2} \times 45^\circ$

$\Rightarrow \quad \angle ACD = \frac{\pi}{8} \quad \left[ 45^\circ = \frac{\pi}{4} \right]$

$∴\quad \sin \frac{\pi}{8} = \frac{AD}{AC}$

$\Rightarrow \quad \sin \frac{\pi}{8} = \frac{y/2}{x} \quad [∵AD = y/2]$

$\Rightarrow \quad \frac{y}{2} = x \sin \frac{\pi}{8}$

$\Rightarrow \quad y = 2x \cdot \sin \frac{\pi}{8}$

Now, on differentiating both sides w.r.t. $t$, we get

$\Rightarrow \quad \frac{dy}{dt} = 2 \cdot \sin \frac{\pi}{8} \frac{dx}{dt}$

$\Rightarrow \quad \frac{dy}{dt} = 2 \cdot \sin \frac{\pi}{8} \cdot v \quad \left[ ∵v = \frac{dx}{dt} \right] \dots (i)$

As we know that $\quad \cos \theta = 1 - 2 \sin^2 \frac{\theta}{2}$

$\Rightarrow \quad 2 \sin^2 \frac{\theta}{2} = 1 - \cos \theta$

$\Rightarrow \quad \sin^2 \frac{\theta}{2} = \frac{1 - \cos \theta}{2}$

$\Rightarrow \sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}}$

Put $\theta = \frac{\pi}{4}$

$\Rightarrow \sin \frac{\pi}{8} = \sqrt{\frac{1 - \cos \frac{\pi}{4}}{2}} = \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{2}} \quad \left[ ∵\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \right]$

$\Rightarrow \sin \frac{\pi}{8} = \frac{\sqrt{2 - \sqrt{2}}}{2} \quad \text{[On multiplying and dividing by } \sqrt{2}\text{]}$

$\Rightarrow \frac{dy}{dt} = 2v \cdot \frac{\sqrt{2 - \sqrt{2}}}{2} \quad \left[ \text{From Eq. (i)} \right] \left[ ∵\sin \frac{\pi}{8} = \frac{\sqrt{2 - \sqrt{2}}}{2} \right]$

$∴= v\sqrt{2 - \sqrt{2}} \text{ unit/s}$

Which is the rate at which $A$ and $B$ are being separated.