Question

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.

Answer 1

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 same velocity v, they will cover same distince in same time.

Therefore , $△ABC$ is an isosceles triangle with AC = BC.
Now draw $CD\perp AB.$
Let, at any instant t, the distance between them be AB.
Let AC = BC = x and AB = y
In $△ACDand△DCB,\angle CAD=\angle CBD\therefore \angle ACD=\angle DCBor\angle ACD=\frac{1}{2}\times \angle ACB\Rightarrow \angle ACD=\frac{1}{2}\times {45}^{\circ }\Rightarrow \angle ACD=\frac{\pi }{8}\therefore sin\frac{\pi }{8}=\frac{AD}{AC}\Rightarrow sin\frac{\pi }{8}=\frac{\frac{y}{2}}{x}\Rightarrow y=2x.Sin\frac{\pi }{8}$
Now, differentiating both sides w.r.t. t, we get
$\frac{dy}{dt}=2.sin\frac{\pi }{8}.\frac{dx}{dt}=2.sin\frac{\pi }{8}v=2v.\frac{\sqrt{2-\sqrt{2}}}{2}=v.\sqrt{2-\sqrt{2}}$
which is the rate at which A and B are being separated.