Question

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

Answer 1

Refer to the figure.

Given $OR$ = distance covered by man A in time t

$OS$ = distance covered by man B in same time t

(Given that both men have started at same time)

Now, As both men are moving with same velocities and time is also same for both men, distance covered will also be same.

Let $OA=OB=x$

Thus, by definition of velocity, we can write,

$v=\frac{dx}{dt}$

Now, by cosine rule, we can write,

$y^2=x^2+x^2+2\times \left(x\right)\times \left(x\right)\times cos\left({45}^0\right)$

$\therefore y^2=2x^2+2x^2\times \frac{1}{\sqrt{2}}$

$\therefore y^2=2x^2+\sqrt{2}{x}^2$

$\therefore y^2=x^2(2+\sqrt{2})$

$\therefore y=x\sqrt{(2+\sqrt{2})}$

Differentiate both sides w.r.t. t, we get,

$\frac{dy}{dt}=\sqrt{(2+\sqrt{2})}\frac{dx}{dt}$

$\therefore \frac{dy}{dt}=\sqrt{(2+\sqrt{2})}v$

This is rate of separation of two men.