How do I find the time taken by dogs to meet each other?

[Thiru07]

You know the original distance, and the relative speed along the line joining two tells you how rapidly that distance diminishes.

Wow. Got it. I can't thank you enough for your step by step guidance.

 

[ehild]

Because of symmetry, the dogs are at the corners of a equilateral triangle all times. The triangle rotates and shrinks with time, and it becomes a point at the end, at the centre of the original triangle. Both the time when the dogs meet and their trajectory can be easily obtained in a polar system of coordinate. The sides of the original triangle are of length a.

The radial and tangential components of the velocity vector of a dog (represented by the red dot now) are v_r=dr/dt=-vcos(30°)= -v√3/2. The tangential component is v_θ=rdθ/dt=vsin(30°)=v/2.
The radial component is constant, so the radius is r=r₀-√3/2 vt. At t=0, r=r₀=a/√3. When the dogs meet, r=0, the time is
t=2/3 a/v.
The equation for θ is dθ/dt=0.5v/r Substituting r=a/√3-√3/2 vt and integrating, we get the time dependence of the angle. The trajectory is obtained by eliminating the time, but also using that dr/dt=dr/dθ dθ/dt, that is,
$-v\frac{\sqrt{3}}{2}=\frac{dr}{d\theta}\frac{v}{2r}$
Integrating, the trajectory is $r(\theta)=\frac{a}{\sqrt{3}}e^{-\sqrt{3} \theta}$