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Pursuit and Chase : Irodov problem 1.12

Hi Brilliantians!

Today I have developed a general formula for the classic Irodov problem-

3 particles are kept at the vertices of an equilateral triangle of side \(a\).The velocity vector of the first particle continually points towards the second,the second towards the third and the third towards the first.After what time will they meet.Each moves with a constant speed \(v\).

Well for a regular \(n\) sided polygon ,

\(T =\) \(\frac{a}{v_{0}(1-cos\frac{2\pi}{n})}\)

Working :

Well the usual stuff,figuring out relative velocities and then dividing the initial distance by the \(v_{rel}\) to get the time.Becuase the trajectory forms logarithmic spirals and maintains its symmetricity all throughout the motion.

Note that each angle of the polygon is given by \(\frac{(n-2)\pi}{n}\) therefore the exterior angle will be \(\frac{2\pi}{n}\).Resolving the velocity component along this direction,we get

\(v_{rel} =\) \(v - v cos \) \(\theta\) where \(\theta\) \(=\) \(\frac{2\pi}{n}\)

hence \(t =\) \(\frac{a}{v_{rel}}\) giving us our desired expression.

Note that for ,

\(n = 3\) we have \(t = 2a/3v\).

\(n = 4\) we have \(t = a/v\).

\(n = 6\) we have \(t = 2a/v\)

\(n = 12\) we have \(t =\) \(\frac{2a}{(2-\sqrt 3)v}\)

\(n = infinity\) we have \(t =\) \(infinity\) because now the path is a circle.

and so on.

Thanks everyone for reading!

Note by Ayon Ghosh
3 weeks, 4 days ago

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