He is Invisible(Original)

For these type of problems(n particles on n-sided polygon) I have derived a direct formula for time when they met at centre

$t = \dfrac{a}{2v\sin^{2} \left( \frac{\pi}{n} \right)}$ ( comment for proof)

Where 'n' is number of sides of polygon

Here $n = 8$

Now we have to find the distance that $9^{th}$ particle has travelled

polygon

See the image above $9^{th}$ particle has to travel the horizontal distance let it be x

$\displaystyle tan\left( \frac{\pi}{8} \right) = \dfrac{\frac{a}{2}}{x}$

$\displaystyle x = \dfrac{a}{2tan\left(\frac{\pi}{8}\right)}$

We know,

$\displaystyle S = ut + \dfrac{Kt^{2}}{2}$

Here $u =0$ , 'K' be the acceleration

$\therefore$

$\displaystyle 2x = Kt^{2}$

Now substitute the value of t & x simplifing to get

$K = \dfrac{4v^{2} sin^{3}\left(\frac{\pi}{8}\right). cos\left(\frac{\pi}{8}\right)}{a}$

Now substituting value of a & v

You will get final result as

$\displaystyle \boxed{\boxed{K = \dfrac{8(\sqrt{2} - 1)}{7}}}$

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The One who reported the problem your answer was dimensionally incorrect.