$\begin{array}{1 1} (A)\;6ms^{-2} \\(B)\;8\;ms^{-2} \\(C)\;7\;ms^{-2} \\(D)\;9\;ms^{-2} \end{array} $

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Let the velocity of the train as it crosses certain pole be u.

Since from this pole the distance of the next pole is $ 90 m$ and next pole is $180m$ , therefore using

$s= ut +\large\frac{1}{2}$$at^2$

$90=u(3)+\large\frac{1}{2}$$a(3)^2$

$180=u(3+2)+\frac{1}{2}a(3+2)^2$

Solving we get , $a=6 ms^{-2}$

Hence A is the correct answer.

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