$\begin{array}{1 1}(A)\;0.30ms^{-1}\\(B)\;0.10ms^{-1} \\(C)\;0.9 ms^{-1} \\(D)\;6.7 ms^{-1} \end{array} $

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Since in $t=2$s, distance covered $s=2m$ we have ,

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

$2= u \times 2+\large\frac{1}{2}$$a \times 2^2$

or $u+a=1$-----(i)

As the distance covered in $t=2$s to $6\;s$ is $2.2 m$ We have

$2.2=u(6-2)+\large\frac{1}{2}$$a (6^2-2^2)$

$2u+8a=1.1$------(ii)

From equation (i) and (ii)

$u=1.15ms^{-1}$

$a= -0.15 ms^{-2}$

From the relation $v=u+at$, We have on putting,

$u=1.15 ms^{-1},a = -0.15 ms^{-2}$ and $ t=7\;s$

$v= 1.15 +(-0.15) \times 7$

$\quad=0.10 ms^{-1}$

Hence B is the correct answer.

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