Explanation

(b) Let the particle moves toward right with velocity $6\ m/s$. Due to retardation after time $t_{1}\ $its velocity becomes zero
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From $v = u - at \Rightarrow 0 = 6 - 2t_{1} \Rightarrow t_{1} = 3sec$
But retardation work on it for $4\ sec.$It means after reaching point $A$ direction of motion get reversed and acceleration works on the particle for next one second.
$S_{OA} = ut_{1} - \frac{1}{2}at_{1}^{2}$=$6 \times 3 - \frac{1}{2}(2)(3)^{2} = 18 - 9 = 9m$
$S_{AB} = \frac{1}{2} \times 2 \times (1)^{2} = 1m$
$\therefore S_{BC} = s_{oa} - s_{aB} = 9 - 1 = 8m$
Now velocity of the particle at pint $B$ in return journey $v = 0 + 2 \times 1 = 2m/s$
In return journey from $B$ to $C,\ $particle moves with constant velocity $2m/s$ to cover the distance $8m.$
$Time\ taken = \frac{Distance}{Velocity} = \frac{8}{2} = 4\ sec$
Total time taken by particle to return at point $O\ is$
$\Rightarrow T = t_{OA} + t_{AB} + t_{BC} = 3 + 1 + 4 = 8\ sec$