Explanation

(b) In this problem point starts moving with uniform acceleration $a$ and after time $t$ (Position $B$) the direction of acceleration get reversed i.e. the retardation of same value works on the point. Due to this velocity of points goes on decreasing and at position $C$ its velocity becomes zero. Now the direction of motion of point reversed and it moves from $C\ $to $A$ under the effect of acceleration $a$.
We have to calculate the total time in this motion.
Starting velocity at position $A$ is equal to zero.
Velocity at position $B \Rightarrow v = at$ [$As\ u = 0$]
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Distance between $A$ and $B,\ S_{AB} = \frac{1}{2}at^{2}$
As same amount of retardation works on a point and it comes to rest therefore
$S_{BC} = S_{AB} = \frac{1}{2}a\ t^{2}$
$\therefore S_{AC} = S_{AB} + S_{BC} = a\ t^{2}\ $and time required to cover this distance is also equal to $t.$
$\therefore Total\ time\ taken\ for\ motion\ between\ A\ and\ C = 2t$
Now for the return journey from $C\ to\ A\ \left( S_{AC} = at^{2} \right)$
$S_{AC} = u\ t + \frac{1}{2}at^{2} \Rightarrow at^{2} = 0 + \frac{1}{2}at_{1}^{2} \Rightarrow t_{1} = \sqrt{2}t$
Hence total time in which point returns to initial point
$T = 2t + \sqrt{2\ }t = (2 + \sqrt{2})t$