(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 $B$ ==> $v = $ at [As $u = 0$]

Distance between $A$ and $B$, ${S_{AB}} = \frac{1}{2}a{t^2}$

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 $A$ $\left( {{S_{AC}} = a\,{t^2}} \right)$

${S_{AC}} = u\;t + \frac{1}{2}a{t^2}$ ==> $a\,{t^2} = 0 + \frac{1}{2}a\,t_1^2$ ==> ${t_1} = \sqrt 2 \;t$

Hence total time in which point returns to initial point

$T = 2t + \sqrt 2 \;t$$ = (2 + \sqrt 2 \;)t$