Let the particle moves toward right with velocity 6 m/s. Due to retardation after time t_{1} its velocity becomes zero.

From v = u - at ⇒ ⇒ \(0 \equiv 6 - 2 \times t_{1}\) ⇒ ⇒ \(t_1 = 3 \text{ sec}\)

But retardation works 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.

\(S_{0A} = ut_1 - \frac{1}{2} a t_1^2\) \(= 6 \times 3 - \frac{1}{2} (2) (3)^2 = 18 - 9 = 9m\)

\(S_{AB} = \frac{1}{2} \times 2 \times (1)^2 = 1 \, m\)

S_{BC} = S_{0A} - S_{AB} = 9 - 1 = 8m

Now velocity of the particle at point B in return journey

\(v = 0 + 2 \times 1\) \(= 2 \mathrm{m/s}\)

In return journey from B to C , particle moves with constant velocity 2 m/s to cover the distance 8 m.

Time taken \(= \frac{\text{Distance}}{\text{Velocity}} = \frac{8}{2} = 4 \text{ sec}\)

Total time taken by particle to return at point 0 is ⇒ ⇒ I = t_{0A} + t_{AB} + t_{BC} = 3 + 1 + 4 = 8 sec .