\((i) The particle is decelerating when the acceleration is negative. Given a = 6t - 2, set 6t - 2 < 0:\)

\(6t - 2 < 0\)

\(6t < 2\)

\(t < \frac{1}{3}\)

\(Thus, the particle is decelerating for 0 < t < \frac{1}{3}.\)

(ii) To find s in terms of t, first find the velocity v by integrating the acceleration:

\(v = \int (6t - 2) \, dt = 3t^2 - 2t + c\)

Next, integrate the velocity to find the displacement s:

\(s = \int (3t^2 - 2t + c) \, dt = t^3 - t^2 + ct + d\)

\(Using the conditions t = 1, s = 7 and t = 3, s = 29, form the equations:\)

\(c + d = 7\)

\(3c + d = 11\)

\(Solving these simultaneous equations gives c = 2 and d = 5.\)

\(Thus, s = t^3 - t^2 + 2t + 5.\)

\((iii) To find the time when the velocity is 10 m s^-1, set v = 10:\)

\(3t^2 - 2t + 2 = 10\)

\(3t^2 - 2t - 8 = 0\)

\(Solving this quadratic equation gives t = 2.\)