(i) For particle A, since it moves with constant speed 10 m s^-1 and passes O two seconds before B, its displacement from O after t seconds is given by:

\(s_A = 20 + 10t\)

For particle B, using the equation of motion with initial speed 16 m s^-1 and deceleration 2 m s^-2, the displacement is:

\(s_B = 16t - \frac{1}{2}(2)t^2 = 16t - t^2\)

\((ii) Particle B comes to rest when its velocity is zero. Using v = u + at, we have:\)

\(0 = 16 - 2t \Rightarrow t = 8\)

\(At t = 8, the distance between A and B is:\)

\(s = s_A - s_B = (20 + 10 \times 8) - (16 \times 8 - 8^2) = 100 - 64 = 36 \text{ m}\)

(iii) To find the minimum distance, consider the function:

\(s = s_A - s_B = (20 + 10t) - (16t - t^2) = t^2 - 6t + 20\)

\(Complete the square: s = (t - 3)^2 + 11\)

\(The minimum value occurs at t = 3, giving s = 11 \text{ m}\)