(i) The acceleration \(a\) is the derivative of the velocity \(v\). So, \(a = \frac{dv}{dt} = 12t - 30\). For the acceleration to be negative, \(12t - 30 < 0\). Solving this inequality gives \(t < 2.5\).

(ii) The particle is at instantaneous rest when \(v = 0\). Solving \(6t^2 - 30t + 24 = 0\) gives \(t = 1\) and \(t = 4\). The distance \(s\) is the integral of \(v\), so \(s = \int (6t^2 - 30t + 24) \, dt = 2t^3 - 15t^2 + 24t\). Evaluating from \(t = 1\) to \(t = 4\), \(s = [2t^3 - 15t^2 + 24t]_1^4\), which gives a distance of 27 m.

(iii) The particle passes through \(O\) when \(s = 0\). Solving \(2t^3 - 15t^2 + 24t = 0\) gives \(t = 0\), \(t = 2.31\), and \(t = 5.19\). The positive values are \(t = 2.31\) and \(t = 5.19\).