(i) For 0 ≤ t ≤ 10, the particle moves with constant acceleration a. At t = 10, the velocity is given by the velocity function: \(v = \frac{800}{10^2} - 2 = 8 - 2 = 6\) m s^-1. Therefore, the speed of P when t = 10 is 6 m s^-1. Since the particle starts from rest, \(v = at\), so \(a = \frac{6}{10} = 0.6\) m s^-2.

(ii) For 10 ≤ t ≤ 20, the acceleration is given by differentiating the velocity function: \(a(t) = \frac{d}{dt} \left( \frac{800}{t^2} - 2 \right) = -\frac{1600}{t^3}\). Set \(a(t) = -0.6\) to find \(t\): \(-\frac{1600}{t^3} = -0.6\) implies \(t^3 = \frac{1600}{0.6}\), so \(t = 13.9\).

(iii) To find the displacement from A when t = 20, integrate the velocity function: \(s = \int (\frac{800}{t^2} - 2) \, dt = -\frac{800}{t} - 2t + C\). Using \(s(10) = 30\) to find \(C\), we have \(-\frac{800}{10} - 20 + C = 30\), so \(C = 130\). Then, \(s(20) = -\frac{800}{20} - 40 + 130 = 50\). Therefore, the displacement from A is 50 m.