(i) The velocity is given by v = qt + rt². We know that v = 4 when t = 1 and t = 2. This gives us the equations:

\(1. q + r = 4 (when t = 1)
2. 2q + 4r = 4 (when t = 2)\)

\(Solving these equations, we find q = 6 and r = -2.\)

The acceleration a is the derivative of v with respect to t:

\(a = \frac{dv}{dt} = q + 2rt = 6 - 4t\)

\(When t = 0.5,\)

\(a = 6 - 4(0.5) = 4 \text{ m s}^{-2}\)

\((ii) The particle is at instantaneous rest when v = 0:\)

\(v = 6t - 2t^2 = 0\)

\(Solving, we get t(6 - 2t) = 0, so t = 0 or t = 3.\)

\((iii) To find the distance from O when t = 0, we integrate the velocity:\)

\(s = \int (6t - 2t^2) \, dt = 3t^2 - \frac{2}{3}t^3 + C\)

\(Given that s = 0 when t = 3,\)

\(0 = 3(3)^2 - \frac{2}{3}(3)^3 + C\)

\(0 = 27 - 18 + C\)

\(C = -9\)

\(Thus, the distance from O when t = 0 is:\)

\(s = 3(0)^2 - \frac{2}{3}(0)^3 - 9 = -9\)

The distance is 9 m.