(i) The velocity is given by differentiating the displacement: \(v(t) = \frac{dx}{dt} = 1.2t - 0.012t^2\).

The acceleration is the derivative of velocity: \(a(t) = \frac{dv}{dt} = 1.2 - 0.024t\).

At \(t = 50\), \(a(50) = 1.2 - 0.024 \times 50 = 0\). Thus, the acceleration is zero.

The speed \(V\) at \(t = 50\) is \(v(50) = 1.2 \times 50 - 0.012 \times 50^2 = 30\) m s^-1.

(ii) For \(t \geq 50\), the motorcyclist travels at constant speed \(V = 30\) m s^-1.

The displacement at \(t = 50\) is \(s_1 = 0.6 \times 50^2 - 0.004 \times 50^3 = 1000\) m.

Let \(t_2\) be the time after \(t = 50\). The total time is \(t = 50 + t_2\).

The total distance is \(1000 + 30t_2\).

The average speed is given by \(\frac{1000 + 30t_2}{50 + t_2} = 27.5\).

Solving \(1000 + 30t_2 = 27.5(50 + t_2)\) gives \(t_2 = 150\).

Thus, \(t = 50 + 150 = 200\) seconds.