(i) To find the displacement \(s\), integrate the velocity function:

\(s = \int (0.6t - 0.03t^2) \, dt = 0.3t^2 - 0.01t^3 + C\).

Since the particle starts from O, \(C = 0\).

Calculate \(s(5)\):

\(s(5) = 0.3 \times 5^2 - 0.01 \times 5^3 = 6.25\).

To find the acceleration, differentiate the velocity function:

\(a = \frac{dv}{dt} = 0.6 - 0.06t\).

Calculate \(a(5)\):

\(a(5) = 0.6 - 0.06 \times 5 = 0.3 \text{ m s}^{-2}\).

(ii) Maximum velocity occurs when \(a = 0\):

\(0.6 - 0.06t = 0 \Rightarrow t = 10\).

Maximum velocity is \(v(10) = 3 \text{ m s}^{-1}\).

Half of maximum velocity is \(1.5 \text{ m s}^{-1}\).

Set \(0.6t - 0.03t^2 = 1.5\):

\(0.03t^2 - 0.6t + 1.5 = 0\).

Solving the quadratic equation:

\(t^2 - 20t + 50 = 0\).

Using the quadratic formula, \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), find:

\(t = 2.93 \text{ s}\) and \(t = 17.07 \text{ s}\).