(i) To find the acceleration, differentiate the velocity function:

\(a(t) = \frac{dv}{dt} = \frac{d}{dt}(0.2t + 0.006t^2) = 0.2 + 0.012t\).

The initial acceleration is \(a(0) = 0.2\).

We are given that \(a(t) = 2.5 \times a(0)\), so:

\(0.2 + 0.012t = 2.5 \times 0.2\).

Simplifying gives:

\(0.2 + 0.012t = 0.5\).

\(0.012t = 0.3\).

\(t = \frac{0.3}{0.012} = 25\).

(ii) To find the displacement, integrate the velocity function:

\(s(t) = \int (0.2t + 0.006t^2) \, dt = 0.1t^2 + 0.002t^3 + C\).

Assuming \(C = 0\) (since the particle starts from O), evaluate from \(t = 0\) to \(t = 25\):

\(s(25) = 0.1 \times 25^2 + 0.002 \times 25^3\).

\(s(25) = 0.1 \times 625 + 0.002 \times 15625\).

\(s(25) = 62.5 + 31.25 = 93.75\).