(i) Using the equation of motion \(v^2 = u^2 + 2as\) where initial velocity \(u = 0\), acceleration \(a = 10 \text{ m/s}^2\), and distance \(s = 1.25 \text{ m}\):

\(v^2 = 2 \times 10 \times 1.25\)

\(v^2 = 25\)

\(v = 5 \text{ m/s}\)

Using \(s = ut + \frac{1}{2}at^2\):

\(1.25 = \frac{1}{2} \times 10 \times t^2\)

\(t^2 = 0.25\)

\(t = 0.5 \text{ s}\)

(ii) The velocity after passing through A is given by integrating the acceleration:

\(v = \int (10 - 0.3t) \, dt = 10t - 0.15t^2 + C\)

At \(t = 0\), \(v = 5 \text{ m/s}\), so \(C = 5\).

\(v = 10t - 0.15t^2 + 5\)

Integrating to find the distance:

\(x = \int (10t - 0.15t^2 + 5) \, dt = 5t^2 - 0.05t^3 + 5t\)

Substitute \(t = 2.5\) (since \(t = 3 - 0.5\)):

\(x = 5 \times 2.5^2 - 0.05 \times 2.5^3 + 5 \times 2.5\)

\(x = 42.97 \text{ m}\)

Total distance \(OP = 1.25 + 42.97 = 44.2 \text{ m}\)