(i) To verify that \(t = 100\) when P is at A, substitute \(t = 100\) into the velocity equation:

\(v(100) = 0.16 \times 1000 - 0.016 \times 10000 = 0\).

Since \(v = 0\), P is at rest, confirming \(t = 100\).

(ii) To find the maximum speed, first find the acceleration \(a\) by differentiating \(v\):

\(a = \frac{dv}{dt} = 1.5 \times 0.16t^{\frac{1}{2}} - 0.032t\).

Set \(a = 0\) to find critical points:

\(1.5 \times 0.16t^{\frac{1}{2}} = 0.032t\).

Solving gives \(t^{\frac{1}{2}} = \frac{0.24}{0.032} \Rightarrow t = 56.25\).

Substitute \(t = 56.25\) into \(v(t)\) to find maximum speed:

\(v_{\text{max}} = 0.16 \times 421.875 - 0.016 \times 3164.0625 = 16.9 \text{ m s}^{-1}\).

(iii) To find the distance OA, integrate \(v\) with respect to \(t\):

\(s = \int v \, dt = \frac{2}{5} \times 0.16t^{\frac{5}{2}} - 0.016 \times \frac{t^3}{3}\).

Evaluate from \(t = 0\) to \(t = 100\):

\(s = 1070 \text{ m}\).

(iv) To find when P passes through O on returning, solve \(s(t) = 0\):

\(\frac{1}{3} t^{\frac{1}{2}} (0.192 - 0.016 \sqrt{t}) = 0\).

Solving gives \(t = 144\).