(i) To find the time taken for the particle to travel from A to B, we set the velocity equation to zero:

\(t^2 (0.009 - 0.0001t) = 0\)

Solving for \(t\), we get \(t = 90\) seconds.

(ii) To find the distance AB, integrate the velocity function:

\(s = \int (0.009t^2 - 0.0001t^3) \, dt = 0.003t^3 - 0.000025t^4 + C\)

Evaluate from \(t = 0\) to \(t = 90\):

\(s = (0.003 \times 90^3 - 0.000025 \times 90^4) - (0 - 0) = 2187 - 1640.25 = 547 \text{ m}\)

(iii) To find the maximum velocity, differentiate the velocity function and set it to zero:

\(\frac{dv}{dt} = 0.018t - 0.0003t^2 = 0\)

Solving for \(t\), we find \(t = 60\) seconds. Substitute back to find the maximum velocity:

\(v(60) = 0.009 \times 60^2 - 0.0001 \times 60^3 = 10.8 \text{ m s}^{-1}\)