(i) To find when the particle is at rest, set \(v = 0\):

\(-0.01t^3 + 0.22t^2 - 0.4t = 0\)

Factor out \(t\):

\(t(-0.01t^2 + 0.22t - 0.4) = 0\)

Solving \(-0.01(t^2 - 22t + 40) = 0\) gives:

\(t^2 - 22t + 40 = 0\)

\((t - 20)(t - 2) = 0\)

Thus, \(t = 2\) or \(t = 20\).

(ii) Acceleration \(a\) is the derivative of velocity:

\(a = \frac{dv}{dt} = -0.03t^2 + 0.44t - 0.4\)

To find when acceleration is greatest, differentiate \(a\) and set to zero:

\(\frac{da}{dt} = -0.06t + 0.44 = 0\)

\(t = \frac{0.44}{0.06} = 7.33\)

(iii) To find the distance, integrate velocity from \(t = 2\) to \(t = 20\):

\(s(t) = \int (-0.01t^3 + 0.22t^2 - 0.4t) \, dt\)

\(s(t) = -\frac{0.01}{4}t^4 + \frac{0.22}{3}t^3 - 0.2t^2 + C\)

Evaluate from \(t = 2\) to \(t = 20\):

\(s(20) - s(2) = \frac{2673}{25} = 106.92\)

\(Distance = 107 m (rounded to nearest whole number).\)