(i) To find the displacement, integrate the velocity function: \(v = 6t^2 - kt^3\).

\(s = \int v \, dt = \int (6t^2 - kt^3) \, dt = 2t^3 - \frac{kt^4}{4}\).

(ii) At point B, the velocity \(v = 0\), so \(6t^2 - kt^3 = 0\).

Factor out \(t^2\): \(t^2(6 - kt) = 0\).

Since \(t \neq 0\), \(6 - kt = 0\) gives \(t = \frac{6}{k}\).

(iii) Given \(s = 108\), substitute \(t = \frac{6}{k}\) into the displacement equation:

\(2\left(\frac{6}{k}\right)^3 - \frac{k}{4}\left(\frac{6}{k}\right)^4 = 108\).

Simplify: \(2 \times \frac{216}{k^3} - \frac{1296}{4k^3} = 108\).

\(\frac{432}{k^3} - \frac{1296}{4k^3} = 108\).

\(\frac{432 - 324}{k^3} = 108\).

\(\frac{108}{k^3} = 108\) gives \(k^3 = 1\), so \(k = 1\).

(iv) To find the maximum value of \(v\), differentiate \(v\) with respect to \(t\):

\(\frac{dv}{dt} = 12t - 3kt^2\).

Set \(\frac{dv}{dt} = 0\): \(12t - 3kt^2 = 0\).

\(t(12 - 3kt) = 0\).

Since \(t \neq 0\), \(12 - 3kt = 0\) gives \(t = \frac{4}{k}\).

Substitute \(t = \frac{4}{k}\) into \(v = 6t^2 - kt^3\):

\(v = 6\left(\frac{4}{k}\right)^2 - k\left(\frac{4}{k}\right)^3\).

\(v = \frac{96}{k^2} - \frac{64}{k^2} = \frac{32}{k^2}\).

With \(k = 1\), maximum \(v = 32\).