(i) The acceleration a is given by the derivative of velocity with respect to time: \(a = \frac{dv}{dt} = 3 \times 2 \times (2t - 5)^2\). Setting \(a = 54\), we have:

\(6(2t - 5)^2 = 54\)

\((2t - 5)^2 = 9\)

\(2t - 5 = \pm 3\)

Solving for \(t\), we get \(t = 1\) or \(t = 4\).

(ii) The displacement s is the integral of velocity with respect to time: \(s = \int v \, dt = \int (2t - 5)^3 \, dt\).

Using the substitution \(u = 2t - 5\), \(du = 2 \, dt\), we have:

\(s = \frac{1}{2} \int u^3 \, du = \frac{1}{2} \times \frac{u^4}{4} + C = \frac{(2t - 5)^4}{8} + C\).

Given \(s = 0\) at \(t = 0\), we find \(C\) by substituting \(t = 0\):

\(0 = \frac{(-5)^4}{8} + C\)

\(C = -\frac{625}{8}\)

Thus, the displacement is \(s = \frac{(2t-5)^4}{8} - \frac{625}{8}\).