Given the acceleration function: \(a = 6t - 18\).

To find the velocity \(v\), integrate the acceleration:

\(v = \int (6t - 18) \, dt = 3t^2 - 18t + C\).

Since the particle starts from rest, \(v = 0\) when \(t = 0\), so \(C = 0\).

Thus, \(v = 3t^2 - 18t\).

To find when the particle comes to rest, set \(v = 0\):

\(3t^2 - 18t = 0\).

Factor the equation: \(3t(t - 6) = 0\).

So, \(t = 0\) or \(t = 6\). The particle comes to rest at \(t = 6\) (ignoring \(t = 0\) as it is the starting point).

To find the distance \(s\), integrate the velocity:

\(s = \int (3t^2 - 18t) \, dt = t^3 - 9t^2 + C\).

Since \(s = 0\) when \(t = 0\), \(C = 0\).

Thus, \(s = t^3 - 9t^2\).

Substitute \(t = 6\) to find the distance:

\(s = 6^3 - 9 \times 6^2 = 216 - 324 = -108\).

Since distance must be positive, \(s = 108\) m.