(i) To find the time when the particle is again at instantaneous rest, we need to find when the velocity \(v = 0\). The acceleration is given by \(a = \frac{dv}{dt} = 6 - 0.24t\). Integrating with respect to \(t\), we get:

\(v = \int (6 - 0.24t) \, dt = 6t - 0.12t^2 + c\)

At \(t = 20\), the particle is at rest, so \(v = 0\):

\(0 = 6 \times 20 - 0.12 \times 20^2 + c\)

\(0 = 120 - 48 + c\)

\(c = -72\)

Now, set \(v = 0\) again to find the next time of rest:

\(0 = 6t - 0.12t^2 - 72\)

\(0.12t^2 - 6t + 72 = 0\)

Solving this quadratic equation gives \(t = 30\).

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

\(s = \int (6t - 0.12t^2 - 72) \, dt = 3t^2 - 0.04t^3 - 72t + k\)

Calculate \(s(30) - s(20)\):

\(s(30) = 3(30)^2 - 0.04(30)^3 - 72(30)\)

\(s(30) = 2700 - 1080 - 2160 = -540\)

\(s(20) = 3(20)^2 - 0.04(20)^3 - 72(20)\)

\(s(20) = 1200 - 320 - 1440 = -560\)

Distance traveled = \(|-540 - (-560)| = 20\).