(i) To find the minimum velocity, differentiate \(v = t^2 - 8t + 12\) to find the acceleration \(a\):

\(a = \frac{dv}{dt} = 2t - 8\)

Set \(a = 0\) to find the critical points:

\(2t - 8 = 0\)

\(t = 4\)

Substitute \(t = 4\) back into the velocity equation:

\(v = 4^2 - 8 \times 4 + 12 = 16 - 32 + 12 = -4 \text{ ms}^{-1}\)

Thus, the minimum velocity is \(-4 \text{ ms}^{-1}\).

(ii) To find the total distance, first find when \(v = 0\):

\(v = t^2 - 8t + 12 = 0\)

\((t - 2)(t - 6) = 0\)

\(t = 2\) or \(t = 6\)

Integrate \(v\) to find the displacement \(s\):

\(s = \int (t^2 - 8t + 12) \, dt = \frac{1}{3}t^3 - 4t^2 + 12t + C\)

Calculate the displacement for each interval:

For \(0 \leq t \leq 2\):

\(s_1 = \left[ \frac{1}{3}t^3 - 4t^2 + 12t \right]_0^2 = \frac{8}{3} - 16 + 24 = \frac{32}{3}\)

For \(2 \leq t \leq 6\):

\(s_2 = \left[ \frac{1}{3}t^3 - 4t^2 + 12t \right]_2^6 = \left( \frac{216}{3} - 144 + 72 \right) - \left( \frac{8}{3} - 16 + 24 \right) = -\frac{32}{3}\)

For \(6 \leq t \leq 8\):

\(s_3 = \left[ \frac{1}{3}t^3 - 4t^2 + 12t \right]_6^8 = \left( \frac{512}{3} - 4 \times 8^2 + 12 \times 8 \right) - \left( \frac{216}{3} - 144 + 72 \right) = \frac{32}{3}\)

Total distance = \(|s_1| + |s_2| + |s_3| = \frac{32}{3} + \frac{32}{3} + \frac{32}{3} = 32 \text{ m}\).