(i) To find the time when the velocity is zero, integrate the acceleration to find the velocity:

\(v = \int (5.4 - 1.62t) \, dt = 5.4t - 0.81t^2 + C\).

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

Set \(v = 0\):

\(5.4t - 0.81t^2 = 0\)

\(t(5.4 - 0.81t) = 0\)

\(t = 0\) or \(t = \frac{5.4}{0.81} = \frac{20}{3}\) s.

Thus, the positive value of \(t\) is \(\frac{20}{3}\) s.

(ii) To find the velocity at \(t = 10\):

\(v(10) = 5.4(10) - 0.81(10)^2 = 54 - 81 = -27 \text{ m/s}\).

The velocity-time graph is an inverted parabola, starting at \(v = 0\) at \(t = 0\), passing through \(\left( \frac{20}{3}, 0 \right)\), and \(v = -27\) at \(t = 10\).

(iii) To find the total distance travelled, integrate the velocity to find the displacement:

\(s = \int (5.4t - 0.81t^2) \, dt = 2.7t^2 - 0.27t^3 + C\).

At \(t = 0\), \(s = 0\), so \(C = 0\).

Calculate the displacement at \(t = \frac{20}{3}\):

\(s = 2.7 \left( \frac{20}{3} \right)^2 - 0.27 \left( \frac{20}{3} \right)^3 = 40 \text{ m}\).

Calculate the displacement at \(t = 10\):

\(s = 2.7(10)^2 - 0.27(10)^3 = 270 - 270 = 0 \text{ m}\).

The total distance travelled is the sum of the absolute displacements: \(40 + 40 = 80 \text{ m}\).