(a) The velocity is given by integrating the acceleration:

\(v = \int a \, dt = \int kt^{1/2} \, dt = \frac{2}{3}kt^{3/2} + c\)

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

At \(t = 9\), \(v = 1.8\):

\(1.8 = \frac{2}{3}k \times 9^{3/2}\)

\(1.8 = \frac{2}{3}k \times 27\)

\(k = \frac{1.8 \times 3}{2 \times 27} = 0.1\)

(b) For \(t > 9\), the velocity is given by:

\(v = 0.2(t - 9)^2 + 1.8\)

The distance travelled from \(t = 0\) to \(t = 9\) is:

\(s_1 = \int_0^9 \frac{2}{3}kt^{3/2} \, dt = \frac{2}{3}k \left[ \frac{2}{5}t^{5/2} \right]_0^9\)

\(s_1 = \frac{4}{15}k \times 243 = \frac{4}{15} \times 0.1 \times 243 = 6.48 \text{ m}\)

The distance travelled from \(t = 9\) to \(t = 18\) is:

\(s_2 = \int_9^{18} (0.2(t - 9)^2 + 1.8) \, dt\)

\(s_2 = \left[ \frac{0.2}{3}(t - 9)^3 + 1.8t \right]_9^{18}\)

\(s_2 = \left[ \frac{0.2}{3} \times 9^3 + 1.8 \times 18 \right] - \left[ 0 + 1.8 \times 9 \right]\)

\(s_2 = \frac{0.2}{3} \times 729 + 32.4 - 16.2 = 64.8 \text{ m}\)

Thus, \(s_1 = \frac{1}{10}s_2\).

(c) For \(t > 9\), the acceleration is:

\(a = \frac{dv}{dt} = \frac{d}{dt}(0.2(t - 9)^2 + 1.8) = 0.4(t - 9)\)

The greatest acceleration occurs at \(t = 10\):

\(a = 0.4(10 - 9) = 0.4 \text{ m s}^{-2}\)