(i) To find the time T when the particle reaches its maximum velocity, set the acceleration \(a = 0\):

\(1.2T^{1/2} - 0.6T = 0\)

Solving for \(T\):

\(1.2T^{1/2} = 0.6T\)

\(T^{1/2} = 0.5T\)

\(T^{1/2} = 2\)

\(T = 4\)

(ii) To find the velocity when the acceleration is maximum, first differentiate \(a\) with respect to \(t\):

\(\frac{da}{dt} = 0.6t^{-1/2} - 0.6\)

Set \(\frac{da}{dt} = 0\) to find \(t\):

\(0.6t^{-1/2} = 0.6\)

\(t^{-1/2} = 1\)

\(t = 1\)

Integrate \(a\) to find the velocity \(v\):

\(v = \int (1.2t^{1/2} - 0.6t) \, dt\)

\(v = 0.8t^{3/2} - 0.3t^2 + C\)

Using the initial condition \(v(0) = 1\):

\(1 = 0.8(0)^{3/2} - 0.3(0)^2 + C\)

\(C = 1\)

Thus, \(v = 0.8t^{3/2} - 0.3t^2 + 1\)

Substitute \(t = 1\) to find the velocity:

\(v = 0.8(1)^{3/2} - 0.3(1)^2 + 1\)

\(v = 0.8 - 0.3 + 1\)

\(v = 1.5\) m s^-1

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

(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 \(t = 10\):

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

(i) To find the velocity \(v\), differentiate the displacement \(s\) with respect to \(t\):

\(v = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 4t^2 + 4t) = 3t^2 - 8t + 4\).

(ii) To find when the particle is at instantaneous rest, set \(v = 0\):

\(3t^2 - 8t + 4 = 0\).

Solving this quadratic equation using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -8\), \(c = 4\):

\(t = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 3 \times 4}}{2 \times 3} = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6}\).

This gives \(t = \frac{12}{6} = 2\) and \(t = \frac{4}{6} = \frac{2}{3}\).

(iii) To find the minimum velocity, differentiate \(v\) and set the derivative to zero:

\(\frac{dv}{dt} = 6t - 8\).

Setting \(\frac{dv}{dt} = 0\) gives \(6t - 8 = 0\), so \(t = \frac{8}{6} = \frac{4}{3}\).

Substitute \(t = \frac{4}{3}\) back into the expression for \(v\):

\(v = 3\left(\frac{4}{3}\right)^2 - 8\left(\frac{4}{3}\right) + 4 = 3 \times \frac{16}{9} - \frac{32}{3} + 4\).

\(v = \frac{48}{9} - \frac{96}{9} + \frac{36}{9} = \frac{48 - 96 + 36}{9} = \frac{-12}{9} = -\frac{4}{3}\).

(i) To find k, integrate the acceleration to get the velocity:

\(v = \int k(3t^2 - 12t + 2) \, dt = k \left( \frac{3t^3}{3} - \frac{12t^2}{2} + 2t \right) + C\)

\(v = k(t^3 - 6t^2 + 2t) + C\)

Given initial velocity \(v = 0.4\) m/s when \(t = 0\), \(C = 0.4\).

At \(t = 1\), \(v = 0.1\):

\(0.1 = k(1 - 6 + 2) + 0.4\)

\(0.1 = k(-3) + 0.4\)

\(-0.3 = -3k\)

\(k = 0.1\)

(ii) Integrate the velocity to find displacement:

\(s = \int (0.1(t^3 - 6t^2 + 2t) + 0.4) \, dt\)

\(s = 0.1 \left( \frac{t^4}{4} - \frac{6t^3}{3} + \frac{2t^2}{2} \right) + 0.4t + C\)

\(s = 0.025t^4 - 0.2t^3 + 0.1t^2 + 0.4t\)

Since the particle starts from the origin, \(C = 0\).

(iii) Verify at \(t = 2\):

\(s = 0.025(2)^4 - 0.2(2)^3 + 0.1(2)^2 + 0.4(2)\)

\(s = 0.4 - 1.6 + 0.4 + 0.8 = 0\)

The particle is at the origin at \(t = 2\).