First, find the velocity \(v\) by differentiating the displacement \(s\) with respect to time \(t\):

\(v = \frac{ds}{dt} = 3t^2 - 12t + 4\).

Next, find the acceleration \(a\) by differentiating the velocity \(v\) with respect to time \(t\):

\(a = \frac{dv}{dt} = 6t - 12\).

Set the acceleration to zero to find the time \(t\) when the acceleration is zero:

\(6t - 12 = 0\)

\(t = 2\).

Substitute \(t = 2\) into the velocity equation to find the velocity at this time:

\(v = 3(2)^2 - 12(2) + 4 = 12 - 24 + 4 = -8 \text{ ms}^{-1}\).

(i) To find the expression for \(s\), we start by integrating the acceleration to find the velocity:

\(v = \int (6t - 12) \, dt = 3t^2 - 12t + C\)

Next, integrate the velocity to find the displacement:

\(s = \int (3t^2 - 12t + C) \, dt = t^3 - 6t^2 + Ct + D\)

Using the conditions \(s = 5\) when \(t = 1\) and \(s = 1\) when \(t = 3\), we set up the equations:

\(5 = 1^3 - 6(1)^2 + C(1) + D\)

\(1 = 3^3 - 6(3)^2 + 3C + D\)

Solving these gives \(C = 9\) and \(D = 1\). Thus, \(s = t^3 - 6t^2 + 9t + 1\).

(ii) The particle is at instantaneous rest when \(v = 0\):

\(3t^2 - 12t + 9 = 0\)

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

\((t - 1)(t - 3) = 0\)

Thus, \(t = 1\) or \(t = 3\).

(iii) To find the total distance travelled, evaluate \(s\) at key points:

For \(0 < t < 1\), \(s = (1 - 6 + 9 + 1) - 1 = 4\)

For \(1 < t < 3\), \(s = (27 - 54 + 27 + 1) - 5 = 4\)

For \(3 < t \leq 4\), \(s = (64 - 96 + 36 + 1) - 1 = 4\)

Total distance is \(4 + 4 + 4 = 12\) m.

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

(i) To find when the particle comes to instantaneous rest, we need to find when the velocity v is zero. The velocity is the integral of acceleration:

\(v = \int (0.4t^3 - 4.8t^{1/2}) \, dt = 0.1t^4 - 3.2t^{3/2} + c\)

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

\(v = 0.1t^4 - 3.2t^{3/2}\)

\(Setting v = 0 gives:\)

\(0.1t^4 = 3.2t^{3/2}\)

Solving for t, we get:

\(t^{5/2} = 32\)

\(t = 4\)

\(Substitute t = 4 into the acceleration equation:\)

\(a = 0.4(4)^3 - 4.8(4)^{1/2} = 16\)

Thus, the acceleration is 16 m s^-2.

(ii) To find the displacement, integrate the velocity:

\(s = \int (0.1t^4 - 3.2t^{3/2}) \, dt = \left[0.02t^5 - 1.28t^{5/2}\right]_0^5\)

Calculate the displacement:

\(s = (0.02(5)^5 - 1.28(5)^{5/2}) - (0.02(0)^5 - 1.28(0)^{5/2})\)

\(s = -9.05\)

Thus, the displacement is -9.05 m.

(i) To find the maximum velocity, set the acceleration to zero:

\(25 - t^2 = 0\)

\(t^2 = 25\)

\(t = 5\)

Integrate the acceleration to find the velocity:

\(v = \int (25 - t^2) \, dt = 25t - \frac{1}{3}t^3 + C\)

Since the particle is initially at rest, \(v = 0\) when \(t = 0\), so \(C = 0\).

Substitute \(t = 5\) to find the maximum velocity:

\(v = 25 \times 5 - \frac{1}{3} \times 5^3 = 125 - \frac{125}{3} = \frac{250}{3} = 83\frac{1}{3} \text{ m s}^{-1}\)

(ii) To find the total distance travelled, integrate the velocity function from \(t = 0\) to \(t = 5\):

\(s = \int_0^5 (25t - \frac{1}{3}t^3) \, dt = \left[ \frac{25}{2}t^2 - \frac{1}{12}t^4 \right]_0^5\)

\(s = \left( \frac{25}{2} \times 25 - \frac{1}{12} \times 625 \right) = 260 \text{ m}\)

(iii) For \(t > 9\), the acceleration is given by \(a = -3t^{-1/2}\). Integrate to find the velocity change from \(t = 9\) to \(t = 25\):

At \(t = 9\), \(v = 25 \times 9 - \frac{1}{3} \times 9^3 = -18 \text{ m s}^{-1}\).

Integrate the acceleration:

\(\Delta v = \int_9^{25} -3t^{-1/2} \, dt = \left[ -6t^{1/2} \right]_9^{25}\)

\(\Delta v = -6 \times 5 + 6 \times 3 = -30 \text{ m s}^{-1}\)

Thus, the velocity at \(t = 25\) is \(-18 - 12 = -30 \text{ m s}^{-1}\).