(a) The velocity-time graph consists of three segments:

1. For \(0 \leq t \leq 5\), the graph is a straight line with equation \(v = 2t + 1\).
2. For \(5 \leq t \leq 7\), the graph is a downward-opening parabola with equation \(v = 36 - t^2\).
3. For \(7 \leq t \leq 13.5\), the graph is a straight line with equation \(v = 2t - 27\).

(b) To find the acceleration at \(t = 6\), differentiate the velocity function for \(5 \leq t \leq 7\):

\(v = 36 - t^2\)

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

At \(t = 6\), \(a = -2(6) = -12\).

(c) The total distance travelled is the integral of the absolute value of velocity over the given intervals:

\(s = \int_{0}^{5} (2t + 1) \, dt + \int_{5}^{6} (36 - t^2) \, dt + \int_{6}^{7} (36 - t^2) \, dt + \int_{7}^{13.5} (2t - 27) \, dt\)

Calculate each integral:

\(\int_{0}^{5} (2t + 1) \, dt = [t^2 + t]_{0}^{5} = 25 + 5 = 30\)

\(\int_{5}^{6} (36 - t^2) \, dt = [36t - \frac{t^3}{3}]_{5}^{6} = (216 - 72) - (180 - \frac{125}{3}) = 36 - 41.67 = -5.67\)

\(\int_{6}^{7} (36 - t^2) \, dt = [36t - \frac{t^3}{3}]_{6}^{7} = (252 - \frac{343}{3}) - (216 - 72) = 36 - 41.67 = -5.67\)

\(\int_{7}^{13.5} (2t - 27) \, dt = [t^2 - 27t]_{7}^{13.5} = (182.25 - 364.5) - (49 - 189) = -182.25 + 140 = -42.25\)

Total distance \(s = 30 + 5.67 + 5.67 + 42.25 = 84.25\).

(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: \(40 + 40 = 80 \text{ m}\).

(i) To find the maximum velocity during the first 2 seconds, differentiate \(v = 12t - 4t^2\) with respect to \(t\):

\(\frac{dv}{dt} = 12 - 8t\).

Set \(\frac{dv}{dt} = 0\) to find the critical points:

\(12 - 8t = 0 \Rightarrow t = 1.5\).

Substitute \(t = 1.5\) into the velocity equation:

\(v = 12 \times 1.5 - 4 \times (1.5)^2 = 9\) m s^-1.

(ii) For \(0 \leq t \leq 2\), the acceleration is \(\frac{dv}{dt} = 12 - 8t\). At \(t = 2\), \(\frac{dv}{dt} = 12 - 8 \times 2 = -4\).

For \(2 \leq t \leq 4\), the acceleration is \(\frac{dv}{dt} = -4\).

Since the acceleration is \(-4\) in both cases, there is no instantaneous change at \(t = 2\).

(iii) The velocity-time graph is a quadratic curve for \(0 \leq t \leq 2\) with a maximum at \(t = 1.5\), and a straight line with a negative gradient from \(t = 2\) to \(t = 4\).

(iv) The distance travelled is the area under the velocity-time graph. For \(0 \leq t \leq 2\), integrate \(12t - 4t^2\):

\(\int (12t - 4t^2) \, dt = 6t^2 - \frac{4}{3}t^3\) from 0 to 2.

\(= 6 \times 2^2 - \frac{4}{3} \times 2^3 = 16\).

For \(2 \leq t \leq 4\), the area of the triangle is \(\frac{1}{2} \times 2 \times 8 = 8\).

\(Total distance = 16 + 8 = 24.\)

Distance travelled is \(\frac{64}{3}\) m or 21.3 m.

(i) The velocity function for \(0 \leq t \leq 10\) is \(v = 4 + 0.2t\). The acceleration is the derivative of velocity with respect to time:

\(a = \frac{dv}{dt} = 0.2\) m s^-2.

(ii) For \(10 \leq t \leq 20\), the velocity function is \(v = -2 + \frac{800}{t^2}\). The acceleration is:

\(a = \frac{dv}{dt} = -\frac{d}{dt}\left(2 - \frac{800}{t^2}\right) = \frac{1600}{t^3}\).

At \(t = 20\), \(a = \frac{1600}{20^3} = -0.2\) m s^-2.

(iii) The velocity-time graph consists of a straight line from \((0, 4)\) to \((10, 6)\) and a curve from \((10, 6)\) to \((20, 0)\) with correct concavity.

(iv) The total distance is the area under the velocity-time graph. For \(0 \leq t \leq 10\), the area is a trapezium:

\(\text{Area} = \frac{1}{2} \times (4 + 6) \times 10 = 50\) m.

For \(10 \leq t \leq 20\), integrate \(v = -2 + \frac{800}{t^2}\):

\(\int_{10}^{20} \left(-2 + \frac{800}{t^2}\right) dt = \left[-2t - \frac{800}{t}\right]_{10}^{20} = (-40 - 40 + 20 + 80) = 20\) m.

\(Total distance = 50 + 20 = 70 m.\)