(i) To find the distance AB, set the derivative of the distance function to zero to find when the car stops:

\(\frac{d}{dt}[0.0000117(400t^3 - 3t^4)] = 0\)

\(0.0000117(1200t^2 - 12t^3) = 0\)

\(1200t^2 = 12t^3\)

\(t = 0, 100\)

Distance \(AB = 0.0000117(400(100)^3 - 3(100)^4) = 1170\) m.

(ii) To find the maximum speed, differentiate the velocity function and set it to zero:

\(v(t) = \frac{d}{dt}[0.0000117(400t^3 - 3t^4)] = 0.0000117(1200t^2 - 12t^3)\)

\(\frac{d}{dt}[v(t)] = 0.0000117(2400t - 36t^2) = 0\)

\(2400t = 36t^2\)

\(t = 0, \frac{200}{3}\)

Maximum speed \(v_{\text{max}} = 0.0000117(1200(\frac{200}{3})^2 - 12(\frac{200}{3})^3) = 20.8\) m/s.

(iii) (a) At A, \(a(t) = 0\).

(b) At B, \(a(t) = 0.0000117(2400 \times 100 - 36 \times 100^2) = -1.40\) m/s².

(iv) The velocity-time graph starts at zero, increases to a maximum, and then decreases back to zero. The maximum is closer to \(t = 100\) than \(t = 0\). The graph has a zero gradient at \(t = 0\) and an inflection point closer to \(t = 0\) than \(t = 100\).

(a) To find the velocity function, integrate the acceleration:

\(v = \int (12 - 2t) \, dt = 12t - \frac{2t^2}{2} + c = 12t - t^2 + c\)

Using the initial condition \(v(0) = -20\), we find \(c = -20\).

Thus, \(v = 12t - t^2 - 20\).

Set \(v = 0\) to find when P is at rest:

\(12t - t^2 - 20 = 0\)

\(t^2 - 12t + 20 = 0\)

Solving this quadratic equation gives \(t = 2\) and \(t = 10\).

The velocity-time graph is an inverted quadratic starting at \((0, -20)\) and ending at \((12, -20)\), with roots at \(t = 2\) and \(t = 10\).

(b) To find the total distance, integrate the velocity function:

\(s = \int_{0}^{12} (12t - t^2 - 20) \, dt\)

\(s = \left[ \frac{12t^2}{2} - \frac{t^3}{3} - 20t \right]_{0}^{12}\)

Calculate the definite integral in segments:

\(s = \left[ \frac{12(2)^2}{2} - \frac{(2)^3}{3} - 20(2) \right] + \left[ \frac{12(10)^2}{2} - \frac{(10)^3}{3} - 20(10) \right] - \left[ \frac{12(0)^2}{2} - \frac{(0)^3}{3} - 20(0) \right]\)

\(s = 18.7 + 85.3 + 18.7 = 123 \text{ m}\)

(a) The acceleration \(a\) is given by differentiating the velocity function:

\(a = \frac{d}{dt}(pt^2 - qt) = 2pt - q.\)

Given that \(a = 0\) when \(t = 2\), we have:

\(2p(2) - q = 0 \Rightarrow 4p = q.\)

At \(t = 6\), the velocity must be continuous, so:

\(p(6)^2 - q(6) = 63 - 4.5(6).\)

Solving these equations:

\(36p - 6q = 36\)

\(4p = q\)

Substitute \(q = 4p\) into the first equation:

\(36p - 6(4p) = 36 \Rightarrow 36p - 24p = 36 \Rightarrow 12p = 36 \Rightarrow p = 3.\)

Then \(q = 4p = 12.\)

(b) The velocity-time graph consists of a quadratic curve from \(t = 0\) to \(t = 6\) and a straight line from \(t = 6\) to \(t = 14\). Key points are \((0, 0)\), \((6, 36)\), and \((14, 0)\).

(c) To find the total distance, integrate the velocity:

For \(0 \leq t \leq 6\):

\(s = \int_0^6 (3t^2 - 12t) \, dt = \left[ t^3 - 6t^2 \right]_0^6 = 216 - 144 = 72.\)

For \(6 \leq t \leq 14\):

\(s = \int_6^{14} (63 - 4.5t) \, dt = \left[ 63t - 2.25t^2 \right]_6^{14} = 144.\)

\(Total distance = 72 + 144 = 216 m.\)

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

\(2t^2 - 5t + 3 = 0\).

Factorize to get \((2t - 3)(t - 1) = 0\), giving \(t = 1\) or \(t = 1.5\).

To find the minimum velocity, complete the square or use calculus. The minimum occurs at \(t = 1.25\):

\(v = 2\left(\frac{5}{4}\right)^2 - 5\left(\frac{5}{4}\right) + 3 = -0.125\) m s\(^{-1}\).

(b) The velocity-time graph is a quadratic curve with roots at \(t = 1\) and \(t = 1.5\). Key points are (1.25, -0.125), (0, 3), (1, 0), (1.5, 0), (3, 6).

(c) To find the distance, integrate the velocity function from \(t = 1\) to \(t = 1.5\):

\(s = \int_{1}^{1.5} (2t^2 - 5t + 3) \, dt\).

Calculate:

\(s = \left[ \frac{2}{3}t^3 - \frac{5}{2}t^2 + 3t \right]_{1}^{1.5}\).

\(s = \left[ \frac{2}{3}(1.5)^3 - \frac{5}{2}(1.5)^2 + 3(1.5) \right] - \left[ \frac{2}{3}(1)^3 - \frac{5}{2}(1)^2 + 3(1) \right]\).

\(s = 0.0417\) m.

(a) To find the velocity, integrate the acceleration function over each interval:

For \(0 \leq t < 2\), \(a = 6 + 4t\). Integrating gives \(v = 6t + 2t^2 + C\).

For \(2 \leq t < 4\), \(a = 14\). Integrating gives \(v = 14t + C\).

For \(4 \leq t \leq 16\), \(a = 16 - 2t\). Integrating gives \(v = 16t - t^2 + C\).

Using the condition \(v = 0\) at \(t = 16\), solve \(0 = 16(16) - 16^2 + C\) to find \(C = 0\).

Set \(v = 55\) and solve \(55 = 16t - t^2\) for \(4 \leq t \leq 16\). This gives \(t = 5\) and \(t = 11\).

(b) The velocity-time graph is a positive quadratic for \(0 \leq t < 2\) and a negative quadratic for \(4 \leq t \leq 16\), passing through the origin and \((16, 0)\).

(c) The distance travelled while decelerating is found by integrating the velocity from \(t = 8\) to \(t = 16\):

\(s = \int_{8}^{16} (16t - t^2) \, dt = \left[ 8t^2 - \frac{1}{3}t^3 \right]_{8}^{16}\)

\(s = \left( 2048 - 1365.33 \right) - \left( 512 - 170.67 \right) = 341 \frac{1}{3} \text{ m}\)