(i) To find the time taken for the motorcyclist to travel from A to B, we set the velocity equation to zero:

\(v = t - 0.01t^2 = 0\)

Solving for t, we get:

\(t(1 - 0.01t) = 0\)

\(Thus, t = 0 or t = 100. Since the motorcyclist starts at t = 0, the time taken is t = 100 seconds.\)

(ii) To find the distance AB, we integrate the velocity function:

\(\int v \, dt = \int (t - 0.01t^2) \, dt\)

This gives:

\(\frac{t^2}{2} - \frac{0.01t^3}{3}\)

\(Evaluating from t = 0 to t = 100:\)

\(\left( \frac{100^2}{2} - \frac{0.01 \times 100^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0.01 \times 0^3}{3} \right)\)

Calculating this gives:

\(5000 - \frac{100000}{3} = 1670 \text{ m}\)

(i) To find the distance travelled by P in the first 3 seconds, integrate the velocity function:

\(s = \int (8t - 2t^2) \, dt = 4t^2 - \frac{2}{3}t^3 + C\)

Using the limits from 0 to 3, we have:

\(s = [4(3)^2 - \frac{2}{3}(3)^3] - [4(0)^2 - \frac{2}{3}(0)^3] = 36 - 18 = 18\) m

(ii) For t > 3, integrate the velocity function:

\(s = \int \frac{54}{t^2} \, dt = -\frac{54}{t} + C\)

Using the condition that \(s(3) = 18\), we find:

\(18 = -\frac{54}{3} + C\)

\(C = 36\)

Thus, the displacement is \(s = 36 - \frac{54}{t}\).

(iii) Set the displacement equal to 27 m:

\(36 - \frac{54}{t} = 27\)

\(\frac{54}{t} = 9\)

\(t = 6\)

Substitute \(t = 6\) into the velocity function:

\(v = \frac{54}{6^2} = \frac{54}{36} = 1.5\) m/s

(i) The velocity of the particle is given by \(v = 0.03t^2\). To find the displacement \(x(t)\), we integrate the velocity function:

\(x(t) = \int v \, dt = \int 0.03t^2 \, dt = 0.01t^3 + C\)

Given that when \(t = 5\), \(x = 2.5\), we substitute to find \(C\):

\(2.5 = 0.01(5)^3 + C\)

\(2.5 = 1.25 + C\)

\(C = 1.25\)

Thus, the expression for displacement is \(x = 0.01t^3 + 1.25\).

(ii) We need to find the velocity when the displacement \(x = 11.25\):

\(0.01t^3 + 1.25 = 11.25\)

\(0.01t^3 = 10\)

\(t^3 = 1000\)

\(t = 10\)

Substitute \(t = 10\) into the velocity equation:

\(v = 0.03(10)^2 = 3 \text{ m/s}\)

(i) To find the time taken for the particle to travel from A to B, we set the velocity equation to zero:

\(t^2 (0.009 - 0.0001t) = 0\)

Solving for \(t\), we get \(t = 90\) seconds.

(ii) To find the distance AB, integrate the velocity function:

\(s = \int (0.009t^2 - 0.0001t^3) \, dt = 0.003t^3 - 0.000025t^4 + C\)

Evaluate from \(t = 0\) to \(t = 90\):

\(s = (0.003 \times 90^3 - 0.000025 \times 90^4) - (0 - 0) = 2187 - 1640.25 = 547 \text{ m}\)

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

\(\frac{dv}{dt} = 0.018t - 0.0003t^2 = 0\)

Solving for \(t\), we find \(t = 60\) seconds. Substitute back to find the maximum velocity:

\(v(60) = 0.009 \times 60^2 - 0.0001 \times 60^3 = 10.8 \text{ m s}^{-1}\)

(i) To find the displacement, integrate the velocity function \(v = 20t - t^3\) with respect to \(t\):

\(s(t) = \int (20t - t^3) \, dt = 10t^2 - 0.25t^4 + C\)

Given that at \(t = 0\), \(s = -36\), we find \(C\):

\(-36 = 10(0)^2 - 0.25(0)^4 + C\)

\(C = -36\)

Thus, the displacement is \(s(t) = 10t^2 - 0.25t^4 - 36\).

(ii) Substitute \(t = 4\) into the displacement expression:

\(s(4) = 10(4)^2 - 0.25(4)^4 - 36\)

\(s(4) = 160 - 64 - 36 = 60\)

Therefore, the displacement is 60 m.

(iii) To find when the particle is at \(O\), set \(s(t) = 0\):

\(10t^2 - 0.25t^4 - 36 = 0\)

Factor the equation:

\((t^2 - 36)(1 - 0.25t^2) = 0\)

Solving \(t^2 - 36 = 0\) gives \(t = 6\) or \(t = -6\).

Solving \(1 - 0.25t^2 = 0\) gives \(t = 2\) or \(t = -2\).

Since \(t\) must be positive, the values are \(t = 2\) and \(t = 6\).