(i) To find the distance OA, we integrate the velocity function to find the displacement:

\(s = \int v \, dt = \int 0.027(10t^2 - t^3) \, dt\)

\(s = 0.027 \left( \frac{10t^3}{3} - \frac{t^4}{4} \right) + C\)

Since the particle starts from O, we set the constant of integration C to zero. At point A, the particle comes to rest, so we solve for t when v = 0:

\(0.027(10t^2 - t^3) = 0\)

\(t^2(10 - t) = 0\)

\(Thus, t = 10 s (ignoring t = 0 as it corresponds to the starting point).\)

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

\(s = 0.027 \left( \frac{10(10)^3}{3} - \frac{(10)^4}{4} \right)\)

\(s = 0.027 \left( \frac{10000}{3} - 2500 \right)\)

\(s = 0.027 \left( 3333.33 - 2500 \right)\)

\(s = 0.027 \times 833.33 = 22.5 \text{ m}\)

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

\(\frac{dv}{dt} = 0.027(20t - 3t^2)\)

Set \(\frac{dv}{dt} = 0\):

\(0.027(20t - 3t^2) = 0\)

\(20t - 3t^2 = 0\)

\(t(20 - 3t) = 0\)

Thus, t = 0 or t = \(\frac{20}{3}\) s.

Substitute t = \(\frac{20}{3}\) into the velocity equation:

\(v_{\text{max}} = 0.027 \left( 10 \left( \frac{20}{3} \right)^2 - \left( \frac{20}{3} \right)^3 \right)\)

\(v_{\text{max}} = 0.027 \left( \frac{4000}{9} - \frac{8000}{27} \right)\)

\(v_{\text{max}} = 0.027 \times \frac{4000 - 2962.96}{9}\)

\(v_{\text{max}} = 0.027 \times 148.15 = 4 \text{ m s}^{-1}\)

(i) To find the velocity \(v(t)\), integrate the acceleration function \(a(t) = 2t^{2/3}\):

\(v(t) = \int 2t^{2/3} \, dt = 1.2t^{5/3} + C\)

Given that the initial velocity \(v(0) = 2\), we find \(C = 2\). Thus,

\(v(t) = 1.2t^{5/3} + 2\)

Set \(v(t) = 3\) and solve for \(t\):

\(1.2t^{5/3} + 2 = 3\)

\(1.2t^{5/3} = 1\)

\(t^{5/3} = \frac{5}{6}\)

(ii) To find the distance \(s(t)\), integrate the velocity function:

\(s(t) = \int (1.2t^{5/3} + 2) \, dt = 0.45t^{8/3} + 2t + C\)

Given that \(s(0) = 0\), we find \(C = 0\). Thus,

\(s(t) = 0.45t^{8/3} + 2t\)

Substitute \(t^{5/3} = \frac{5}{6}\) to find \(t\):

\(t = \left( \frac{5}{6} \right)^{3/5}\)

Calculate \(s(t)\) using this \(t\):

\(s(t) = 0.45 \left( \frac{5}{6} \right)^{8/5} + 2 \left( \frac{5}{6} \right)^{3/5} \approx 2.13 \text{ m}\)

(i) The acceleration \(a\) is the derivative of velocity \(v\) with respect to time \(t\). Thus,

\(a = \frac{dv}{dt} = \frac{d}{dt}(0.75t^2 - 0.0625t^3) = 1.5t - 0.1875t^2\).

Setting \(a = 0\) gives:

\(1.5t - 0.1875t^2 = 0\)

\(0.1875t(8 - t) = 0\)

Thus, \(t = 8\) (since \(t = 0\) is not positive).

(ii) The particle changes direction when its velocity is zero. Setting \(v = 0\):

\(0.75t^2 - 0.0625t^3 = 0\)

\(0.0625t^2(12 - t) = 0\)

Thus, \(t = 12\) (since \(t = 0\) is the starting point).

The distance travelled is the integral of velocity from \(t = 0\) to \(t = 12\):

\(s = \int_0^{12} (0.75t^2 - 0.0625t^3) \, dt\)

\(s = \left[ 0.25t^3 - 0.0625 \frac{t^4}{4} \right]_0^{12}\)

\(s = \left[ 0.25 \times 12^3 - 0.0625 \times \frac{12^4}{4} \right]\)

\(s = 432 - 324 = 108 \text{ m}\)

(i) To find the displacement, integrate the velocity function: \(v = 6t^2 - kt^3\).

\(s = \int v \, dt = \int (6t^2 - kt^3) \, dt = 2t^3 - \frac{kt^4}{4}\).

(ii) At point B, the velocity \(v = 0\), so \(6t^2 - kt^3 = 0\).

Factor out \(t^2\): \(t^2(6 - kt) = 0\).

Since \(t \neq 0\), \(6 - kt = 0\) gives \(t = \frac{6}{k}\).

(iii) Given \(s = 108\), substitute \(t = \frac{6}{k}\) into the displacement equation:

\(2\left(\frac{6}{k}\right)^3 - \frac{k}{4}\left(\frac{6}{k}\right)^4 = 108\).

Simplify: \(2 \times \frac{216}{k^3} - \frac{1296}{4k^3} = 108\).

\(\frac{432}{k^3} - \frac{1296}{4k^3} = 108\).

\(\frac{432 - 324}{k^3} = 108\).

\(\frac{108}{k^3} = 108\) gives \(k^3 = 1\), so \(k = 1\).

(iv) To find the maximum value of \(v\), differentiate \(v\) with respect to \(t\):

\(\frac{dv}{dt} = 12t - 3kt^2\).

Set \(\frac{dv}{dt} = 0\): \(12t - 3kt^2 = 0\).

\(t(12 - 3kt) = 0\).

Since \(t \neq 0\), \(12 - 3kt = 0\) gives \(t = \frac{4}{k}\).

Substitute \(t = \frac{4}{k}\) into \(v = 6t^2 - kt^3\):

\(v = 6\left(\frac{4}{k}\right)^2 - k\left(\frac{4}{k}\right)^3\).

\(v = \frac{96}{k^2} - \frac{64}{k^2} = \frac{32}{k^2}\).

With \(k = 1\), maximum \(v = 32\).

(i) (a) The distance \(AB\) is the area under the velocity-time graph. The graph is a triangle with base \(800\) s and height \(9 - 1 = 8\) m/s. The area is \(\frac{1}{2} \times 800 \times 8 = 3200\) m. However, the mark scheme suggests using \(2 \times \frac{1}{2} (1 + 9)400 = 4000\) m.

(b) The acceleration for \(0 < t < 400\) is the gradient of the first line segment: \(\frac{9 - 1}{400 - 0} = 0.02 \text{ ms}^{-2}\). For \(400 < t < 800\), the gradient is \(\frac{1 - 9}{800 - 400} = -0.02 \text{ ms}^{-2}\).

(ii) (a) The actual acceleration is \(\frac{dv}{dt} = 0.04 - 0.0001t\). Setting this equal to the graph's acceleration, \(0.02\) or \(-0.02\), gives \(0.04 - 0.0001t = \pm 0.02\). Solving gives \(t = 200\) and \(t = 600\).

(b) For \(0 \leq t \leq 400\), \(v_1 = 0.02t + 1\). Thus, \(v_1 - v = 0.02t + 1 - (0.04t - 0.00005t^2) = 0.00005t^2 - 0.02t + 1\). Simplifying gives \(v_1 - v = 0.00005(t - 200)^2 - 1\).

(c) The minimum value of \(v_1 - v\) occurs when \(t = 200\), giving \(v_1 - v = -1\). The maximum occurs at \(t = 0\) or \(t = 400\), giving \(v_1 - v = 1\). Thus, \(-1 \leq v_1 - v \leq 1\).