(i) The acceleration during the first 10 seconds is calculated using the formula:

\(a = \frac{\Delta v}{\Delta t} = \frac{5 - 0}{10} = 0.5 \text{ m s}^{-2}\)

(ii) The total distance travelled is the area under the velocity-time graph. This consists of several segments:

- Triangle from 0 to 10 s: \(\frac{1}{2} \times 10 \times 5 = 25 \text{ m}\)

- Rectangle from 10 to 30 s: \(20 \times 5 = 100 \text{ m}\)

- Trapezium from 30 to 40 s: \(\frac{1}{2} \times 10 \times (5 + V) = 5(5 + V) \text{ m}\)

- Rectangle from 40 to 70 s: \(30 \times V = 30V \text{ m}\)

- Triangle from 70 to 90 s: \(\frac{1}{2} \times 20 \times V = 10V \text{ m}\)

Total distance: \(25 + 100 + 5(5 + V) + 30V + 10V = 150 + 45V\)

Given total distance is 465 m:

\(150 + 45V = 465\)

\(45V = 315\)

\(V = 7 \text{ m s}^{-1}\)

(iii) Using energy methods, the change in kinetic energy (KE) and work done against friction are considered:

- Change in KE: \(\frac{1}{2} \times 80 \times 7^2 - \frac{1}{2} \times 80 \times 5^2 = 960 \text{ J}\)

- Work done against friction: \(20 \times \frac{(5 + 7)}{2} \times 10 = 1200 \text{ J}\)

Potential energy (PE) loss: \(80gh = 960 + 1200\)

\(80gh = 2160\)

\(gh = 27\)

\(h = 2.7 \text{ m}\)

(i) The change in gravitational potential energy (PE) is given by the formula:

\(\text{PE gain} = mgh\)

where \(m = 8 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(h = 20 \sin 30^{\circ}\).

Calculate \(h\):

\(h = 20 \sin 30^{\circ} = 20 \times 0.5 = 10 \text{ m}\)

Substitute into the PE formula:

\(\text{PE gain} = 8 \times 9.8 \times 10 = 784 \text{ J}\)

According to the mark scheme, the change in PE is 800 J.

(ii) The total work done against gravity and friction is given by:

\(8g \times 20 \sin 30^{\circ} + 20F = 1146\)

Substitute \(g = 9.8 \text{ m/s}^2\):

\(8 \times 9.8 \times 10 + 20F = 1146\)

Simplify:

\(784 + 20F = 1146\)

Rearrange to solve for \(F\):

\(20F = 1146 - 784\)

\(20F = 362\)

\(F = \frac{362}{20} = 18.1 \text{ N}\)

According to the mark scheme, the frictional force is 17.3 N.

(i) The change in gravitational potential energy (PE) is given by the formula:

\(\Delta PE = mgh\)

where \(h = x \sin \alpha\). Given \(\sin \alpha = \frac{5}{13}\), we have:

\(\Delta PE = 8 \times g \times x \times \frac{5}{13}\)

Assuming \(g = 9.8 \text{ m/s}^2\), the change in PE is:

\(\Delta PE = \frac{400}{13} x = 30.8x\)

(ii) Using the energy method, the work done against friction is:

\(WD = 15 \times x\)

The initial kinetic energy (KE) is:

\(KE = \frac{1}{2} \times 8 \times 5^2 = 100\)

Using the energy conservation principle:

\(KE_{\text{initial}} = \Delta PE + WD\)

Substitute the values:

\(100 = \frac{400}{13} x + 15x\)

Simplifying gives:

\(100 = \left( \frac{400}{13} + 15 \right) x\)

\(100 = \frac{260}{13} x\)

Solving for \(x\):

\(x = \frac{260}{119} = 2.18\)

(i)(a) For the smooth plane, apply Newton's second law parallel to the plane:

\(200 - 30g \sin 20^{\circ} = 30a\)

\(a = \frac{200 - 30 \times 9.8 \times \sin 20^{\circ}}{30} = 3.25 \text{ ms}^{-2}\)

(i)(b) Using the work-energy principle, the change in kinetic energy \(\Delta KE\) is:

\(\Delta KE = \frac{1}{2} \times 30 \times v^2\)

Using \(v^2 = 2as\), where \(s = 12 \text{ m}\):

\(v^2 = 2 \times 3.25 \times 12 = 77.9\)

\(\Delta KE = \frac{1}{2} \times 30 \times 77.9 = 1170 \text{ J}\)

(ii)(a) For the rough plane, resolve forces perpendicular to the plane to find the normal force \(N\):

\(N = 30g \cos 20^{\circ}\)

The frictional force \(F = \mu N = 0.12 \times 30g \cos 20^{\circ}\)

Apply Newton's second law parallel to the plane:

\(200 - 30g \sin 20^{\circ} - F = 30a\)

\(a = \frac{200 - 30 \times 9.8 \times \sin 20^{\circ} - 0.12 \times 30 \times 9.8 \times \cos 20^{\circ}}{30} = 2.12 \text{ ms}^{-2}\)

(ii)(b) When the force acts at 10° above the line of greatest slope, resolve forces perpendicular to the plane:

\(N + 200 \sin 10^{\circ} = 30g \cos 20^{\circ}\)

\(N = 30g \cos 20^{\circ} - 200 \sin 10^{\circ}\)

The frictional force \(F = 0.12 \times N\)

Apply Newton's second law parallel to the plane:

\(200 \cos 10^{\circ} - 30g \sin 20^{\circ} - F = 30a\)

\(a = \frac{200 \cos 10^{\circ} - 30 \times 9.8 \times \sin 20^{\circ} - 0.12 \times (30g \cos 20^{\circ} - 200 \sin 10^{\circ})}{30} = 2.16 \text{ ms}^{-2}\)

(a) The driving force \(F\) is given by \(P = Fv\), where \(P = 16000\) W and \(v = 20\) m/s. Thus, \(F = \frac{16000}{20} = 800\) N.

Using Newton's second law, \(F - 500 = 1200a\), we have:

\(800 - 500 = 1200a\)

\(300 = 1200a\)

\(a = 0.25\) m/s²

(b) For steady speed, \(F = 500\) N, so \(\frac{16000}{v} = 500\).

Solving for \(v\), we get:

\(v = \frac{16000}{500} = 32\) m/s

(c) Work done by the engine is \(16000 \times 15 = 240000\) J.

Change in kinetic energy is \(\frac{1}{2} \times 1200 \times v^2 - \frac{1}{2} \times 1200 \times 20^2\).

Change in potential energy is \(1200 \times g \times 316 \times \frac{1}{60} = 63200\) J.

Using the work-energy principle:

\(240000 - 128400 = \frac{1}{2} \times 1200 \times v^2 - 240000 + 63200\)

\(240000 - 128400 = 600v^2 - 240000 + 63200\)

Solving for \(v\), we find \(v = 21.9\) m/s.