(a) For the engine, apply Newton's second law:

\(125000 - 120000g \times 0.02 - 22000 - T = 120000a\)

\(125000 - 24000 - 22000 - T = 120000a\)

\(79000 - T = 120000a\)

For the coach:

\(T - 60000g \times 0.02 - 13000 = 60000a\)

\(T - 12000 - 13000 = 60000a\)

\(T - 25000 = 60000a\)

Combine the equations:

\(125000 - 120000g \times 0.02 - 22000 - 13000 = (120000 + 60000)a\)

\(125000 - 24000 - 22000 - 13000 = 180000a\)

\(66000 = 180000a\)

\(a = 0.3 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T - 25000 = 60000 \times 0.3\)

\(T = 43000 \text{ N}\)

(b) Given power \(P = 4500000 \text{ W}\) and speed \(v = 30 \text{ m/s}\), find the driving force \(DF\):

\(DF = \frac{P}{v} = \frac{4500000}{30} = 150000 \text{ N}\)

Resolve forces for the system:

\(150000 - 120000g \sin \beta - 60000g \sin \beta - 22000 - 13000 = 0\)

\(150000 - 180000g \sin \beta - 35000 = 0\)

\(115000 = 180000g \sin \beta\)

\(\sin \beta = \frac{115000}{180000g}\)

\(\beta = 3.7^\circ\)

(a)(i) The work done against the resisting force is calculated using the formula:

\(\text{Work Done} = \text{Force} \times \text{Distance}\)

The distance covered in 8 seconds at 36 m/s is:

\(36 \times 8 = 288 \text{ m}\)

Thus, the work done is:

\(1250 \times 288 = 360000 \text{ J} = 360 \text{ kJ}\)

(a)(ii) The power developed by the engine is:

\(\text{Power} = \frac{\text{Work Done}}{\text{Time}} = \frac{360000}{8} = 45000 \text{ W} = 45 \text{ kW}\)

(a)(iii) With the power increased by 12 kW, the new power is:

\(57000 \text{ W}\)

Using the formula for power:

\(P = DF \times v\)

We find the driving force (DF):

\(DF = \frac{57000}{36} = 1583.3 \text{ N}\)

Using Newton's second law:

\(1583.3 - 1250 = 1400a\)

Solving for acceleration:

\(a = \frac{333.3}{1400} = 0.238 \text{ m s}^{-2}\)

(b) For the inclined section, using the power equation:

\(64000 = 1250 \times 32 + 1400g \sin \theta \times 32\)

Solving for \(\theta\):

\(64000 = 40000 + 44800 \sin \theta\)

\(24000 = 44800 \sin \theta\)

\(\sin \theta = \frac{24000}{44800} = 0.5357\)

\(\theta = \arcsin(0.5357) \approx 3.1^\circ\)

(a) The power exerted by the cyclist is given by \(P = Fv\), where \(F\) is the forward force and \(v\) is the velocity. Thus, \(F = \frac{150}{4} = 37.5 \text{ N}\).

Using Newton's second law, \(F - 20 = m \times 0.25\).

Substitute \(F = 37.5\):

\(37.5 - 20 = m \times 0.25\)

\(17.5 = m \times 0.25\)

\(m = \frac{17.5}{0.25} = 70 \text{ kg}\).

(b) Resolving forces up the plane, the equation is:

\(\frac{150}{3} - 20 - 70g \sin \theta = 0\).

Solving for \(\theta\):

\(50 - 20 = 70g \sin \theta\)

\(30 = 70g \sin \theta\)

\(\sin \theta = \frac{30}{70g}\)

\(\theta = \sin^{-1} \left( \frac{3}{m} \right)\)

\(\theta = 2.5^\circ\) to 1 d.p.

(a) The driving force \(DF\) is given by \(DF = \frac{22500}{v}\). Applying Newton's second law for constant speed (\(a = 0\)), we have:

\(DF - 1400g \times 0.1 - 600 = 0\)

Solving for \(v\), we find \(v = 11.25 \text{ m s}^{-1}\).

(b) For the section inclined at 2°, using Newton's second law:

\(DF - 1400g \sin 2^{\circ} - 600 = 1400a\)

Substitute \(DF = \frac{22500}{11.25}\) and solve for \(a\):

\(\frac{22500}{11.25} - 1400g \sin 2^{\circ} - 600 = 1400a\)

\(a = 0.651 \text{ m s}^{-2}\) (3sf)

(a)(i) The total resistance force is the sum of the resistances on the car and caravan:

\(F = 400 + 250 = 650 \text{ N}\)

The power \(P\) is given by \(P = Fv\), where \(v = 25 \text{ m s}^{-1}\). Thus,

\(P = 650 \times 25 = 16250 \text{ W} = 16.25 \text{ kW}\)

(a)(ii) The engine's power is increased to 39 kW, so the driving force \(DF\) is:

\(DF = \frac{39000}{25} = 1560 \text{ N}\)

Using Newton's second law for the system:

\(1560 - 650 = 2400a\)

Solving for \(a\):

\(a = \frac{1560 - 650}{2400} = 0.379 \text{ m s}^{-2}\)

For the tension \(T\) in the tow-bar:

\(T - 250 = 800a\)

Substituting \(a\):

\(T = 250 + 800 \times 0.379 = 553 \text{ N}\)

(b) The car and caravan travel up a hill inclined at \(\sin^{-1}(0.05)\). The driving force is:

\(DF = 650 + 2400 \times 10 \times 0.05\)

Using \(P = Fv\):

\(32500 = (650 + 2400 \times 0.05)v\)

Solving for \(v\):

\(v = \frac{32500}{650 + 120} = 17.6 \text{ m s}^{-1}\)