First, calculate the driving force (DF) for each case using the formula \(DF = \frac{P}{v}\), where \(P\) is the power and \(v\) is the speed.

Case 1: Horizontal road

\(DF = \frac{36000}{18} = 2000 \text{ N}\)

The resistance force is \(Av + B\), so:

\(18A + B = 2000\)

Case 2: Inclined road

\(DF = \frac{21000}{12} = 1750 \text{ N}\)

The resistance force is \(Av + B\) plus the component of weight along the incline:

\(12A + B + 1000g \times 0.05 = 1750\)

\(12A + B + 500 = 1750\)

\(12A + B = 1250\)

Now solve the simultaneous equations:

1. \(18A + B = 2000\)

2. \(12A + B = 1250\)

Subtract equation 2 from equation 1:

\((18A + B) - (12A + B) = 2000 - 1250\)

\(6A = 750\)

\(A = 125\)

Substitute \(A = 125\) into equation 1:

\(18(125) + B = 2000\)

\(2250 + B = 2000\)

\(B = -250\)

(i) The driving force (DF) required to move the lorry up the hill is given by the sum of the resistance to motion and the component of the weight of the lorry acting down the slope. The component of the weight down the slope is \(12,000 \times g \times 0.08\), where \(g\) is the acceleration due to gravity (approximately 9.8 m s^-2). Therefore,

\(\text{DF} = 1500 + 12,000 \times 9.8 \times 0.08 = 11,100 \text{ N}\)

The power \(P\) of the engine is given by \(P = \text{DF} \times v\), where \(v = 5 \text{ m s}^{-1}\). Thus,

\(P = 11,100 \times 5 = 55,500 \text{ W} = 55.5 \text{ kW}\)

(ii) Given that the resistance to motion is \(kv^2\) and at \(v = 5 \text{ m s}^{-1}\), the resistance is 1500 N, we have:

\(k \times 5^2 = 1500\)

\(25k = 1500\)

\(k = 60\)

(iii) On a straight level road, the driving force is equal to the resistance, which is \(60v^2\). The power is still 55.5 kW, so:

\(55,500 = 60v^2 \times v = 60v^3\)

\(v^3 = \frac{55,500}{60}\)

\(v^3 = 925\)

\(v = 9.74 \text{ m s}^{-1}\)

(i) The driving force \(F\) is given by \(F = \frac{P}{v}\), where \(P = 6000 \text{ W}\) and \(v = 20 \text{ m s}^{-1}\). Thus, \(F = \frac{6000}{20} = 300 \text{ N}\).

The net force on the system is zero since the speed is constant: \(300 - R - 80 = 0\).

Solving for \(R\), we get \(R = 220 \text{ N}\).

(ii) The new driving force \(DF\) is \(\frac{12500}{25} = 500 \text{ N}\).

For the car: \(DF - T - R = 1500a\).

For the trailer: \(T - 80 = 300a\).

For the system: \(DF - 80 - R = 1800a\).

Using \(DF = 500 \text{ N}\) and \(R = 220 \text{ N}\), solve the equations:

1. \(500 - T - 220 = 1500a\)

2. \(T - 80 = 300a\)

Solving these simultaneously gives \(a = 0.111 \text{ m s}^{-2}\) and \(T = 113 \text{ N}\).

(i) The driving force is given by the power equation: \(Fv = 36000\). Therefore, \(F = \frac{36000}{20} = 1800 \text{ N}\). Using Newton's Second Law: \(1800 - R = 3200 \times 0.2\). Solving for \(R\), we get \(R = 1160 \text{ N}\).

(ii) The driving force required to maintain the speed on the incline is \(F = 3200g\sin(1.5^{\circ}) + 1160\). The power required is \(P = (3200g\sin(1.5^{\circ}) + 1160) \times 30\). Calculating gives \(P = 59900 \text{ W}\).

(iii) Using Newton's Second Law: \(- (3200g\sin(1.5^{\circ}) + 1160) = 3200a\). Solving for \(a\), we find \(a = -0.62426 \text{ m/s}^2\). Using the equation \(0^2 = 30^2 + 2as\), solve for \(s\) to get \(s = 721 \text{ m}\).

(i) The power of the car's engine is given by the formula:

\(\text{Power} = \text{Force} \times \text{Velocity}\)

Substituting the given values:

\(\text{Power} = 350 \times 15 = 5250 \text{ W}\)

(ii) When the car descends the hill, the forces acting along the slope are:

\(\text{Driving Force (DF)} + 1200g \sin 1^{\circ} - 350 = 1200 \times 0.12\)

Solving for DF:

\(\text{DF} = 1200 \times 0.12 - 1200g \sin 1^{\circ} + 350\)

The power is then:

\(P = \text{DF} \times 15 = 4270 \text{ W}\)

(iii) Using Newton's second law down the slope:

\(1200g \sin 1^{\circ} - 350 = 1200a\)

Using the constant acceleration formula:

\(18^2 = 20^2 + 2as\)

Solving for \(s\):

\(s = 324 \text{ m}\)

Alternatively, using energy methods:

Potential energy loss: \(1200g \times s \times \sin 1^{\circ}\)

Kinetic energy loss: \(\frac{1}{2} \times 1200 \times (20^2 - 18^2)\)

Equating energy losses to work done against resistance:

\(1200g \times s \times \sin 1^{\circ} + \frac{1}{2} \times 1200 \times (20^2 - 18^2) = 350s\)

Solving gives \(s = 324 \text{ m}\).