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

\(P = F \times v\)

where \(F\) is the total resistance force and \(v\) is the velocity. The total resistance force \(F = 650 + 150 = 800\) N. Therefore,

\(P = 800 \times 24 = 19200\) W or 19.2 kW.

(a)(ii) When the engine’s power is increased to 40 kW, we have:

\(40000 = DF \times 24\)

Using Newton’s Second Law for the system:

\(\frac{40000}{24} - 800 = 2250a\)

Solving for \(a\):

\(a = \frac{40000}{24 \times 2250} - \frac{800}{2250} \approx 0.385\) m s^-2

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

\(T - 150 = 500a\)

\(T = 500 \times 0.385 + 150 = 343\) N

(b) Resolving forces up the hill, the total resistance force \(DF = 800 + 2250 \times g \times 0.14\).

Using \(P = DF \times v\):

\(31000 = (800 + 2250 \times 9.8 \times 0.14) \times v\)

Solving for \(v\):

\(v = \frac{31000}{800 + 2250 \times 9.8 \times 0.14} \approx 7.85\) m s^-1

To find the values of \(P\) and \(R\), we use the formula for power: \(P = Fv\), where \(F\) is the driving force and \(v\) is the velocity. The driving force \(F\) can be expressed as \(F = ma + R\), where \(m\) is the mass, \(a\) is the acceleration, and \(R\) is the resistance.

For the first condition (speed = 14 m s\(^{-1}\), acceleration = 1.4 m s\(^{-2}\)):

\(1000 \frac{P}{14} - R = 800 \times 1.4\)

For the second condition (speed = 25 m s\(^{-1}\), acceleration = 0.33 m s\(^{-2}\)):

\(1000 \frac{P}{25} - R = 800 \times 0.33\)

Solving these two equations simultaneously:

1. \(1000 \frac{P}{14} - R = 1120\)

2. \(1000 \frac{P}{25} - R = 264\)

Subtract equation 2 from equation 1:

\(1000 \left( \frac{P}{14} - \frac{P}{25} \right) = 1120 - 264\)

\(1000 \left( \frac{25P - 14P}{350} \right) = 856\)

\(1000 \frac{11P}{350} = 856\)

\(11P = 299.6\)

\(P = 27.2 \text{ kW}\)

Substitute \(P = 27.2\) back into one of the original equations to find \(R\):

\(1000 \frac{27.2}{14} - R = 1120\)

\(1942.857 - R = 1120\)

\(R = 1942.857 - 1120\)

\(R = 822.857 \approx 825 \text{ N}\)

(i) The driving force \(F\) provided by the engine is given by \(F = \frac{1000P}{25}\).

Using Newton's second law, the net force is also given by \(F - 600 = 1000 \times 0.2\).

Substituting the expression for \(F\), we have:

\(\frac{1000P}{25} - 600 = 200\)

\(40P - 600 = 200\)

\(40P = 800\)

\(P = 20\)

(ii) For steady speed, the acceleration \(a = 0\). Therefore, the driving force equals the resistance:

\(\frac{1000P}{v_{\text{max}}} = 600\)

Substituting \(P = 20\), we get:

\(\frac{20000}{v_{\text{max}}} = 600\)

\(v_{\text{max}} = \frac{20000}{600}\)

\(v_{\text{max}} = 33.3 \text{ m s}^{-1}\)

(i) The driving force (DF) is given by the formula for power: \(P = Fv\), where \(P = 1500,000\) W and \(v = 37.5\) m/s. Thus, \(DF = \frac{1500,000}{37.5} = 40,000\) N.

Using Newton's second law, \(DF - R = ma\), where \(R = 30,000\) N and \(m = 400,000\) kg.

Substitute the values: \(40,000 - 30,000 = 400,000a\).

Solve for \(a\): \(10,000 = 400,000a\) gives \(a = 0.025\) m/s².

(ii) For steady speed, acceleration \(a = 0\). Thus, \(DF - R = 0\).

\(\frac{1500,000}{v} - 30,000 = 0\).

Solve for \(v\): \(\frac{1500,000}{v} = 30,000\) gives \(v = 50\) m/s.

(i) The driving force (DF) provided by the engine is calculated using power and speed:

\(DF = \frac{17280}{12} = 1440 \text{ N}\).

Using Newton's second law, \(DF - R = ma\), where \(R = 960 \text{ N}\):

\(1440 - 960 = 1200a\).

Solving for \(a\), we get \(a = 0.4 \text{ m/s}^2\).

(ii) For constant speed \(V\), the net force is zero:

\(\frac{17280}{V} - 960 = 0\).

Solving for \(V\), we find \(V = 18 \text{ m/s}\).

(iii) After the engine is switched off, the car decelerates due to resistance:

\(-960 = 1200a\), giving \(a = -0.8 \text{ m/s}^2\).

Using the equations of motion for the segment \(BC\):

\(0 = 18 + (-0.8)t_{BC}\), solving gives \(t_{BC} = 22.5 \text{ s}\).

\(0 = 18^2 + 2(-0.8)s_{BC}\), solving gives \(s_{BC} = 202.5 \text{ m}\).

The distance \(AB\) is \(18 \times (52.5 - 22.5) = 540 \text{ m}\).

Thus, the total distance \(AC = AB + BC = 540 + 202.5 = 742.5 \text{ m}\).