(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 \(AC = AB + BC = 540 + 202.5 = 742.5 \text{ m}\).