(i) The power provided by the engine is 18 kW, which is 18000 W. At maximum speed, the power is used to overcome the resistance force R. The formula for power is given by:

\(P = F \cdot v\)

where \(P = 18000\) W and \(v = 30\) m/s. Solving for \(F\):

\(18000 = R \cdot 30\)

\(R = \frac{18000}{30} = 600\) N

(ii) Using Newton's second law, the net force \(F\) is given by:

\(F = ma\)

At 20 m/s, the power is still 18000 W, so the driving force \(DF\) is:

\(DF = \frac{18000}{20} = 900\) N

The net force is the driving force minus the resistance:

\(900 - 600 = 1200a\)

\(300 = 1200a\)

\(a = \frac{300}{1200} = 0.25\) m/s²