(i) At maximum speed, the driving force (DF) equals the resistance (R), so:

\(DF = R = 600 \text{ N}\)

The power (P) is given by:

\(P = DF \times v\)

Substituting the given power:

\(24000 = 600 \times v\)

Solving for \(v\):

\(v = \frac{24000}{600} = 40 \text{ m s}^{-1}\)

Thus, the speed cannot exceed 40 m s^-1.

(ii) Using Newton's second law, the net force is:

\(DF - R = ma\)

Substituting the given values:

\(\frac{24000}{15} - 600 = 1250a\)

Solving for \(a\):

\(1600 - 600 = 1250a\)

\(1000 = 1250a\)

\(a = \frac{1000}{1250} = 0.8 \text{ m s}^{-2}\)

Thus, the acceleration is 0.8 m s^-2.