(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