(i) The work done by the car's engine is calculated using the formula for power:

\(P = \frac{\text{Work Done}}{\Delta t}\)

Given \(P = 14,000 \text{ W}\) and \(\Delta t = 25 \text{ s}\),

\(\text{Work Done} = 14,000 \times 25 = 350,000 \text{ J}\).

(ii) To find the gain in kinetic energy, first calculate the velocities at A and B using the formula:

\(\frac{P}{v} - 235 = 1600 \times a\)

For point A: \(\frac{14,000}{v_A} - 235 = 1600 \times 0.5\)

\(v_A = \frac{2800}{207} \approx 13.53 \text{ m/s}\)

For point B: \(\frac{14,000}{v_B} - 235 = 1600 \times 0.25\)

\(v_B = \frac{2800}{127} \approx 22.05 \text{ m/s}\)

The gain in kinetic energy is:

\(\text{KE gain} = \frac{1}{2} m (v_B^2 - v_A^2)\)

\(= \frac{1}{2} \times 1600 \times (22.05^2 - 13.53^2)\)

\(= 242,500 \text{ J}\) or \(242.5 \text{ kJ}\).

(iii) The distance \(AB\) is found using the work-energy principle:

\(350,000 = 242,500 + 235 \times AB\)

\(235 \times AB = 350,000 - 242,500\)

\(AB = \frac{107,500}{235} \approx 457 \text{ m}\).