First, calculate the driving force (DF) using the power formula:

\(P = F \cdot v\), where \(P = 400 \text{ W}\) and \(v = 4 \text{ m s}^{-1}\).

\(F = \frac{P}{v} = \frac{400}{4} = 100 \text{ N}\).

For uphill motion, apply Newton's second law:

\(DF - 80g \sin 2^{\circ} = 80a\).

Substitute \(DF = 100 \text{ N}\) and \(g = 9.8 \text{ m s}^{-2}\):

\(100 - 80 \times 9.8 \times \sin 2^{\circ} = 80a\).

Calculate \(\sin 2^{\circ} \approx 0.0349\):

\(100 - 80 \times 9.8 \times 0.0349 = 80a\).

\(100 - 27.352 = 80a\).

\(72.648 = 80a\).

\(a = \frac{72.648}{80} \approx 0.9 \text{ m s}^{-2}\).

For downhill motion, apply Newton's second law:

\(DF + 80g \sin 2^{\circ} = 80a\).

\(100 + 80 \times 9.8 \times 0.0349 = 80a\).

\(100 + 27.352 = 80a\).

\(127.352 = 80a\).

\(a = \frac{127.352}{80} \approx 1.6 \text{ m s}^{-2}\).

Using Newton's second law, the net force on the car is given by:

\(F - R = ma\)

At point A, the force equation is:

\(F_A - 800 = 600a_A\)

The power of the engine is 40 kW, so:

\(F_A = \frac{40000}{25} = 1600 \text{ N}\)

Substituting into the force equation at A:

\(1600 - 800 = 600a_A\)

\(800 = 600a_A\)

\(a_A = \frac{800}{600} = \frac{4}{3} \text{ m s}^{-2}\)

At point B, the acceleration is half of that at A:

\(a_B = \frac{1}{2}a_A = \frac{2}{3} \text{ m s}^{-2}\)

Using the force equation at B:

\(\frac{40000}{v_B} - 800 = 600 \left(\frac{2}{3}\right)\)

\(\frac{40000}{v_B} - 800 = 400\)

\(\frac{40000}{v_B} = 1200\)

\(v_B = \frac{40000}{1200} = 33.3 \text{ m s}^{-1}\)

First, calculate the driving force (DF) using the power formula:

\(\text{Power} = \text{Force} \times \text{Velocity}\)

\(35000 = \text{DF} \times 16\)

\(\text{DF} = \frac{35000}{16} = 2187.5 \text{ N}\)

Apply Newton's second law to find the acceleration:

\(\text{Net force} = \text{mass} \times \text{acceleration}\)

The net force is given by:

\(\text{DF} - 1150g \sin(1.2^{\circ}) - 975 = 1150a\)

Substitute the known values:

\(2187.5 - 1150 \times 9.8 \times \sin(1.2^{\circ}) - 975 = 1150a\)

Calculate the gravitational component:

\(1150 \times 9.8 \times \sin(1.2^{\circ}) \approx 235.8 \text{ N}\)

Substitute back:

\(2187.5 - 235.8 - 975 = 1150a\)

\(976.7 = 1150a\)

Solve for \(a\):

\(a = \frac{976.7}{1150} \approx 0.849 \text{ m s}^{-2}\)

Rounding to three significant figures gives:

\(a = 0.845 \text{ m s}^{-2}\)

(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.

(i) Using Newton's second law, the net force on the car is given by:

\(F - 900 = 1200a\)

where \(F\) is the force provided by the engine, and \(a\) is the acceleration (or deceleration in this case).

The power \(P\) of the engine is related to the force and speed by:

\(F = \frac{P}{v}\)

Substituting the given values:

\(F = \frac{18000}{25} = 720 \text{ N}\)

Substitute \(F\) back into the equation:

\(720 - 900 = 1200a\)

\(-180 = 1200a\)

\(a = -0.15 \text{ m s}^{-2}\)

Thus, the deceleration is 0.15 m s^-2.

(ii) To find the least speed, set the net force to zero:

\(\frac{18000}{v} - 900 = 0\)

Solve for \(v\):

\(\frac{18000}{v} = 900\)

\(v = \frac{18000}{900} = 20 \text{ m s}^{-1}\)

Therefore, the least speed is 20 m s^-1.