(i) For \(0 < t < 0.5\), the average acceleration \(a_{\text{ave}}\) is calculated as:

\(a_{\text{ave}} = \frac{6 - 4}{0.5 - 0} = \frac{2}{0.5} = 4 \text{ m/s}^2\)

For \(16.3 < t < 24.5\), the average acceleration \(a_{\text{ave}}\) is:

\(a_{\text{ave}} = \frac{21 - 19}{24.5 - 16.3} = \frac{2}{8.2} \approx 0.244 \text{ m/s}^2\)

(ii) Using the accelerations from part (i), assume \(a_1 = 4 \text{ m/s}^2\) at \(v = 5\) and \(a_2 = 0.244 \text{ m/s}^2\) at \(v = 20\).

Using Newton's second law, \(F = ma\), and the given power and resistance:

For \(v = 5\):

\(\frac{P}{5} - R = 1230 \times 4\)

For \(v = 20\):

\(\frac{P}{20} - R = 1230 \times 0.244\)

Solving these equations simultaneously:

\(P = 30800 \text{ W}\)

\(R = 1240 \text{ N}\)

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

\(F - 700 = 880 \times 0.625\)

Solving for \(F\):

\(F = 880 \times 0.625 + 700 = 1250 \text{ N}\)

The power \(P\) is given by the formula:

\(P = F \times v\)

Substituting the values:

\(P = 1250 \times 16 = 20,000 \text{ W}\)

(i) The power supplied by the engine is 30 kW, which is 30000 W. The work done by the engine is given by:

\(\text{Work Done} = \text{Power} \times \text{Time} = 30000 \times 100 = 3000000 \text{ J}\)

The work done against the resistance is:

\(\text{Work Done} = \text{Resistance} \times \text{Distance} = 750 \times AB\)

Equating the two expressions for work done:

\(750 \times AB = 3000000\)

\(AB = \frac{3000000}{750} = 4000 \text{ m}\)

(ii) Using the equation of motion \(v^2 = u^2 + 2as\), where \(u = 40 \text{ m/s}\), \(v = 20 \text{ m/s}\), and \(a\) is the acceleration:

\(20^2 = 40^2 + 2(-1.25)BC\)

Solving for \(BC\):

\(400 = 1600 - 2.5BC\)

\(2.5BC = 1200\)

\(BC = \frac{1200}{2.5} = 480 \text{ m}\)

(iii) The work done by the engine is:

\(30000 \times 14 = 420000 \text{ J}\)

The gain in kinetic energy is:

\(\frac{1}{2} \times 600 \times (30^2 - 20^2) = 150000 \text{ J}\)

The work done against resistance is:

\(750 \times CD = 420000 - 150000\)

\(750 \times CD = 270000\)

\(CD = \frac{270000}{750} = 360 \text{ m}\)

(i) The power \(P\) is given by \(P = Fv\), where \(F\) is the force and \(v\) is the velocity. Thus, \(F = \frac{P}{v} = \frac{720}{12} = 60 \, \text{N}\).

Using Newton's second law, \(F - R = ma\), where \(m = 75 \, \text{kg}\) and \(a = 0.16 \, \text{m/s}^2\). Therefore, \(60 - R = 75 \times 0.16\).

Solving for \(R\), we get \(R = 60 - 12 = 48\).

(ii) Given \(a > 0\), we have \(\frac{P}{v} - R = ma\). Since \(a > 0\), it implies \(\frac{P}{v} > R\).

Substituting \(R = 48\), we have \(\frac{720}{v} > 48\).

Solving for \(v\), \(v < \frac{720}{48} = 15\). Thus, the speed is less than \(15 \, \text{m/s}\).

Using the formula for power, \(P = Fv\), where \(F\) is the force exerted by the engine, we have:

\(\frac{P}{v} - R = ma\)

For \(v = 19\) m/s and \(a = 0.6\) m/s\(^{-2}\):

\(\frac{P}{19} - R = 1250 \times 0.6\)

For \(v = 30\) m/s and \(a = 0.16\) m/s\(^{-2}\):

\(\frac{P}{30} - R = 1250 \times 0.16\)

We have two equations:

1. \(\frac{P}{19} - R = 750\)
2. \(\frac{P}{30} - R = 200\)

To eliminate \(R\), subtract the second equation from the first:

\(\frac{P}{19} - \frac{P}{30} = 750 - 200\)

\(\frac{30P - 19P}{570} = 550\)

\(11P = 550 \times 570\)

\(P = 28500\)

Substitute \(P = 28500\) back into the first equation:

\(\frac{28500}{19} - R = 750\)

\(1500 - R = 750\)

\(R = 750\)