Using Newton's second law, the driving force \(DF\) is given by:

\(DF - 600 - 1250 \times 0.02g = 1250 \times 0.5\)

Solving for \(DF\):

\(DF = 600 + 1250 \times 0.02 \times 9.8 + 1250 \times 0.5\)

\(DF = 600 + 250 + 625 = 1475 \text{ N}\)

The power provided by the engine is \(23 \text{ kW} = 23000 \text{ W}\). Therefore,

\(v = \frac{23000}{1475}\)

\(v \approx 15.6 \text{ m/s}\)

Thus, the speed of the car is 15.6 m/s.

(i) The power \(P\) of the engine is given as \(22.5 \, \text{kW} = 22500 \, \text{W}\). The force \(F\) exerted by the engine is given by \(F = \frac{P}{v}\), where \(v = 18 \, \text{m/s}\). Thus, \(F = \frac{22500}{18} = 1250 \, \text{N}\).

Using Newton's second law, the net force is \(F - R = ma\), where \(m = 600 \, \text{kg}\) and \(a = 1.4 \, \text{m/s}^2\). Therefore, \(1250 - R = 600 \times 1.4\).

Solving for \(R\), we get \(R = 1250 - 840 = 410 \, \text{N}\).

(ii) When the car is moving at a constant speed of \(15 \, \text{m/s}\), the power \(P\) is given by \(P = R \times v\). Thus, \(P = 410 \times 15 = 6150 \, \text{W}\).

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

\(P = F \times V\)

where \(F\) is the force (resistance) and \(V\) is the speed.

Given:

• \(P = 1330 \text{ kW} = 1330000 \text{ W}\)
• \(F = 28 \text{ kN} = 28000 \text{ N}\)

Substitute these values into the formula:

\(1330000 = 28000 \times V\)

Solve for \(V\):

\(V = \frac{1330000}{28000} = 47.5 \text{ m s}^{-1}\)

(i) Using the formula for power, \(P = Fv\), where \(F\) is the driving force and \(v\) is the velocity, we have:

\(144000 = F \times 16\)

\(F = \frac{144000}{16} = 9000 \text{ N}\)

Using Newton's second law, \(F - 4800 = 12500a\), where \(a\) is the acceleration at A:

\(9000 - 4800 = 12500a\)

\(4200 = 12500a\)

\(a = \frac{4200}{12500} = 0.336 \text{ m s}^{-2}\)

To find the speed at B, use the equation \(v^2 = u^2 + 2as\):

\(v^2 = 16^2 + 2 \times 0.096 \times s\)

Given \(a = 0.096 \text{ m s}^{-2}\) at B, solve for \(v\):

\(v = 24 \text{ m s}^{-1}\)

(ii) Work done by the driving force is \(5800 \times 500\).

Work done against resistance is \(4800 \times 500\).

Loss in kinetic energy is \(\frac{1}{2} \times 12500 \times (24^2 - 16^2)\).

Using energy conservation:

\(5800 \times 500 = 12500gh + \frac{1}{2} \times 12500 \times (24^2 - 16^2) + 4800 \times 500\)

Solving for \(h\):

\(5800 \times 500 = 12500gh + 4800 \times 500 + \frac{1}{2} \times 12500 \times (24^2 - 16^2)\)

\(2900000 = 12500gh + 2400000 + 12500 \times 80\)

\(2900000 = 12500gh + 2400000 + 1000000\)

\(2900000 = 12500gh + 3400000\)

\(12500gh = 2900000 - 3400000\)

\(12500gh = -500000\)

\(h = \frac{-500000}{12500 \times 9.8}\)

\(h = 20 \text{ m}\)

(i) The power exerted by the runner is given by \(P = Fv\), where \(F\) is the force. On horizontal ground, the resistance force \(F = kv\). Therefore, \(P = kv^2\).

Given \(P = 100 \text{ W}\) and \(v = 4 \text{ m s}^{-1}\), we have:

\(100 = k \times 4^2\)

\(100 = 16k\)

\(k = \frac{100}{16} = 6.25\)

(ii) When running uphill, the forces acting on the runner include the resistance \(kv\), gravitational component \(70g \sin \alpha\), and the power equation \(P = Fv\).

The equation becomes:

\(\frac{100}{v} = 70g \times 0.05 + 6.25v\)

\(\frac{100}{v} = 35 + 6.25v\)

Multiplying through by \(v\) gives:

\(100 = 35v + 6.25v^2\)

Rearranging gives the quadratic equation:

\(6.25v^2 + 35v - 100 = 0\)

Solving this quadratic equation using the quadratic formula \(v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6.25\), \(b = 35\), and \(c = -100\), we find:

\(v = \frac{-35 \pm \sqrt{35^2 - 4 \times 6.25 \times (-100)}}{2 \times 6.25}\)

\(v = \frac{-35 \pm \sqrt{1225 + 2500}}{12.5}\)

\(v = \frac{-35 \pm \sqrt{3725}}{12.5}\)

Calculating gives the maximum speed \(v \approx 2.08 \text{ m s}^{-1}\).