(i) The power \(P\) is given by \(P = Fv\), where \(F\) is the force and \(v\) is the speed. Given \(P = 20000\) W and \(F = 650\) N, we have:

\(20000 = 650v\)

Solving for \(v\), we get:

\(v = \frac{20000}{650} = 30.8 \text{ m s}^{-1}\)

(ii) The force required to move up the hill is the sum of the resistive force and the component of the gravitational force along the hill. This is given by:

\(F = 650 + 1400g \times \frac{1}{7}\)

Where \(g = 9.8 \text{ m s}^{-2}\). The power \(P\) is then:

\(\frac{P}{10} = 650 + 1400 \times 9.8 \times \frac{1}{7}\)

Solving for \(P\), we find:

\(P = 26500 \text{ W}\)

(iii) The power when descending is 80% of the power found in part (ii):

\(P = 0.8 \times 26500 = 21200 \text{ W}\)

Using \(P = Fv\), where \(v = 20 \text{ m s}^{-1}\), we have:

\(21200 = F \times 20\)

\(F = 1060 \text{ N}\)

The net force is given by:

\(1060 - 650 - 1400 \times 9.8 \times \frac{1}{7} = 1400a\)

Solving for \(a\), we get:

\(a = 1.72 \text{ m s}^{-2}\)

(i) (a) The power developed by the engine is given by the formula:

\(P = Fv\)

where \(F = 1550 \text{ N}\) and \(v = 40 \text{ m s}^{-1}\).

\(P = 1550 \times 40 = 62000 \text{ W} = 62 \text{ kW}\).

(b) The new power is \(62 \text{ kW} - 22 \text{ kW} = 40 \text{ kW} = 40000 \text{ W}\).

Let \(DF\) be the new driving force. Then:

\(40000 = DF \times 40\)

\(DF = 1000 \text{ N}\).

Using Newton's second law:

\(DF - 1550 = 1100a\)

\(1000 - 1550 = 1100a\)

\(-550 = 1100a\)

\(a = -0.5 \text{ m s}^{-2}\) or \(d = 0.5 \text{ m s}^{-2}\).

(ii) The driving force \(DF\) is given by:

\(DF = 1100g \sin 8^{\circ} + 1550\)

\(DF = 3081 \text{ N}\).

The power equation is:

\(80000 = 3081v\)

\(v = \frac{80000}{3081} \approx 26 \text{ m s}^{-1}\).

(i) The driving force required to overcome the resistance is 300 N. The power \(P\) is given by the formula \(P = Fv\), where \(F\) is the force and \(v\) is the velocity.

Substituting the given values, \(P = 300 \times 40 = 12000 \text{ W} = 12 \text{ kW}\).

(ii) The power at 90% of the value found in part (i) is \(0.9 \times 12000 = 10800 \text{ W}\).

Using the formula \(P = Fv\), we find the driving force \(F\) when the speed is 25 m s^-1: \(F = \frac{10800}{25} = 432 \text{ N}\).

Applying Newton's second law, \(F - 300 = 1000a\), where 300 N is the resistance.

Thus, \(432 - 300 = 1000a\).

Solving for \(a\), we get \(a = \frac{132}{1000} = 0.132 \text{ ms}^{-2}\).

(i) The power \(P\) required to overcome the resistance is given by \(P = F \times v\), where \(F = 1350\) N and \(v = 32\) m/s.

\(P = 1350 \times 32 = 43200 \text{ W} = 43.2 \text{ kW}\).

(ii) When the car travels up the hill, the driving force \(DF\) must overcome both the resistance and the component of the weight down the slope.

The component of the weight down the slope is \(1200g \sin \theta\), where \(g = 9.8 \text{ m/s}^2\).

\(DF - 1350 - 1200 \times 9.8 \times 0.1 = 0\)

\(DF = 1350 + 1200 \times 9.8 \times 0.1 = 2550 \text{ N}\)

The power \(P\) is given by \(P = DF \times v\).

\(76500 = 2550 \times v\)

\(v = \frac{76500}{2550} = 30 \text{ m/s}\)

(i) To show that \(R = \frac{420}{v} + 43.56\):

Using Newton's second law, the acceleration \(a\) is given by:

\(a = \frac{5^2 - 3^2}{2 \times 500} = 0.016\)

The force equation is:

\(DF + 90g \times 0.05 - R = 90 \times 0.016\)

Using \(DF = \frac{P}{v}\), we have:

\(R = \frac{420}{v} + 43.56\)

(ii) To find the cyclist’s speed at the mid-point of the hill:

\(v_M^2 = 3^2 + 2 \times 0.016 \times 250\)

The speed at the mid-point is 4.12 m s^-1.

Decrease in \(R\) from top to mid-way:

\(420 \left( \frac{1}{3} - \frac{1}{\sqrt{17}} \right)\)

Decrease in \(R\) from midway to bottom:

\(420 \left( \frac{1}{\sqrt{17}} - \frac{1}{5} \right)\)

The decreases are 38.1 and 17.9 respectively.