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(a)(i)
Acceleration is defined as the change in velocity per unit time:
\(a = \frac{v - u}{t}\)
Given \(u = 0\text{ m/s}\), \(v = 12\text{ m/s}\), and \(t = 8.0\text{ s}\):
\(a = \frac{12 - 0}{8.0} = 1.5\text{ m/s}^2\).

(a)(ii)
First, find the total mass of the system:
\(m = \text{mass of cyclist} + \text{mass of bicycle} = 65\text{ kg} + 15\text{ kg} = 80\text{ kg}\).
Now, use the kinetic energy formula:
\(E_k = \frac{1}{2} m v^2\)
\(E_k = \frac{1}{2} \times 80\text{ kg} \times (12\text{ m/s})^2 = 40 \times 144 = 5760\text{ J}\) (or \(5.76\text{ kJ}\)).

(b)(i)
When traveling at a constant speed in a straight line, there is no acceleration. Therefore, the resultant force must be \(0\text{ N}\).

(b)(ii)
As the cyclist moves down the slope, their height decreases, so their gravitational potential energy decreases.
Since their speed is constant, their kinetic energy remains constant.
The lost gravitational potential energy is transferred to thermal energy in the surroundings and bicycle components due to work done against air resistance and friction.

(a)(i) [Total: 3 marks]
• Recall and use of \(a = \frac{\Delta v}{t}\) [1]
• Correct calculation: \(1.5\) [1]
• Correct unit: \(m/s^2\) (or \(m\ s^{-2}\)) [1]

(a)(ii) [Total: 2 marks]
• Addition of masses to find total mass (\(80\text{ kg}\)) AND substitution into \(E_k = \frac{1}{2} m v^2\) [1]
• Correct evaluation: \(5760\text{ J}\) (or \(5.76\text{ kJ}\)) [1]

(b)(i) [Total: 1 mark]
• \(0\text{ N}\) (accept zero, ignore missing units but reject incorrect units) [1]

(b)(ii) [Total: 2.88 marks]
• Identifies that gravitational potential energy decreases [1]
• Identifies that kinetic energy remains constant / does not change [1]
• States that gravitational potential energy is transferred to thermal/heat energy (due to friction or air resistance) [0.88]