(i) The runner accelerates from rest at 1.2 m/s² for 5 s, reaching a velocity of:

\(v = 1.2 \times 5 = 6 \text{ m/s}\)

The runner then moves at this constant speed for 12 s, and finally decelerates uniformly to rest over 3 s.

The velocity-time graph consists of three segments: a linearly increasing segment, a constant segment, and a linearly decreasing segment.

The total distance is the area under the velocity-time graph, which is a trapezium:

\(\text{Area} = \frac{1}{2} \times (12 + 20) \times 6 = 96 \text{ m}\)

(ii) The cyclist accelerates uniformly for 10 s and decelerates uniformly to rest at Q at t = 30. The total time for the cyclist is 20 s, and the distance is 96 m.

Using the formula for the area of a triangle:

\(\frac{1}{2} \times 20 \times v = 96\)

\(v = 9.6 \text{ m/s}\)

Using the equation \(s = ut + \frac{1}{2} a t^2\) with \(u = 0\), \(s = 48\), and \(t = 10\):

\(48 = \frac{1}{2} a (10)^2\)

\(a = \frac{9.6}{10} = 0.96 \text{ m/s}^2\)