Solution:

Let the bus's position be \(x_b = (1/2)at^2 = t^2\). The cyclist's position (starting 96m behind) is \(x_c = 20t - 96\). For meeting, \(t^2 = 20t - 96 \Rightarrow t^2 - 20t + 96 = 0\). Factoring gives \((t-8)(t-12) = 0\), so \(t=8\)s or \(t=12\)s. At \(t=8\)s, cyclist is faster than bus. At \(t=12\)s, bus is faster, so it overtakes.