Let after a time t, the cyclist overtake the bus. Then \(96+\frac{1}{2} \times 2 \times t^2=20 \times t\) or \(t^2-20 t+\) \(96=0\)
\(\therefore t =\frac{20 \pm \sqrt{400-4 \times 96}}{2 \times 1}\)
\(=\frac{20 \pm 4}{2}=8\,sec . \text { and } 12\,sec .\)