Old O level additional mathematics question on linear motion

Solution 1:

• Taking the time that the cyclist start his journey as my datum

  No, you've actually taken the point when the cyclist bypassed the train, rather than when they started their journey, as your datum.

• You assumed (against the given information) that the train stopped accelerating after the $15$ mins.

Here's a simpler (and clearer) approach:

Let the required time be $t$ after the train starts. Then both the cyclist and driver will have travelled the same distance from its starting point:

\begin{align}\text{cyclist's headstart} &+u_\text{bicycle}\,t &&=\frac12\,a_\text{train}\,t^2\\ 10 \textrm{ mph}\times 5\textrm{ mins} &+ 10 \textrm{ mph}\times t &&=\frac12\times\frac{20\textrm{ mph}-0\textrm{ mph}}{15\textrm{ mins} }\times t^2\\t&=18.956\textrm{ mins} \\\text{distance from train's starting point}&=4.0\textrm{ miles}\end{align}