Question reads as follows:
A road runs alongside a railway line and a cyclist travels along the road at a uniform speed of $10 ~\text{m.p.h.}$ The driver of a goods train starts his train $5$ minutes after the cyclist passes him and the train accelerates uniformly, attaining a speed of $20 ~\text{m.p.h.}$ in the next $15$ minutes. Find when and where the driver overtakes the cyclist.
Solution gives: $19$ minutes after train starts and $4$ miles. Try as I might I keep getting $20$ minutes after train starts and $4\frac{1}{6}$ miles.
My working is as follows:
Taking the time that the cyclist start his journey as my datum I have:
For cyclist: $~\text {v}=10~\text{m.p.h.}$ and $~\text {s}_1=10~\text{t}$
For train: acceleration $$~\text{a}=\frac{~\text{v-u}}{~\text{t}-\frac{1}{12}}~\text{where} ~\text{t}=\frac{1}{3}~\text{hr and v}=20~\text{m.p.h.} $$
i.e. $$a=\frac{20}{\frac{1}{4}}=80~\text{m.p.h}^2$$
During the first $20$ minutes the train travels:
$$s_2=\frac{80(t-\frac{1}{12})^2}{2}=\frac{40}{16}=\frac{5}{2}~\text{miles,}~\text{again where} ~\text{t}=\frac{1}{3}$$
Thereafter, train travels:
$$~\text{s}_3=20(t-\frac{1}{3})+\frac{5}{2}= 20~\text{t}-\frac{25}{6}$$
We require:
$$~\text{s}_1=~\text{s}_3$$ $$10~\text{t}=20~\text{t}-\frac{25}{6}$$ i.e.$$~\text{t}=\frac{25}{60}~\text{hr or 25}~\text{min}$$
but this $~\text{t}$ is the value for the bicycle. Therefore,$~\text{t}$ for train is $20$ minutes
As $$~\text {s}_1=10~\text{t}$$
we have $${s}_1=\frac{25}{6}$$
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}