A train starts from rest at S=0 and is subjected to an acceleration as shown in figure. Then

$a-s$ equation corresponding to given graph is,
$a=6-\frac{s}{5}$ 

$\therefore v·\frac{dv}{ds}=\left(6-\frac{s}{5}\right)$

or ${\int }_0^{v}vdv={\int }_0^s\left(6-\frac{s}{5}\right)ds$ 

or  $v=\sqrt{12s-\frac{s^2}{5}}$

At $s=10m,v=10m/s$

Maximum values of $v$ is obtained when $\frac{dv}{ds}=0$ which gives $s=30 \mathrm{m}$ 

$\therefore v_{\max }=\sqrt{12\times 30-\frac{(30{)}^2}{5}}$ 

$=\sqrt{180} \mathrm{m}/\mathrm{s}$