Given, $v=108 \mathrm{~km} / \mathrm{h}=30 \mathrm{~m} / \mathrm{s}$
For first equation of motion
$v=u+a t$
$\therefore \quad 30=0+a \times 5 \quad(\because u=0)$
or $\quad a=6 \mathrm{~m} / \mathrm{s}^2$
So, distance travelled by metro train in 5 s
$s_1=\frac{1}{2} a t^2=\frac{1}{2} \times(6) \times(5)^2=75 \mathrm{~m}$
Distance travelled before coming to rest
$=45 \mathrm{~m}$
So, from third equation of motion
$\begin{aligned} & 0^2=(30)^2-2 a^{\prime} \times 45 \\ & \text { or } \quad a^{\prime}=\frac{30 \times 30}{2 \times 45}=10 \mathrm{~m} / \mathrm{s}^2 \\ & \end{aligned}$
Time taken in travelling 45 m is
$t_3=\frac{30}{10}=3 \mathrm{~s}$
Now, total distance $=395 \mathrm{~m}$
ie, $\quad 75+s^{\prime}+45=395 \mathrm{~m}$
or $s^{\prime}=395-(75+45)=275 \mathrm{~m}$
$\therefore \quad t_2=\frac{275}{30}=9.2 \mathrm{~s}$
Hence, total time taken in whole journey
$\begin{aligned} & =t_1+t_2+t_3 \\ & =5+9.2+3 \\ & =17.2 \mathrm{~s}\end{aligned}$