A metro train starts from rest and in 5 s achieves 108 ~km / h ....

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 $$\begin{aligned} & =t_1+t_2+t_3 \\ & =5+9.2+3 \\ & =17.2 \mathrm{~s}\end{aligned}$$