A metro train starts from rest and in $$5s$$  achieves $$108\,km/h.$$   After that it moves with constant velocity and comes to rest after travelling $$45\,m$$  with uniform retardation. If total distance travelled is $$395\,m,$$  find total time of travelling.

Solution :
Given : $$u = 0,t = 5\,\sec ,v = 108\,km/hr = 30\,m/s$$
By equation of motion $$v = u + at$$
$$\eqalign{ & {\text{or}}\,\,a = \frac{v}{t} = \frac{{30}}{5} = 6\,m/{s^2}\left[ {\because u = 0} \right] \cr & {S_1} = \frac{1}{2}a{t^2} = \frac{1}{2} \times 6 \times {5^2} = 75\,m \cr} $$
Distance travelled in first $$5\,\sec $$  is $$75\,m.$$
Distance travelled with uniform speed of $$30\,m/s$$  is $${S_2}$$
$$\eqalign{ & 395 = {S_1} + {S_2} + {S_3} \Rightarrow 395 = 75 + {S_2} + 45 \cr & \Rightarrow {S_2} = 275\,m \cr} $$
Time taken to travel $$275\,m = \frac{{275}}{{30}} = 9.2\,\sec $$
For retarding motion, we have
$$\eqalign{ & {0^2} - {32^2} = 2\left( { - a} \right) \times 45,\,{\text{we}}\,{\text{get}}\,a = 10\,m/{s^2} \cr & S = ut + \frac{1}{2}a{t^2} \Rightarrow 45 = 30t + \frac{1}{2}\left( { - 10} \right){t^2} \cr & \Rightarrow 45 = 30t - 5{t^2} \cr} $$
On solving we get, $$t = 3\,\sec $$
Total time taken $$ = 5 + 9.2 + 3 = 17.2\,\sec .$$