A metro train starts from rest and in 5 s 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.

• 12.2 s

• 15.3 s

• 9 s

• 17.2 s

D.

17.2 s

Given:- v= 108 km/h = 30 m/s

For first equation of motion

          v = u +  $\mathrm{\alpha }$ t

         30 = 0 + $\mathrm{\alpha }$ × 5                   ( $\because$ u = 0)

⇒          a = 6 m/s²

So, distance travelled by metro train in 5s

             ${\mathrm{s}}_1 = \frac{1}{2}\mathrm{\alpha } {\mathrm{t}}^2$  

                 = $\frac{1}{2} \times \left(6\right) \times {\left(5\right)}^2$

             ${\mathrm{s}}_{1 }= 75 \mathrm{m}$

Distance travelled before coming to rest = 45 m

So, from third equation of motion

            ${\left(0\right)}^2 = {\left(30\right)}^2 - 2 \mathrm{\alpha }\text{'} \times 45$

⇒          $\mathrm{\alpha }\text{'} = \frac{30 \times 30}{2 \times 45}$

⇒           $\mathrm{\alpha }\text{'} = 10 \mathrm{m}/{\mathrm{s}}^2$

Time taken in travelling 45 m is

             ${\mathrm{t}}_3 = \frac{30}{10}$

               ${\mathrm{t}}_3$ = 3s

Now, total distance = 395 m

i.e       75 + s' + 45 = 395 m

i.e        s'  = 395 - (75 + 45)

⇒          s' = 275 m             

∴           t₂ =  $\frac{275}{30}$

             t₂  = 9.2 s

Hence, total time taken in whole journey

              = t₁ + t₂ + t₃

              = 5 + 9.2 + 3

               = 17.2 s