a train starts from rest from a station with accleration 0.2 on a staight track and then comes to rest after attaning maximum speed on another station due to retardation 0.4 ..if total time spent is half an hour, then distance between two stations is {neglect length of train}...

Train accelerates with a = 0.2 m/s² , t = t₁and v= v₁

v=u + at and v₁=0.2t₁ or t₁= 0.2 v₁ /2 …(1)

s=ut+(1/2)at²

Case II:

v₂=v₁+a₂t₂(train is retarding and s =s₁)

s₁=1/2 x 0.2 x t₁²

v₂=v₁+a₂t₂

0 = v₁+0.4t₂

t₂=–v1/0.4 …(2)

s₂=v₁t₂ +(1 /2) x a₂t₂²

t=t₁+t₂ = 30 min = 1800

t₁=1200 secs and t₂ = 600 secs

substituting a, v and t in (1) and (3), we get

s₁ = 144 km and s₂= 72 km

Total distance s= s₂+ s₂

s = 216km

Regards

Arun (askIITians forum expert)