a train starts from rest from a station with acceleration 0.2ms^-2 on astraight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4ms^-2 . If the total time spent is 30 minutes, then the distance betwen the two stations is(neglect the length of train):

1)216km

2)512km

3)728km

4)1296km

Train accelerates with a = 0.2 m/s² , t = t₁ and 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

Thanks & Regards,

Vasantha Sivaraj,

askIITians faculty

1. Then I will get the correct answerSimple tricks for this is distance = 1/2(AB/A+B)t^2. Where A and B are the value of acceleration and t is the time so plz try believe on Mahbub Azhari