a train starts from a station with acceleration 0.2 m/s² on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s². if total time spent is half an hour, then distance between two stations is {neglect length of train}

1. 216 km

2. 512 km

3. 728 km

4. 1296 km

In the first case when train is accelerating

v = u + at

here u = 0

a = 0.2 m/s²

t = t₁

v = v₁

so,

v¹ = 0.2t₁

thus,

t₁ = v₁/0.2    (1)

similarly,

s = ut + (1/2)at²

here s = s₁

and by substituting the other values, we have

s1 = (1/2) x 0.2 x t₁²  (2)

now, when the train is retarding

v₂ = v₁ + a₂t₂

here as u = v₁

a₂ = -0.4 m/s²

and v₂ = 0

thus,

0 = v₁ + 0.4t₂

or

t2 = -v₁/0.4    (3)

also

s₂ = v₁t₂ + (1/2)a₂t₂²  

so

s₂ = v₁t₂ - (1/2)x0.4xt₂²   (4)

now as t = t₁ + t₂ = 30min = 1800

by adding equations (1) and (3), we get

1800 = v₁[1/0.2 + 1/0.4]

or

v₁ = 1800/7.5

thus,

v₁ = 240 m/s

so, t₁ = v₁/0.2 = 240/0.2

or

t₁ = 1200 secs.

similarly

t₂ = 240/0,4 = 600 secs.

now by substituting appropriate values of a,v and t in equations (2) and (4), we get

s₁ = (1/2) x 0.2 x (1200)²

or

s₁ = 144 km

similarly,

s₂ = (240 x 600) - [(1/2) x 0.4 x (600)²]

so, the total distance travelled will be

s = s₁ + s₂ = 144km + 72km

or

s = 216km

which is option (1)