# 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 km2. 512 km3. 728 km4. 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)

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