a train starts from a station with acceleration 0.2 m/s^{2} on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s^{2}. 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/s^{2}

t = t_{1}

v = v_{1}

so,

v^{1} = 0.2t_{1}

thus,

t_{1 }= v_{1}/0.2 ** (1)**

similarly,

s = ut + (1/2)at^{2}

here s = s_{1}

and by substituting the other values, we have

s1 = (1/2) x 0.2 x t_{1}^{2} **(2)**

now, when the train is retarding

v_{2} = v_{1} + a_{2}t_{2}

here as u = v_{1}

a_{2} = -0.4 m/s^{2}

and v_{2} = 0

thus,

0 = v_{1} + 0.4t_{2}

or

t2 = -v_{1}/0.4 ** (3)**

also

s_{2} = v_{1}t_{2 }+ (1/2)a_{2}t_{2}^{2 }

so

s_{2} = v_{1}t_{2 }- (1/2)x0.4xt_{2}^{2 }** (4)**

now as t = t_{1} + t_{2} = 30min = 1800

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

1800 = v_{1}[1/0.2 + 1/0.4]

or

v_{1} = 1800/7.5

thus,

v_{1} = 240 m/s

so, t_{1} = v_{1}/0.2 = 240/0.2

or

t_{1} = 1200 secs.

similarly

t_{2} = 240/0,4 = 600 secs.

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

s_{1} = (1/2) x 0.2 x (1200)^{2}

or

s_{1} = 144 km

similarly,

s_{2} = (240 x 600) - [(1/2) x 0.4 x (600)^{2}]

so, the total distance travelled will be

s = s_{1 }+ s_{2} = 144km + 72km

or

**s = 216km**

which is **option (1)**

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