A train stopping at 2 stations 4 km apart takes 4 min on the journey from one of the station to the other.Assuming that it first accelerates with uni.acceleration x and then that of uniform retardation y, prove that (1/x)+(1/y)=2

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Badiuddin askIITians.ismu Expert · Answer

Dear Chilukuri

let t1 is time of acceleration and s1 is distance covered in this time .v is the velocity after time t

let t2 is time of retardation and s2 is distance covered in this time .

v²=0²-2xs1

s1=v²/2x..............1

v=0+xt1

v/x=t1 ..................2

and 0=v-yt2

t2=v/y .................3

0²=v²-2ys2

s2=v²/2y..............4

add 2 and 3

1/x +1/y =(t1+t2)/v

=4/v..........5

add 1 and 4

s1+s2 =v²/2(1/x +1/y)

4=v²/2(1/x +1/y)

or (1/x +1/y) =8/v²

put value of v from equation 5

1/x +1/y =2

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A train stopping at 2 stations 4 km apart takes 4 min on the journey from one of the station to the other.Assuming that it first accelerates with uni.acceleration x and then that of uniform retardation y, prove that (1/x)+(1/y)=2

A train stopping at 2 stations 4 km apart takes 4 min on the journey from one of the station to the other.Assuming that it first accelerates with uni.acceleration x and then that of uniform retardation y, prove that (1/x)+(1/y)=2

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