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`        A train starts from rest with acceleration n and describes distance S1 , S2 , S3  in 1st ,2nd and 3rd of the journey. The ratio of S1/S2/S3 is a》 1:2:3                           b》1:3:5c》 1:1:1                           d》 1:4:9`
one year ago

## Answers : (3)

```							Your question is not clear."1st 2nd 3rd" what "of the journey"?If it 1st,2nd,3rd hour,then the solution is as follows.Otherwise let me know.We break the time segment in 0 to t,t to 2t,2t to 3t,all of equal interval t.Now for the 1st segment initial speed u=0.s1=ut+(1/2)ft2=(1/2)nt2 [as acceleration f=n given]and final velocity v=u+ft=nt.This v will serve as initial velocity in the 2nd time segment.So in the 2nd time segment initial velocity u=nt.s2=ut+(1/2)ft2=nt2+(1/2)nt2=(3/2)nt2and final velocity v=u+ft=nt+nt=2nt.This velocity will act as initial velocity in 3rd time segment.So in 3rd time segment initial velocity u=2nt.s3=ut+(1/2)ft2=2nt2+(1/2)nt2=(5/2)nt2so the ratio is1:3:5Please approve if helped.
```
one year ago
```							I am damn sure u remember the distance called by a freely falling particle in 1st 2nd and 3rd second is in ratio1:3:5Infact this is true for any uniformly acclerated particle with velocity =0 at t =0
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one year ago
```							Oh I am sorry.I wrote the formula for free fall.Here they ask about train.So you have to go withv2=u2+2fsFor the 1st segment as u=0,v=nt(nt)2=2ns1or,s1=(nt2)/2In the second segment,u=nt,v=2nt,so(2nt)2=(nt)2+2ns2or,s2=(3nt2)/2In the 3rd segment,u=2nt and v=3nt
```
one year ago
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