ADNAN MUHAMMED

Grade 12,

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》1:3:5 c》 1:1:1 d》 1:4:9

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》1:3:5
c》 1:1:1                           d》 1:4:9

Grade:11

3 Answers

Susmita
425 Points
5 years ago
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)nt[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)nt2
and 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)nt2
so the ratio is1:3:5
Please approve if helped.
Saurav
224 Points
5 years ago
I am damn sure u remember the distance called by a freely falling particle in 1st 2nd and 3rd second is in ratio1:3:5
Infact this is true for any uniformly acclerated particle with velocity =0 at t =0
 
Susmita
425 Points
5 years ago
Oh I am sorry.I wrote the formula for free fall.Here they ask about train.So you have to go with
v2=u2+2fs
For the 1st segment as u=0,v=nt
(nt)2=2ns1
or,s1=(nt2)/2
In the second segment,u=nt,v=2nt,so
(2nt)2=(nt)2+2ns2
or,s2=(3nt2)/2
In the 3rd segment,u=2nt and v=3nt

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