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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》1:3:5 c》 1:1:1 d》 1:4:9

2 years ago

Susmita
425 Points

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
2 years ago
Saurav
224 Points

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

2 years ago
Susmita
425 Points

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
2 years ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions