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Grade: 11
        
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
10 months ago

Answers : (3)

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
Please approve if helped.
10 months 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
 
10 months ago
Susmita
425 Points
							
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
10 months ago
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