Mechanics> A train starts from rest with acceleratio...

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

Technical Nafis , 7 Years ago

Grade 11

3 Answers

Susmita

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.

s_2⁼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.

Approved

Last Activity: 7 Years ago

Saurav

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

Approved

Last Activity: 7 Years ago

Susmita

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

Last Activity: 7 Years ago

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Mechanics> A train starts from rest with acceleratio...

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

Technical Nafis , 7 Years ago

Susmita

Saurav

Susmita

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