dude this isnt a question on power-its pure kinematics.

I'd asked this question earlier but couldn't get the correct answer.Here goes:

Initial velocity=v...say

Pull of engine=resistance(no acc.)=KM=>K=a

When uncoupled,extra pull=KM-K(M-m)

                                        =Km

Acc of train=Km/(M-m)

Retardation for bogey=K

Distance travelled by train=L

                                     & L=vt+0.5(Kmt^2/(M-m))....2nd eqn of motion..(remember this relation)

                                =>2L/t=2v+Kmt/(M-m)

Once steam shuts off=s..say

V^2-U^2=2as

It stops,so V=0

So,U^2=2Ks

2Ks=(v+Kmt/(M-m))^2.....1st eq of motion

=>s=[(v+kmt/(M-m))^2]/2K....remember this

Distance travelled by last bogey=s2...say

s2=v^2/2k....3rd eqn of motion

Distance between them=l+s-s2

Simplify it...its slightly tricky and you'll get your answer..

Thats quite some typing..

Rate if it helped!!