The speed of a train drops 48 `"km.h"^(-1)` to 24 `"km.h"^(-1)` after moving through a distance of 108 m with uniform retardation. How much farther would it move with the same retardation before coming to rest?

Explanation

`v^(2) = u^(2)+2as,` s= 108m = 0.108 km
`:. " " a = (v^(2)-u^(2))/(2s)`
`((24)^(2)-(48)^(2))/(2xx0.108)=-8000 "km.h"^(-2)`
For the second part of motion,
`u = 24 "km.h"^(-1)` and v=0
`:. " " s =(v^(2)-u^(2))/(2a)`
`=(0-(24)^(2))/(2xx(-8000))=0.036`km = 36 m