a body moving with uniform retardation covers 3km before its speed is reduced to half of its initial value. it comes to rest in another distance of.......i want the solution. the answer is 2km

Let initial v be V
so final = v/2 
from eq of motion = v ka square - u ka square = 2as
v/2 ka whole square - v ka square = 2a * 3
after it come to rest final be v = 0
initial become v/2
so from eq of motion = v ka square - u ka square =2as
0 ka square - v /2 ka square = 2as 
0 ka square - v/2 ka whole square =  2as
divide both eq 
u will get s/3 = 1/3 
so s= 1 km