According to the given information for the first 4cm, u=v0 and v=3v0. Applying equation of motion for constant acceleration,
v2=u2+2as
⇒(3v0)2=v2+2a(4)⇒a=−9(8)8v02=−9v02
When the body stops, v=0
⇒0=v02+2a(4+x)
⇒v02=2(9v02)(4+x)
⇒4.5=4+x
⇒x=0.5cm