*Answer credited to @Asafmen*

If it got reduced from a constant speed and they reduced in the other two seconds, by the second t=4 the speed is zero.

That means we need to divide the movement into 4 parts -

1st half: constant speed Vo.

2nd half: speed reduced by negative constant acceleration, a = -1/2 Vo

Why? Since in 2 seconds we went from Vo to 0.

Displacement for the 1st half then can be marked as 2X.

2nd half, X

3X =18

X=6

Sections are 6-6-3-0 in terms of how much distance the mobile moved at

The formula is:

1st half,

D=tVo

(Displacement at constant speed)

Second half,

D=Vot+1/2at^2

(Displacement at constants acceleration)

When we combine formulas -

D total is 18

Vo is what we want to find

Acceleration- negative half speed

t is 2 (each section is 2sec)

Combine them like on here:

Don’t forget the +1/2 turns to a minus due to negative acceleration.