From Mechanics

For the second stage where the lift slows down we use \displaystyle s=\frac{1}{2}(u+v)t. This gives the distance travelled during this stage is

\displaystyle s=\frac{1}{2}\left( 0 \textrm{.}6 \right)\times 5=1 \textrm{.}5\ \text{m}

The total distance travelled using part b) is

\displaystyle 37.5+1 \textrm{.}5=39\ \text{m}