From Mechanics

For the second stage where the lift slows down we first must calculate the maximum speed which is reached after 10 seconds using \displaystyle v=u+at.

From part a) we have the acceleration is \displaystyle 0\textrm{.}075 \text{ m}{{\text{s}}^{-2}}, giving the maximum sped reached is \displaystyle 0+0\textrm{.}075\times 10=0\textrm{.}75\ \text{m}{{\text{s}}^{-1}}

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+ 0 \textrm{.}75 \right)\times 5=1 \textrm{.}875\ \text{m}

The total distance travelled using part b) is

\displaystyle 3\textrm{.}75+1 \textrm{.}875=5\textrm{.}625\ \text{m}