From Mechanics

Using \displaystyle v=u+at\ \ gives

\displaystyle v=5+0\textrm{.}2\times 5=6\text{ m}{{\text{s}}^{-1}}

The distance travelled during this stage using \displaystyle s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \ is

\displaystyle s1=5\times 5+\frac{1}{2}0\textrm{.}2\times {{5}^{2}}=27\textrm{.}5\ \text{m}

Together with the first stage the total distance travelled is

\displaystyle 25+27\textrm{.}5=52\textrm{.}5\ \text{m}