From Mechanics

We investigate the distance after \displaystyle 10\ \text{s}.

Using \displaystyle s=ut+\frac{1}{2}a{{t}^{\ 2}} gives

\displaystyle s=0+\frac{1}{2}\times \left( 0 \textrm{.}075 \right)\times {{10}^{2}}=3\textrm{.}75\ \text{m}