From Mechanics

Using the result for the velocity obtained in part a)

\displaystyle \begin{align} & s(10)=\int_{0}^{10}{v}dt \\ & \\ & =\int_{0}^{10}{\left( 8t-\frac{2{{t}^{2}}}{5} \right)dt} \\ & \\ & =\left[ 4{{t}^{2}}-\frac{2{{t}^{3}}}{15} \right]_{0}^{10} \\ & \\ & =\left( 4\times {{10}^{2}}-\frac{2\times {{10}^{3}}}{15} \right)-0 \\ & \\ & =267\text{ m} \end{align}