From Mechanics

Using the result for the velocity obtained in part a)

\displaystyle \begin{align} & s(15)=\int_{0}^{15}{v}dt \\ & \\ & =\int_{0}^{15}{( t-\frac{{{t}^{2}}}{30} )dt} \\ & \\ & =\left[ \frac{{{t}^{2}}}{2}-\frac{{{t}^{3}}}{90} \right]_{0}^{15} \\ & \\ & =\left( \frac{{{15}^{2}}}{2}-\frac{{{15}^{3}}}{90} \right)-0 \\ & \\ & =75\text{ m} \end{align}