Answer
$$s\left( t \right) = - \cos 2t + 1$$
Work Step by Step
$$\eqalign{
& v\left( t \right) = 2\sin 2t;{\text{ }}s\left( 0 \right) = 0 \cr
& {\text{Calculating the position function}} \cr
& s\left( t \right) = \int {v\left( t \right)} dt \cr
& s\left( t \right) = \int {\left( {2\sin 2t} \right)} dt \cr
& s\left( t \right) = - \cos 2t + C \cr
& {\text{Calculating the function for the initial position }}s\left( 0 \right) = 0 \cr
& 0 = - \cos 2\left( 0 \right) + C \cr
& C = 1 \cr
& {\text{Thenerefore,}} \cr
& s\left( t \right) = - \cos 2t + 1 \cr
& {\text{Graphing the velocity and position functions}} \cr} $$