Answer
$$v\left( t \right) = 2t + 4$$
$$s\left( t \right) = {t^2} + 4t$$
Work Step by Step
$$\eqalign{
& v\left( t \right) = 2t + 4;{\text{ }}s\left( 0 \right) = 0 \cr
& {\text{Calculating the position function}} \cr
& s\left( t \right) = \int {v\left( t \right)} dt \cr
& s\left( t \right) = \int {\left( {2t + 4} \right)} dt \cr
& s\left( t \right) = {t^2} + 4t + C \cr
& {\text{Calculating the function for the initial position }}s\left( 0 \right) = 0 \cr
& 0 = {\left( 0 \right)^2} + 4\left( 0 \right) + C \cr
& C = 0 \cr
& s\left( t \right) = {t^2} + 4t \cr
& {\text{Graphing the velocity and position functions}} \cr} $$
