Answer
$$s\left( t \right) = 2{t^3} + 2{t^2} - 10t$$
Work Step by Step
$$\eqalign{
& v\left( t \right) = 6{t^2} + 4t - 10;{\text{ }}s\left( 0 \right) = 0 \cr
& {\text{Calculating the position function}} \cr
& s\left( t \right) = \int {v\left( t \right)} dt \cr
& s\left( t \right) = \int {\left( {6{t^2} + 4t - 10} \right)} dt \cr
& s\left( t \right) = 2{t^3} + 2{t^2} - 10t + C \cr
& {\text{Calculating the function for the initial position }}s\left( 0 \right) = 0 \cr
& 0 = 2{\left( 0 \right)^3} + 2{\left( 0 \right)^2} - 10\left( 0 \right) + C \cr
& C = 0 \cr
& {\text{Thenerefore,}} \cr
& s\left( t \right) = 2{t^3} + 2{t^2} - 10t \cr
& {\text{Graphing the velocity and position functions}} \cr} $$