Answer
$$s\left( t \right) = \frac{1}{{30}}{t^3} + 1$$
Work Step by Step
$$\eqalign{
& a\left( t \right) = 0.2t;\,\,\,v\left( 0 \right) = 0,\,\,\,\,s\left( 0 \right) = 1 \cr
& v'\left( t \right) = a\left( t \right),{\text{ then}} \cr
& v\left( t \right) = \int {a\left( t \right)dt} \cr
& v\left( t \right) = \int {0.2tdt} \cr
& v\left( t \right) = 0.2\left( {\frac{{{t^2}}}{2}} \right) + C \cr
& v\left( t \right) = \frac{1}{{10}}{t^2} + C \cr
& {\text{Use the initial condition }}v\left( 0 \right) = 0 \cr
& 0 = \frac{1}{{10}}{\left( 0 \right)^2} + C \cr
& C = 0 \cr
& {\text{Thus}}{\text{, }} \cr
& v\left( t \right) = \frac{1}{{10}}{t^2} \cr
& s'\left( t \right) = v\left( t \right),{\text{ then}} \cr
& s\left( t \right) = \int {v\left( t \right)dt} \cr
& s\left( t \right) = \int {\frac{1}{{10}}{t^2}dt} \cr
& s\left( t \right) = \frac{1}{{10}}\left( {\frac{{{t^3}}}{3}} \right) + C \cr
& s\left( t \right) = \frac{1}{{30}}{t^3} + C \cr
& {\text{Use the initial condition }}s\left( 0 \right) = 1 \cr
& 1 = \frac{1}{{30}}{\left( 0 \right)^3} + C \cr
& C = 1 \cr
& {\text{Thus}}{\text{, }} \cr
& s\left( t \right) = \frac{1}{{30}}{t^3} + 1 \cr} $$
