Answer
$$s\left( t \right) = 20t - 4.9{t^2}{\text{ and }}v\left( t \right) = 20 - 9.8t$$
Work Step by Step
$$\eqalign{
& a\left( t \right) = - 9.8,\,\,\,\,v\left( 0 \right) = 20,\,\,\,\,s\left( 0 \right) = 0 \cr
& \cr
& {\text{Use theorem 6}}{\text{.2 }} \cr
& {\text{Given the acceleration }}a\left( t \right){\text{ of an object moving along a line and its initial velocity }}v\left( 0 \right){\text{, }} \cr
& {\text{the velocity of the object for future time }}t \geqslant 0{\text{ is }}v\left( t \right) = v\left( 0 \right) + \int_0^t {a\left( x \right)} dx \cr
& {\text{let }}a\left( x \right) = - 9.8{\text{ and }}v\left( 0 \right) = 20.{\text{ then }}v\left( t \right){\text{ is}} \cr
& v\left( t \right) = 20 + \int_0^t {\left( { - 9.8} \right)} dx \cr
& {\text{integrate}} \cr
& v\left( t \right) = 20 + \left( { - 9.8x} \right)_0^t \cr
& {\text{evaluate the limits}} \cr
& v\left( t \right) = 20 + \left( { - 9.8\left( t \right) - 9.8\left( 0 \right)} \right) \cr
& {\text{simplify}} \cr
& v\left( t \right) = 20 - 9.8t \cr
& \cr
& {\text{Use theorem 6}}{\text{.1 }} \cr
& {\text{Given the velocity }}v\left( t \right){\text{ of an object moving along a line and its initial position }}s\left( 0 \right){\text{, }} \cr
& {\text{the position function of the object for future time }}t \geqslant 0{\text{ is }}s\left( t \right) = s\left( 0 \right) + \int_0^t {v\left( x \right)} dx \cr
& {\text{let }}v\left( x \right) = 20 - 9.8x{\text{ and }}s\left( 0 \right) = 0.{\text{ then }}s\left( t \right){\text{ is}} \cr
& s\left( t \right) = 0 + \int_0^t {\left( {20 - 9.8x} \right)} dx \cr
& {\text{integrate}} \cr
& s\left( t \right) = \left( {20x - \frac{{9.8{x^2}}}{2}} \right)_0^t \cr
& s\left( t \right) = \left( {20x - 4.9{x^2}} \right)_0^t \cr
& {\text{evaluate the limits}} \cr
& s\left( t \right) = \left( {20\left( t \right) - 4.9{{\left( t \right)}^2}} \right) - \left( {20\left( 0 \right) - 4.9{{\left( 0 \right)}^2}} \right) \cr
& {\text{simplify}} \cr
& s\left( t \right) = 20t - 4.9{t^2} \cr} $$
