To solve this problem we will apply the conventions related to the cinematic movement theorem, for which the kinematic equations of linear motion are equally detached. We will use the speed and position equations to determine the general formula according to the given values.
To velocity function we have
[tex]v(t) = v(0)+\int^t_0 a(x) dx[/tex]
Our values are,
[tex]a(x) = -9.8[/tex]
[tex]v(0) = 20[/tex]
[tex]s(0) = 0[/tex]
Replacing at this equation and solving we have that the equation for the velocity would be,
[tex]v(t) = 20+\int^t_0 -9.8 dx[/tex]
[tex]v(t) = 20-\int^t_0 9.8 dx[/tex]
[tex]v(t) = 20-9.8x|^t_0[/tex]
[tex]v(t) = 20-9.8(t-0)[/tex]
[tex]v(t) = 20-9.8t[/tex]
Therefore the velocity function is [tex]v(t) = 20-9.8t[/tex]
At the same time for the position function:
[tex]s(t) = s(0) +\int^t_0 v(x) dx[/tex]
Replacing we have that
[tex]s(t) = s(0) +\int^t_0 20-9.8x dx[/tex]
[tex]s(t) = 0 +20x-9.8\frac{x^2}{2}|^t_0[/tex]
[tex]s(t) = (20t-9.8\frac{t^2}{2})-(20\cdot 0 -9.8 \frac{0^2}{2})[/tex]
[tex]s(t) = 20t-9.8\frac{t^2}{2}[/tex]
Therefore the position function is
[tex]s(t) = 20t-9.8\frac{t^2}{2}[/tex]
