Let [tex]a(t),v(t),s(t)[/tex] be the acceleration, velocity, and position functions, respectively. Then
[tex]v(t)=s'(t)[/tex]
[tex]a(t)=v'(t)=s''(t)[/tex]
Since the acceleration is a constant 30 ft/s^2, you have
[tex]a(t)=s''(t)=30[/tex]
[tex]\displaystyle\int s''(t)\,\mathrm dt=\int30\,\mathrm dt[/tex]
[tex]\displaystyle s'(t)=v(t)=30t+C_1[/tex]
You're told that the initial velocity is [tex]v(0)=-10[/tex], so you get
[tex]-10=30(0)+C_1\implies C_1=-10[/tex]
Now [tex]s'(t)=30t-10[/tex]. Integrating once more, you find
[tex]\displaystyle\int s'(t)\,\mathrm dt=\int(30t-10)\,\mathrm dt[/tex]
[tex]s(t)=15t^2-10t+C_2[/tex]
and since [tex]s(0)=4[/tex], you get
[tex]4=15(0)^2-10(0)+C_2\implies C_2=4[/tex]
which means the exact position function is
[tex]s(t)=15t^2-10t+4[/tex]
