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