Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 357: 64

$s(t)=\frac{1}{12}t^4-\frac{2}{3}t^3+3t^2+\frac{211}{12}t$

Note that $a(t)=s''(t)$. Then $$v(t)=\frac{1}{3}t^3-2t^2+6t+C$$ $$s(t)=\frac{1}{12}t^4-\frac{2}{3}t^3+3t^2+Ct+D$$ Using the first initial condition, it follows that $0=D$. Then using the second initial condition, we have that $20=\frac{1}{12}-\frac{2}{3}+3+C \Rightarrow C=\frac{211}{12}$