Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 363: 72

$s=\frac{1}{2}at^2+v_0t+s_0$ (see work)

Note that $s''=a$. Since $a$ is constant, $a(t)=a.$ Then $v'=a$, so $v=at+C.$ But we also know that $v(0)=v_0,$ hence $v=at+v_0.$ Then $s'=v$, so $s=\frac{1}{2}at^2+v_0t+D$, and again we know that $s(0)=s_0$, so it follows that $s=\frac{1}{2}at^2+v_0t+s_0$.