Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 224: 48

$s(t)=4.9t^2-3t$

Here, $a=9.8; v(0)=-3;s(0)=0$ Since the derivative of velocity is acceleration, $v(t)=9.8t+C$ Therefore, $v(0)=-3$ Then $-3=9.8(0)+C \implies C=-3$ Thus, $v(t)=9.8t-3$ Since the derivative of position is velocity, $s(t)=4.9t^2-3t+C'$ Therefore, $s(0)=0$ Then $0=4.9(0)^2-3(0)+C' \implies C'=0$ Hence, $s(t)=4.9t^2-3t$