Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 198: 46

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

Here, $a=9.8; v(0)=-3;s(0)=0$ Since, the derivative of velocity is acceleration, thus $v(t)=9.8t+C$ Since, we have $v(0)=-3$ This implies that $-3=9.8(0)+C$ or, $C=-3$ So, $v(t)=9.8t-3$ Since,we know that the derivative of position $s(t)$ is velocity, so we have: $s(t)=4.9t^2-3t+C'$ Since, we have, $s(0)=0$ Therefore, $0=4.9(0)^2-3(0)+C'$ or, $C'=0$ Hence, $s(t)=4.9t^2-3t$