Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 198: 45

$s(t)=16t^2+20t+5$

Here, we are given that $a=32; v(0)=-20;s(0)=5$ Since,we know that the derivative of velocity is acceleration so, $v(t)=32t+C$ and, $v(0)=20$ This implies $(32)(0)+C=20 $ or, $C=20$ So, $v(t)=32t+20$ Now, as we know the derivative of position ; $s(t)$ is velocity. and $s(t)=16t^2+20t+C'$ Since, we have $s(0)=5$ This implies that $(16)(0)+(20)(0)+C'=5$ or, $C'=5$ Hence, $s(t)=16t^2+20t+5$