Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 223: 44

$s(t)=16t^2-2t+1$

Here, $s=\dfrac{ds}{dt}=32t-2$, thus the general form of the function is: $s(t)=16t^2-2t+C$ Since we have $s(0.5)=4$ Therefore, $s(0.5)=4 \implies 4=16(0.5)^2-2(0.5)+C$ Thus, $C=1$ Hence, $s(t)=16t^2-2t+1$