Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 198: 42

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

Since, the derivative of the function$s=\dfrac{ds}{dt}=32t-2$ is $s(t)=16t^2-2t+C$ As we are given that $s(0.5)=4$ This implies that $s(0.5)=4$ or, $4=16(0.5)^2-2(0.5)+C \implies C=1$ Hence, $s(t)=16t^2-2t+1$