Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 198: 51

$7\dfrac{2}{3}$; it is greater than $7.5$

Suppose $s(t)$ is the distance covered by the trucker at time $t$. Thus, we get $s(0)=0$ and $s(24)=184$ Need to use The Mean Value Theorem to calculate the velocity $v(t)$ at time $t$ such that $|v(t)|=|s'(t)|=|\dfrac{f(b)-f(a)}{b-a}|$ or, $|v(t)|=|\dfrac{s(24)-s(0)}{(24-0)}|=\dfrac{184}{24}=7\dfrac{2}{3}$; it is greater than $7.5$