SOLUTION 11: We have that $s(t)$ is distance (miles) at time $t$ (hours). We will assume that $s(t)$ is differentiable, where $s'(t)$ is the velocity (mph) at time $t$ (hrs.). Let's apply the Mean Value Theorem to function $s$ on the closed interval $[1/3, 1.2]$. Then $$s'(c) = \displaystyle{ s(1.2) - s(1/3) \over 1.2 - 1/3 } \ \ \ \ \longrightarrow$$ (for some time $t=c$ in the open interval $(1/3, 1.2)$) $$s'(c) = \displaystyle{ 82 - 10 \over 6/5 - 1/3 } \ \ \ \ \longrightarrow$$ $$s'(c) = \displaystyle{ 72 \over 18/15 - 5/15 } \ \ \ \ \longrightarrow$$ $$s'(c) = \displaystyle{ 72 \over 13/15 } \ \ \ \ \longrightarrow$$ $$s'(c) = \displaystyle{ 72 \cdot { 15 \over 13 } } \ \ \ \ \longrightarrow$$ $$s'(c) = \displaystyle{ 1080 \over 13 } \approx 83.1 \ mph !!!$$ Thus, the Mean Value Theorem is proof that you were speeding at time $t=c$ hours !