Answer
Acceleration is the derivative of velocity.
The time interval (in hours) we observe is $[a, b]=[0,1/30]$
By the Mean Value Theorem, a $t_{0}\displaystyle \in(0, \frac{1}{30})$ exists, such that
$v^{\prime}(t_{0})=a(t_{0})=1500$ miles/h$r^{2}$
Work Step by Step
The Mean Value Theorem
If $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ ,
then there exists a number $c$ in $(a, b)$ such that$ f^{\prime}(c)=\displaystyle \frac{f(b)-f(a)}{b-a}.$
---
Acceleration is the derivative of velocity.
Setting 9:13 A.M. to be $t=0$ hours,
two minutes later will be $\displaystyle \frac{1}{30}$ hours.
$[a, b]=[0,1/30]$
We have
$\displaystyle \frac{v(1/30)-v(0)}{\frac{1}{30}-0}=\frac{30(85-35)}{1}=1500$ miles/h$r^{2}$
By the Mean Value Theorem, a $t_{0}\displaystyle \in(0, \frac{1}{30})$ exists, such that
$v^{\prime}(t_{0})=a(t_{0})=1500$ miles/h$r^{2}$
