# 34 mnt - Application of the mean value theorem

#### karush

##### Well-known member
$10 min = \dfrac{h}{6}$
So
$a(t)=v'(t) =\dfrac{\dfrac{(50-30)mi}{h}}{\dfrac{h}{6}} =\dfrac{20 mi}{h}\cdot\dfrac{6}{h}=\dfrac{120 mi}{h^2}$
Hopefully Last edited by a moderator:

#### Prove It

##### Well-known member
MHB Math Helper

$10 min = \dfrac{h}{6}$
So
$a(t)=v'(t) =\dfrac{\dfrac{(50-30)mi}{h}}{\dfrac{h}{6}} =\dfrac{20 mi}{h}\cdot\dfrac{6}{h}=\dfrac{120 mi}{h^2}$
Hopefully The average acceleration is $\displaystyle \frac{20}{\frac{1}{6}} = 120 \,\textrm{mi}/\textrm{h}^2$

Since the function is continuous and smooth, there must be some value $c \in \left[ 0, \frac{1}{6} \right]$ such that $a\left( c \right) = 120$ by the Mean Value Theorem.

• topsquark and karush

#### HallsofIvy

##### Well-known member
MHB Math Helper
Karush, what you said was simply that the average acceleration was120. In order to correctly answer the question, you have to appeal to the "Mean Value Theorem" as Prove It did.

• topsquark