MHB 34 MVT - Application of the mean value theorem

[karush]

​

$10 min = \dfrac{h}{6}$
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 🕶

 

[Prove It]

​

$10 min = \dfrac{h}{6}$
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.

 

[HallsofIvy]

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.