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[SOLVED] 34 MVT - Application of the mean value theorem

karush

Well-known member
Jan 31, 2012
2,678
2020_03_29_18.14.50_edit.jpg

$10 min = \dfrac{h}{6}$
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
Jan 26, 2012
1,403


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

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.