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- Thread starter karush
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- Thread starter
- #1

The average acceleration is $\displaystyle \frac{20}{\frac{1}{6}} = 120 \,\textrm{mi}/\textrm{h}^2 $$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

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.

- Jan 29, 2012

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