- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,728

**At \(\displaystyle 2:10\) it reads \(\displaystyle 50 \text { mph}\).**

Show that at some time between \(\displaystyle 2:00\) and \(\displaystyle 2:10\)

the acceleration is exactly \(\displaystyle 120\text{ mi} / h^2\)

This is in section of the

*so since*

**Mean Value Theorem**$$

f'(c)=\frac{f(b)-f(a)}{b-a}

\Rightarrow

\frac{50-30}{10-0}

$$

I don't see that $120 mi/ h^2$ is going to fit into this also why is there $h^2$