Show that sometimes the acceleration is exactly 120mi/h^2

[karush]

At \(\displaystyle 2:00\text{ pm}\) a car’s speedometer reads \(\displaystyle 30\text { mph}\).
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 Mean Value Theorem so since
$$
f'(c)=\frac{f(b)-f(a)}{b-a}
\Rightarrow
\frac{50-30}{10-0}
$$

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

 

[Evgeny.Makarov]

20 mph in 10 minutes is exactly 120 mph in an hour.

The letter "h" should be in an upright font, just like "mi" and "mph" because it is not the name of a variable that has a numerical value (like $x$), but a contraction for "hour".

 

[karush]

if you look at "h" in the text commands it shows an upright "h"
the slash may of changed it
I was wondering tho why there is an $h^2$

the latex seem to want to change the h to italics

 

[Evgeny.Makarov]

I was wondering tho why there is an $h^2$

As opposed to what: h, s (for seconds), s² or something else?

the latex seem to want to change the h to italics

LaTeX makes characters in math mode italic. To make them upright, use \text{} or \mathrm{}. Of course, this distinction between upright/italic is not really important.

 

[karush]

As opposed to what: h, s (for seconds), s² or something else?

LaTeX makes characters in math mode italic. To make them upright, use \text{} or \mathrm{}. Of course, this distinction between upright/italic is not really important.

ok h^2 in latex is $h^2$

\text {h}^2 in latex $\text {h}^2$

I can see that we get

$$\frac{120 \text{ mi}}{\text{h}}$$

but what is

$$\frac{120 \text{ mi}}{\text{h}^2}$$

 

[Evgeny.Makarov]

If you changed your location, you would have a change of 120 miles over the course of 1 hour: 120 mi/h, or 120 mph. As it is, you are hanging speed, which itself is measured in mph. Thus, you changed 120 mph in 1 hour, or $\dfrac{120\,\mathrm{mi/h}}{\mathrm{h}}$, or $\dfrac{120\,\mathrm{mi}} {\mathrm{h}^2}$.

All acceleration is measured in distance per time squared, whether it's m/s² or mi/h².

 

[karush]

ok, that makes sense...

they didn't have any examples of this type of exercise.