Can the Mean Value Theorem Determine Instantaneous Acceleration?

[Asphyxiated]

Homework Statement

At 2:00PM a car's speedometer reads 50km/h. At 2:10PM it reads 65km/h. Show that at some time between 2:00 and 2:10 the acceleration is exactly 90km/h^2.

Homework Equations

Mean Value Theorem

If f is continuous on [a,b] and f is differentiable on (a,b) then there exists an x value, c, in (a,b) so that:

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

Rolle's Theorem

If f is continuous on [a,b] and f is differentiable on (a,b) and f(a)=f(b) then there is an x value, c, in (a,b) where f'(c)=0

The Attempt at a Solution

So i am fairly certain that we are suppose to use the MVT here for this problem but I am not really sure where to go. I wrote down Rolle's Theorem just because it is the other theorem we are using in this section.

To start I stated that x will be minutes after 2:00PM and y will be the velocity measured in km/h so at x = 0, y = 50 and when x = 10, y = 65 so using MVT:

f'(c)= \frac {65-50}{10-0} = \frac {15}{10}

but I am not sure where that gets me... is that the correct way to go about it or any I doing it wrong?

thanks!

 

[Sniperman724]

The (15/10) that you got means that on the graph of the velocity of the car, on at least one point, the derivative of that function will equal (15/10 m/s^2)
http://en.wikipedia.org/wiki/Mean_value_theorem

I hope that helped

 

[Asphyxiated]

Well I just realized that if i set x to hours instead of minutes and use x = 1/6 y = 65km/h

using MVT

f'(c) = \frac {65-50}{\frac{1}{6}-0} = \frac {15}{\frac{1}{6}} = 15*6=90 \frac{km}{h^{2}}

so its pretty simple, using the right units, lol.