Bob is driving through town. Two traffic radar detectors are spaced 0.75 miles apart. He passes the first and is clocked at 30 mph. One minute later, he passes the second and is clocked at 34mph. Show mathematically that somewhere between the two that he exceeded the 35mph speed limit. Fully justify your answer, citing precise facts.

Question

anonymous · Answer

This is can be proved using the mean value theorem for integral. There exist a time t between (0,1) so such that
f(t) (1) = 0.75
f(t) = 0.75 mi/min

converting to mile/hr and you get 45 mi/hr. This mean at some time between 0 and 1min, his speed equals to the average speed.

anonymous · Answer

so the fact that his speed was 30 mi/hr at t = 0 and 40 mi/hr at t = 1 is irrelevant.

anonymous · Answer

but wasnt he at 34mi/hr at t = 1?

anonymous · Answer

that doesn't mean he didn't speed up sometime between t = 0min and t = 1min, which is garanteed by the mean value theorem for integral

anonymous · Answer

okay so then f'(c) = f(1)-f(0)/1 = 45mph?

anonymous · Answer

what do you mean by f'(c) ?

anonymous · Answer

I am just trying to relate what you said to what I have in my notes. I have:

Suppost f(x) is continuous on [a, b], and f(x) if differentiable on (a, b).
Then there is at least one number c in (a, b) where the average slope = the exact slope

formula:
f'(c)= f(b)-f(a)/b-a

anonymous · Answer

ok, that works. The important thing is you need understand what f represents. 
f(1) - (0) is the change in distance which is 0.75 miles. And a = 0 , b = 1,
0.75 / (1- 0) = 0.75 mi/min (which is 45 mi/hr)

So the the Mean Value theorem, there is a c in (0,1) such that 
f'(c) = 0.75 mi/hr

anonymous · Answer

Now do you see why his initial speed and final speed recorded by the speed camera are irrelevant? The average speed is 45 mi/hr. We just need to show at some point his speed equals the average speed 45, which is greater than speed limit

anonymous · Answer

Yes thanks you but just clarifying just in case for f'(c) you meant .75 mi/min in the very last line of your response.

anonymous · Answer

right?

anonymous · Answer

I don't get what you just said

anonymous · Answer

in your response relating your answer to my notes at the end you said f'(c) = 0.75 mi/hr .

I am just wondering if that was a typo and it should be f'(c) = 0.75 mi/min? or is it suppose to be as you put it?

anonymous · Answer

ah yes. That was my typo. Should be 0.75 miles/min

anonymous · Answer

Alright thank you very much!

anonymous · Answer

no prob :)