How does this speed limit question work?

Question

Now I might be heavily misguided and probably flat out wrong here, but this is an example question I got for using Rolle's Theorem and Mean Value Theorem:

How does this work? We can't determine if the speed ever surpasses 55 even using the theorems mostly because no equation of the speed nor distance nor acceleration of the car is given. I was just thinking of a counter example (correct me if I'm wrong), a straight line from 55 to 50 in [0, 4] minutes (or [0, 1/15] hours if using the correct units) never surpasses 55, another would be a negative parabola(correct this term if possible). My teacher says that since the average rate of change is 75mph (shouldn't it be -75mph since it DECREASES 5 miles in 4 minutes), that means the car must have reached 75mph at max. Now take my counterexample, it does indeed surpass 55...... at negative time, but the question specifically asks for the domain within the 4 minutes from the first station to the second station, what happened outside shouldn't have to matter.

Again I don't understand how this works, maybe there is something I'm missing. Any help is greatly appreciated.

Answer 1

Firstly - if you forget about maths and just use common sense, it is clear that you cannot travel 5 miles in 4 minutes while staying under 75MPH. Credit (+1) to your teacher (or whoever came up with it) for coming up with a question that relates this intuitive real world situation to the Mean Value Theorem.

Let $y(t)$ denote distance travelled in miles from the first station, at time $t$ measured in hours, with $t=0$ corresponding to the moment when the truck passed the first station.

Then $y(0)=0, y(\frac1{15})=5$.

If we assume that the truck's motion was continuous and differentiable on $[0,\frac1{15}]$, then we may apply the Mean Value Theorem, to get that at some $t\in [0,\frac1{15}]$, the speed of the truck $y'(t)$ is given by: $$ y'(t)=\frac{y(\frac1{15})-y(0)}{\frac1{15}-0}=\frac 5{\frac1{15}}=75{\rm MPH} $$

Answer 2

We assume that its distance $f(t)$ from the first station is a differentiable function of time $t$ over the stretch in question. We know that $f(0)=0$ and $f(1/15)=5$, where times is measured in hours and distance in miles, so the average speed over the stretch in miles per hour is

$$\frac{5-0}{\frac1{15}-0}=75\,,$$

the slope of the secant line between the endpoints of the curve. The mean value theorem says that there is some $t_0$ between $0$ and $\frac1{15}$ such that $f'(t_0)=75$, i.e., some moment at which the truck’s speed was $75$ mph.