At $2:00$pm car's speedometer reads $30$mph, and at $2:10$pm it reads $35$mph. Use the Mean Value Theorem to find acceleration the car must achieve.

Question

I'm only assuming that $f(a)$ and $f(b)$ are assigned to each respective velocity, but I'm not sure how the mean value theorem can be applied to distance rate and time.

Answer 1

Hint: the acceleration at time $t$ is given by $v'(t)$ where $v(t)$ is the speed. The mean value theorem now states that for all $a,b$ there exists a $t_0\in(a,b)$ such that $$ a(t_0)= v'(t_0) = \frac{v(b)-v(a)}{b-a}. $$

Answer 2

The mean value theorem states that for a differentiable function $f$, there exists a $c \in (a,b)$ such that the following equality holds: $$f'(c) = \frac{f(b)-f(a)}{b-a}$$ Our function $f(x)$ is the velocity as a function of time. I.e. for some time $x$, $f(x)$ outputs a velocity.

For the sake of convenience, let's reduce the units of the question, and the times to fractions of hours. Note that 10 minutes is $10/60 = 1/6$ of an hour: $$f(0/6) = 30, f(1/6) = 35$$ Then we get that $$f'(c) = \frac{35-30}{1/6-0/6} = \frac{5}{1/6} = \frac 5 1 \cdot \frac 6 1 = 30$$ We note that since $f(x)$ denotes the speed of the vehicle at some point in time $x$, the derivative $f'(x)$ denotes the velocity of the vehicle. This makes sense because the derivative is a rate of change, and acceleration is the rate of change of velocity.

So $f'(c)$ is the acceleration at some point $c$. This acceleration is $30$. We note that we measured time in hours and miles per hour, so we can just put the units back in: the rate of change is $30$ miles per hour per hour, i.e. the acceleration is $30\mathrm{mph}^2$.

(To be absolutely clear: $\mathrm{mph}^2$ is just the unit for acceleration.)