Question

at 2:00 p.m. a cars speedometer reads 30 mi/h. at 2:30 p.m. it reads 50 mi/h. show that at some time between 2:00 and 2:30 the acceleration is exactly 40 mi/h². let v(t) be the velocity of the car t hours after 2:00 p.m. then $\frac{v(1/2)-v(0)}{1/2 - 0}=square$. by the mean value theorem, there is a number c such that 0 < c < $square$ with v(c) = $square$. since v(t) is the acceleration at time t, the acceleration c hours after 2:00 p.m. is exactly 40 mi/h².

Explanation:

Step1: Calculate the average rate of change

We know that $v(0) = 30$ mi/h and $v(\frac{1}{2})=50$ mi/h. The formula for the average rate of change of $v(t)$ over the interval $[0,\frac{1}{2}]$ is $\frac{v(\frac{1}{2})-v(0)}{\frac{1}{2}-0}$. Substituting the values, we get $\frac{50 - 30}{\frac{1}{2}-0}=\frac{20}{\frac{1}{2}} = 40$ mi/h².

Step2: Apply the Mean - Value Theorem

The Mean - Value Theorem states that if $y = v(t)$ is continuous on the closed interval $[a,b]=[0,\frac{1}{2}]$ and differentiable on the open interval $(a,b)=(0,\frac{1}{2})$, then there exists a number $c\in(0,\frac{1}{2})$ such that $v^{\prime}(c)=\frac{v(b)-v(a)}{b - a}$. Here, $a = 0$, $b=\frac{1}{2}$, and $\frac{v(\frac{1}{2})-v(0)}{\frac{1}{2}-0}=40$ mi/h². So $v^{\prime}(c) = 40$ mi/h².

Answer:

First blank: $40$; Second blank: $\frac{1}{2}$; Third blank: $40$