Question

at 2:00 p.m, a cars speedometer reads 30 mi/h. at 2:20 p.m. it reads 50 mi/h. show that at some time between 2:00 and 2:20 the acceleration is exactly 60 mi/h². let ( v(t) ) be the velocity of the car ( t ) hours after 2:00 p.m. then ( \frac{v(1 / 3)-v(0)}{1 / 3-0}=) by the mean value theorem, there is a number ( c ) such that ( 0 < c < ) with ( v^{prime}(c)=) since ( v^{prime}(t) ) is the acceleration at time ( t ), the acceleration ( c ) hours after 2:00 p.m. is exactly ( 60 mathrm{mi} / mathrm{h}^{2} ).

Explanation:

Step1: Calculate the difference quotient

We know that \(v(0) = 30\) (velocity at \(t = 0\), which is 2:00 p.m.) and \(v(\frac{1}{3})=50\) (since 20 minutes \(=\frac{20}{60}=\frac{1}{3}\) hours after 2:00 p.m.).
The difference quotient \(\frac{v(\frac{1}{3})-v(0)}{\frac{1}{3}-0}=\frac{50 - 30}{\frac{1}{3}}\).

$$ \frac{50 - 30}{\frac{1}{3}}=\frac{20}{\frac{1}{3}}=20\times3 = 60 $$

Step2: Apply the Mean Value Theorem

The Mean Value Theorem states that if \(y = v(t)\) is continuous on the closed interval \([a,b]=[0,\frac{1}{3}]\) and differentiable on the open interval \((a,b)=(0,\frac{1}{3})\), then there exists a number \(c\in(0,\frac{1}{3})\) such that \(v^{\prime}(c)=\frac{v(b)-v(a)}{b - a}\).

Answer:

\(\frac{v(\frac{1}{3})-v(0)}{\frac{1}{3}-0}=60\), \(0\lt c\lt\frac{1}{3}\), \(v^{\prime}(c) = 60\)