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…

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} ).

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} ).

Answer

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)