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.\nlet ( 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}= ) ( cdot ) by the mean value theorem, there is a number ( c ) such that ( 0 < c < ) ( ) with ( v^{prime}(c)= ) \nsince ( v^{prime}(t) ) is the acceleration at time ( t ), the acceleration ( c ) hours after 2:00 p.m. is exactly 60 mi/h².

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.\nlet ( 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}= ) ( cdot ) by the mean value theorem, there is a number ( c ) such that ( 0 < c < ) ( ) with ( v^{prime}(c)= ) \nsince ( v^{prime}(t) ) is the acceleration at time ( t ), the acceleration ( c ) hours after 2:00 p.m. is exactly 60 mi/h².

Answer

Explanation:

Step1: Calculate the time interval

The time from 2:00 p.m. ((t = 0)) to 2:20 p.m. ((t=\frac{1}{3}) hours, since (20) minutes (=\frac{20}{60}=\frac{1}{3}) hours).

Step2: Apply the Mean - Value Theorem formula

The Mean - Value Theorem states that (\frac{v(b)-v(a)}{b - a}=v^{\prime}(c)), where (a = 0), (b=\frac{1}{3}), (v(0)=30) mi/h, and (v(\frac{1}{3}) = 50) mi/h. Substitute the values into the formula: (\frac{v(\frac{1}{3})-v(0)}{\frac{1}{3}-0}=\frac{50 - 30}{\frac{1}{3}}). [ \begin{align*} \frac{50 - 30}{\frac{1}{3}}&=\frac{20}{\frac{1}{3}}\ &=20\times3\ & = 60 \end{align*} ]

Answer:

(\frac{v(\frac{1}{3})-v(0)}{\frac{1}{3}-0}=60), and (c) is in the interval ((0,\frac{1}{3}))