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}= ) ( cdot ) 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 mi/h².

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: \(\frac{v(\frac{1}{3})-v(0)}{\frac{1}{3}-0}=\frac{50 - 30}{\frac{1}{3}}\).

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Answer:

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