At 2: 00 pm a car's speedometer reads 30 mi/h. at 2: 20 pm 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 pm. then v(1/3) − v(0)/1/3 − 0 = . by the mean value theorem, there is a number c such that 0 < c < with v'(c) = . since v'(t) is the acceleration at time t, the acceleration c hours after 2: 00 pm is exactly 60 mi/h2.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then .  By the Mean Value Theorem, there is a number c such that with . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

Then there is a number c in (a, b) such that

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then and (note that 20 minutes is of an hour), so the average rate of change of v on the interval is

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in at which .

c is a time time between 2:00 and 2:20 at which the acceleration is .

alot

step-by-step explanation:

its a wood chuck that can chuck wood

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square