Question

At 2:00 PM a car's speedometer reads 30 mi/h. At 2:15 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:15 the acceleration is exactly 80 mi/h^2. Let v(f) 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 0 Since v'(t) is the acceleration at time t.______ the acceleration c hours after 2:00 PM is exactly 80 mi/h^2.

Answer 1

Here is the correct format for the question

At 2:00 PM a car's speedometer reads 30 mi/h. At 2:15 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:15 the acceleration is exactly 80 mi/h².Let v(f) be the velocity of the car t hours after 2:00 PM.Then $\dfrac{v(1/4)-v(0)}{1/4 -0} = \Box$. By the Mean Value Theorem, there is a number c such that 0 < c < $\Box$  with v'(c) = $\Box$. Since v'(t)  is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 80 mi/h^2.

Answer:

Step-by-step explanation:

From the information given :

At  2:00 PM ;

a car's speedometer v(0) = 30 mi/h

At 2:15 PM;

a car's speedometer v(1/4) = 50 mi/h

Given that:

v(f) should be the velocity of the car t hours after 2:00 PM

Then $\dfrac{v(1/4)-v(0)}{1/4 -0} = \Box$ will be:

$= \dfrac{50-30}{1/4 -0}$

$= \dfrac{20}{1/4 }$

= 20 × 4/1

= 80 mi/h²

By the Mean value theorem; there is a number c such that :

$\mathbf{0 < c< \dfrac{1}{4}}$     with $\mathbf{v'(c) = \dfrac{v(1/4)-v(0)}{1/4 -0}} \mathbf{ = 80 \ mi/h^2}$