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

Answer 2

By the mean value, theorem a number $C$ is $0 < C < \frac{1}{4}$.

The velocity of the car is $80 \ mi/h^{2}$.

Speed:

Speed is defined as The rate of change of position of an object in any direction. Speed is measured as the ratio of distance to the time in which the distance was covered. Speed is a scalar quantity as it has only direction and no magnitude.

Given that,

at 2:00 pm $v(0)=30 \ mi/h$

at 2:15 pm $v(1/4)=50 \ mi/h$

Then,

$=\frac{v(1/4)-v(0)}{1/4-0} \\=\frac{50-30}{1/4} \\=20\times4\\=80 \ mi/h^{2}$

By the mean value theorem a number $C$ such that express as,

$0 < C < \frac{1}{4}$.

Now with,

${v}'(c)=\frac{v\left ( \frac{1}{4} \right )-v\left ( 0 \right )}{\frac{1}{4}-0} \\ =80 \ mi/h^{2}$

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