Question

please look at the image, thank you!!if it helps im supposed to round 4 decimal places

Answer 1

Given:

The function represents the position of an object for time x.

$f(x)=\sqrt{16-x^2}$

To find:

The average velocity over the interval [2, 4].

Solution:

It is known that the average velocity over the interval [a, b] is given by:

$\frac{f(b)-f(a)}{b-a}$

For the given position function, the average velocity over the interval [2, 4] is:

$\begin{gathered} Avg=\frac{f(4)-f(2)}{4-2} \\ =\frac{\sqrt{16-4^2}-\sqrt{16-2^2}}{2} \\ =\frac{\sqrt{16-16}-\sqrt{16-4}}{2} \\ =\frac{0-\sqrt{12}}{2} \\ =-\frac{2\sqrt{3}}{2} \\ =-\sqrt{3} \\ =-1.7321 \end{gathered}$

Thus, the average velocity is -1.7321.