Question

an object dropped from rest from the top of a tall building on planet x falls a distance d(t)=20t^2 foot in the first t seconds. find the average rate of change of distance with respect to time as t changes from t_1 = 6 to t_2 = 10. this rate is known as the average velocity, or speed. the average velocity as t changes from 6 to 10 seconds is \frac{foot}{sec} (simplify your answer.)

Explanation:

Step1: Find distance at $t_1 = 6$

Substitute $t = 6$ into $d(t)=20t^{2}$.
$d(6)=20\times6^{2}=20\times36 = 720$ feet.

Step2: Find distance at $t_2 = 10$

Substitute $t = 10$ into $d(t)=20t^{2}$.
$d(10)=20\times10^{2}=20\times100 = 2000$ feet.

Step3: Calculate average rate of change

The formula for average rate of change of a function $y = f(x)$ from $x_1$ to $x_2$ is $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Here, the average velocity $v_{avg}=\frac{d(t_2)-d(t_1)}{t_2 - t_1}$.
$v_{avg}=\frac{2000 - 720}{10 - 6}=\frac{1280}{4}=320$ feet/sec.

Answer:

320