Question

an object dropped from rest from the top of a tall building on planet x falls a distance d(t)=5t^2 feet 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 feet/sec (simplify your answer)

Explanation:

Step1: Calculate $d(t_1)$

Given $d(t)=5t^{2}$, when $t_1 = 6$, $d(6)=5\times6^{2}=5\times36 = 180$ feet.

Step2: Calculate $d(t_2)$

When $t_2 = 10$, $d(10)=5\times10^{2}=5\times100 = 500$ feet.

Step3: Use average - rate - of - change formula

The average rate of change of a function $y = f(x)$ from $x=a$ to $x = b$ is $\frac{f(b)-f(a)}{b - a}$. Here, the average velocity $v_{avg}=\frac{d(t_2)-d(t_1)}{t_2 - t_1}$. Substitute $d(t_1)=180$, $d(t_2)=500$, $t_1 = 6$ and $t_2 = 10$ into the formula: $\frac{500 - 180}{10 - 6}=\frac{320}{4}=80$ feet/sec.

Answer:

80