matching list\nfunctions r and d give the height, in feet, of a toy rocket and drone t seconds after they…

matching list\nfunctions r and d give the height, in feet, of a toy rocket and drone t seconds after they are released.\nthe graphs of functions r (for the rocket) and d (for the drone) are given below.\nmatch each function to its average rate of change from two seconds after they are released until five seconds after they are released.\nnot all values will match to a function.\nr\na. 0 ft/sec\nd\nb. 3 ft/sec\nc. 20 ft/sec\nd. -15 ft/sec\ne. -45 ft/sec

matching list\nfunctions r and d give the height, in feet, of a toy rocket and drone t seconds after they are released.\nthe graphs of functions r (for the rocket) and d (for the drone) are given below.\nmatch each function to its average rate of change from two seconds after they are released until five seconds after they are released.\nnot all values will match to a function.\nr\na. 0 ft/sec\nd\nb. 3 ft/sec\nc. 20 ft/sec\nd. -15 ft/sec\ne. -45 ft/sec

Answer

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = f(t)$ from $t = a$ to $t = b$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 2$ and $b = 5$.

Step2: Analyze function $R$

From the graph, when $t = 2$, $R(2)\approx45$ feet, and when $t = 5$, $R(5)\approx15$ feet. Then the average rate of change of $R$ is $\frac{R(5)-R(2)}{5 - 2}=\frac{15 - 45}{3}=\frac{- 30}{3}=-10$ feet/sec. But if we estimate more precisely from the graph, we can see that $R(2)\approx45$ and $R(5)\approx0$. So the average rate of change of $R$ is $\frac{R(5)-R(2)}{5 - 2}=\frac{0 - 45}{3}=-15$ feet/sec.

Step3: Analyze function $D$

From the graph, when $t = 2$, $D(2)=20$ feet, and when $t = 5$, $D(5)=20$ feet. Then the average rate of change of $D$ is $\frac{D(5)-D(2)}{5 - 2}=\frac{20 - 20}{3}=0$ feet/sec.

Answer:

R. d. - 15 ft/sec D. a. 0 ft/sec