Question

bill launches a rocket straight up into the air. the table below gives the height h(t) of the rocket (in meters) at a few times t (in seconds) during its flight.
(a) find the average rate of change for the height from 0 seconds to 2.2 seconds.
□ meters per second
(b) find the average rate of change for the height from 6.6 seconds to 13.2 seconds.
□ meters per second

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = f(x)$ over the interval $[x_1,x_2]$ is $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Here, the height function is $H(t)$ and the average rate of change of height with respect to time $t$ over the interval $[t_1,t_2]$ is $\frac{H(t_2)-H(t_1)}{t_2 - t_1}$.

Step2: Solve part (a)

For the interval from $t_1 = 0$ to $t_2=2.2$ seconds, $H(0)=0$ and $H(2.2) = 110$. Then the average rate of change is $\frac{H(2.2)-H(0)}{2.2 - 0}=\frac{110 - 0}{2.2}=50$ meters per second.

Step3: Solve part (b)

For the interval from $t_1 = 6.6$ to $t_2 = 13.2$ seconds, $H(6.6)=198$ and $H(13.2)=44$. Then the average rate of change is $\frac{H(13.2)-H(6.6)}{13.2 - 6.6}=\frac{44 - 198}{6.6}=\frac{- 154}{6.6}=-\frac{1540}{66}=-\frac{770}{33}\approx - 23.33$ meters per second.

Answer:

(a) 50 meters per second
(b) $-\frac{770}{33}\approx - 23.33$ meters per second