jim launches a rocket straight up into the air. the table below gives the height h(t) of the rocket (in…

jim 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.\ntime t (seconds) height h(t) (meters)\n0 0\n2.2 110\n6.6 198\n8.8 44\n13.2 0\n(a) find the average rate of change for the height from 2.2 seconds to 6.6 seconds.\n meters per second\n(b) find the average rate of change for the height from 8.8 seconds to 13.2 seconds.\n meters per second

jim 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.\ntime t (seconds) height h(t) (meters)\n0 0\n2.2 110\n6.6 198\n8.8 44\n13.2 0\n(a) find the average rate of change for the height from 2.2 seconds to 6.6 seconds.\n meters per second\n(b) find the average rate of change for the height from 8.8 seconds to 13.2 seconds.\n meters per second

Answer

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = f(x)$ from $x = a$ to $x = b$ is $\frac{f(b)-f(a)}{b - a}$. For the height function $H(t)$, the average rate of change from $t=a$ to $t = b$ is $\frac{H(b)-H(a)}{b - a}$.

Step2: Solve part (a)

We have $a = 2.2$, $b = 6.6$, $H(2.2)=110$, and $H(6.6)=198$. $\frac{H(6.6)-H(2.2)}{6.6 - 2.2}=\frac{198 - 110}{6.6-2.2}=\frac{88}{4.4}=20$

Step3: Solve part (b)

We have $a = 8.8$, $b = 13.2$, $H(8.8)=44$, and $H(13.2)=0$. $\frac{H(13.2)-H(8.8)}{13.2 - 8.8}=\frac{0 - 44}{13.2-8.8}=\frac{-44}{4.4}=- 10$

Answer:

(a) 20 (b) -10