Question

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.

time t (seconds)	height h(t) (meters)
2.3	92
4.6	138
9.2	23
11.5	0

(a) find the average rate of change for the height from 0 seconds to 2.3 seconds.
meters per second
(b) find the average rate of change for the height from 9.2 seconds to 11.5 seconds.
meters per second

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 $\frac{f(b)-f(a)}{b - a}$.

Step2: Solve part (a)

For the time interval from $t = 0$ to $t=2.3$ seconds, $a = 0$, $b = 2.3$, $H(0)=0$, and $H(2.3)=92$.
The average rate of change is $\frac{H(2.3)-H(0)}{2.3 - 0}=\frac{92 - 0}{2.3}=40$ meters per second.

Step3: Solve part (b)

For the time interval from $t = 9.2$ to $t = 11.5$ seconds, $a = 9.2$, $b = 11.5$, $H(9.2)=23$, and $H(11.5)=0$.
The average rate of change is $\frac{H(11.5)-H(9.2)}{11.5 - 9.2}=\frac{0 - 23}{2.3}=- 10$ meters per second.

Answer:

(a) 40 meters per second
(b) - 10 meters per second