Question

the height (in feet) of a falling object with an initial velocity of 48 feet per second launched straight upward from the ground is given by h(t)=-16t² + 48t, where t is time (in seconds). what is the average rate of change of the height as a function of time from t = 1 to t = 2? what is the average rate of change of the height as a function of time from t = 1 to t = 3? interpret the average rate of change for the interval t = 1 to t = 3

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 $\frac{f(b)-f(a)}{b - a}$. Here, $h(t)=-16t^{2}+48t$, $a = 1$, and $b = 3$.

Step2: Calculate $h(1)$ and $h(3)$

$h(1)=-16(1)^{2}+48(1)=-16 + 48=32$.
$h(3)=-16(3)^{2}+48(3)=-16\times9 + 48\times3=-144 + 144 = 0$.

Step3: Calculate the average rate of change

$\frac{h(3)-h(1)}{3 - 1}=\frac{0 - 32}{2}=-16$ feet/sec.

Answer:

-16 feet/sec