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)=-16t^{2}+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, $h(t)=-16t^{2}+48t$, $a$ and $b$ are values of $t$.

Step2: Calculate average rate of change from $t = 1$ to $t = 2$

First, find $h(1)$ and $h(2)$.
$h(1)=-16\times1^{2}+48\times1=-16 + 48=32$.
$h(2)=-16\times2^{2}+48\times2=-16\times4 + 96=-64 + 96 = 32$.
The average rate of change is $\frac{h(2)-h(1)}{2 - 1}=\frac{32 - 32}{1}=0$.

Step3: Calculate average rate of change from $t = 1$ to $t = 3$

Find $h(3)$: $h(3)=-16\times3^{2}+48\times3=-16\times9+144=-144 + 144 = 0$.
The average rate of change is $\frac{h(3)-h(1)}{3 - 1}=\frac{0 - 32}{2}=-16$.

Step4: Interpret the average rate of change from $t = 1$ to $t = 3$

The average rate of change of $-16$ means that, on average, the height of the object is decreasing at a rate of 16 feet per second over the time - interval from $t = 1$ second to $t = 3$ seconds.

Answer:

What is the average rate of change of the height as a function of time from $t = 1$ to $t = 2$? 0
What is the average rate of change of the height as a function of time from $t = 1$ to $t = 3$? -16
Interpret the average rate of change for the interval $t = 1$ to $t = 3$: The height of the object is decreasing at an average rate of 16 feet per second over the interval from $t = 1$ to $t = 3$ seconds.