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 $y = f(t)$ from $t = a$ to $t = b$ is $\frac{f(b)-f(a)}{b - a}$.

Step2: Calculate $h(1)$

Given $h(t)=-16t^{2}+48t$, when $t = 1$, $h(1)=-16\times1^{2}+48\times1=-16 + 48=32$.

Step3: Calculate $h(2)$

When $t = 2$, $h(2)=-16\times2^{2}+48\times2=-16\times4 + 96=-64 + 96 = 32$.

Step4: Find average rate of change from $t = 1$ to $t = 2$

$\frac{h(2)-h(1)}{2 - 1}=\frac{32-32}{1}=0$.

Step5: Calculate $h(3)$

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

Step6: Find average rate of change from $t = 1$ to $t = 3$

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

Step7: 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:

The average rate of change from $t = 1$ to $t = 2$ is $0$.
The average rate of change from $t = 1$ to $t = 3$ is $-16$.
The interpretation for the interval $t = 1$ to $t = 3$ is that the height of the object is decreasing at an average rate of 16 feet per second.