Question

the height (in feet) of a falling object with an initial velocity of 40 feet per second launched straight upward from the ground is given by h(t)=-16t² + 40t, 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 $y = h(t)$ over the interval $[a,b]$ is $\frac{h(b)-h(a)}{b - a}$.

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

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

Step5: Calculate $h(3)$

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

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

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

Step7: Interpret the average rate of change from $t = 1$ to $t = 3$

A negative average rate of change of $-16$ feet/sec means the object is falling 16 feet for every 1 second.

Answer:

What is the average rate of change of the height as a function of time from $t = 1$ to $t = 2$? 0 feet/sec
What is the average rate of change of the height as a function of time from $t = 1$ to $t = 3$? -16 feet/sec
Interpret the average rate of change for the interval $t = 1$ to $t = 3$: The object is falling 16 feet for every 1 second