heather is driving a car. the table below gives the distance d(t) (in meters) she has driven at a few times…

heather is driving a car. the table below gives the distance d(t) (in meters) she has driven at a few times t (in seconds) after she starts. time t (seconds) distance d(t) (meters) 0 0 2 78.6 5 151.5 7 205.1 9 255.9 (a) find the average rate of change for the distance driven from 0 seconds to 2 seconds. meters per second (b) find the average rate of change for the distance driven from 7 seconds to 9 seconds. meters per second

heather is driving a car. the table below gives the distance d(t) (in meters) she has driven at a few times t (in seconds) after she starts. time t (seconds) distance d(t) (meters) 0 0 2 78.6 5 151.5 7 205.1 9 255.9 (a) find the average rate of change for the distance driven from 0 seconds to 2 seconds. meters per second (b) find the average rate of change for the distance driven from 7 seconds to 9 seconds. meters per second

Answer

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 $\frac{f(b)-f(a)}{b - a}$. For the distance function $D(t)$, the average rate of change from $t=a$ to $t = b$ is $\frac{D(b)-D(a)}{b - a}$.

Step2: Calculate average rate of change from $t = 0$ to $t = 2$

We have $a = 0$, $b = 2$, $D(0)=0$ and $D(2)=78.6$. Then $\frac{D(2)-D(0)}{2 - 0}=\frac{78.6-0}{2}=\frac{78.6}{2}=39.3$ meters per second.

Step3: Calculate average rate of change from $t = 7$ to $t = 9$

We have $a = 7$, $b = 9$, $D(7)=205.1$ and $D(9)=255.9$. Then $\frac{D(9)-D(7)}{9 - 7}=\frac{255.9 - 205.1}{2}=\frac{50.8}{2}=25.4$ meters per second.

Answer:

(a) 39.3 (b) 25.4