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

ashley is driving a racecar. 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 5 seconds to 9 seconds. meters per second

ashley is driving a racecar. 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 5 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 for part (a)

We have $a = 0$, $b = 2$, $D(0)=0$, and $D(2)=78.6$. Then the average rate of change is $\frac{D(2)-D(0)}{2 - 0}=\frac{78.6-0}{2}=\frac{78.6}{2}=39.3$.

Step3: Calculate average rate of change for part (b)

We have $a = 5$, $b = 9$, $D(5)=151.5$, and $D(9)=255.9$. Then the average rate of change is $\frac{D(9)-D(5)}{9 - 5}=\frac{255.9 - 151.5}{4}=\frac{104.4}{4}=26.1$.

Answer:

(a) 39.3 (b) 26.1