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

martina 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 71.2 4 166.4 6 172.8 10 239.6 (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 4 seconds to 10 seconds. meters per second

martina 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 71.2 4 166.4 6 172.8 10 239.6 (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 4 seconds to 10 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 $D(2)=71.2$. Using the formula $\frac{D(2)-D(0)}{2 - 0}=\frac{71.2-0}{2}=\frac{71.2}{2}=35.6$.

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

We have $a = 4$, $b = 10$, $D(4)=166.4$, and $D(10)=239.6$. Using the formula $\frac{D(10)-D(4)}{10 - 4}=\frac{239.6 - 166.4}{6}=\frac{73.2}{6}=12.2$.

Answer:

(a) 35.6 (b) 12.2