a cars distance, $d(t)$, in meters, from a stop sign after $t$ seconds is shown in the table.\n|$t$|$d(t)$|\n…

a cars distance, $d(t)$, in meters, from a stop sign after $t$ seconds is shown in the table.\n|$t$|$d(t)$|\n|----|----|\n|0|0|\n|1|2.25|\n|2|9|\n|3|20.25|\n|4|36|\n|5|56.25|\nwhat is the average rate of change of the cars distance, in meters per second, from 0 to 5 seconds?

a cars distance, $d(t)$, in meters, from a stop sign after $t$ seconds is shown in the table.\n|$t$|$d(t)$|\n|----|----|\n|0|0|\n|1|2.25|\n|2|9|\n|3|20.25|\n|4|36|\n|5|56.25|\nwhat is the average rate of change of the cars distance, in meters per second, from 0 to 5 seconds?

Answer

Explanation:

Step1: Recall average - rate - of - change formula

The formula for the average rate of change of a function $y = f(x)$ from $x = a$ to $x = b$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 0$, $b = 5$, $f(t)=d(t)$, $d(0)=0$, and $d(5)=56.25$.

Step2: Substitute values into formula

$\frac{d(5)-d(0)}{5 - 0}=\frac{56.25-0}{5}$

Step3: Calculate the result

$\frac{56.25}{5}=11.25$

Answer:

$11.25$