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

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 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 the formula

The average rate of change of $d(t)$ from $t = 0$ to $t = 5$ is $\frac{d(5)-d(0)}{5 - 0}$. Substitute $d(0)=0$ and $d(5)=56.25$ into the formula: $\frac{56.25-0}{5}=\frac{56.25}{5}=11.25$.

Answer:

$11.25$