Question

A ball is kicked up in the air from the ground. The height of the ball can be modeled as a function of time in seconds.
This function is represented on the graph.


Enter the average rate of change for the height of the ball, measured seconds
feet per second, between 0 seconds and 2.​

Answer 1

Answer:

  2 units per second

Step-by-step explanation:

The vertical (height) units are not shown on the graph or given in the problem statement. We'll call them "units".

The ball goes from 0 units high to 4 units high in 2 seconds. The average rate of change in height is ...

  ((4 - 0) units)/(2 seconds) = 2 units/second

Answer 2

Answer:

The slope of the function between 0 seconds and 2 is 2 feet per second.

Step-by-step explanation:

The rate of change of a function f(x) on [a,b] is

$m=\frac{f(b)-f(a)}{b-a}$

We need to find the average rate of change for the height of the ball, measured seconds feet per second, between 0 seconds and 2.​

From the given graph it is clear that the value of function is 0 at x=0 and 4 at x=2. So,

$f(0)=0, f(2)=4$

The average rate of change between 0 seconds and 2 is

$m=\frac{f(2)-f(0)}{2-0}$

$m=\frac{4-0}{2}$

$m=\frac{4}{2}$

$m=2$

Therefore, the slope of the function between 0 seconds and 2 is 2 feet per second.