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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.​

A ball is kicked up in the air from the ground. The height of the ball can be modeled-example-1
User Elric
by
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2 Answers

5 votes

Answer:

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

Explanation:

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


m=(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=(f(2)-f(0))/(2-0)


m=(4-0)/(2)


m=(4)/(2)


m=2

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

User Nikunj Chaklasiya
by
7.5k points
4 votes

Answer:

2 units per second

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

User Willeke
by
9.0k points

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