Question

A rock is dropped from a hot air balloon at a height of 200 meters. The rock's height from the ground in meters, h(t), is approximated by the formula h(t) = -5tᵗ² + 200, where t is the time in seconds. What is the average rate of change in m/s of the height of the rock between 2 and 5 seconds?

Answer 1

The average rate of change in m/s of the height of the rock between 2 and 5 seconds is -35 m/s.

Step-by-step explanation:

The average rate of change of the height of the rock between 2 and 5 seconds can be found by calculating the difference in height at those two time points and dividing it by the difference in time:

Average rate of change = (h(5) - h(2)) / (5 - 2)

Substituting the values into the equation, we have:

Average rate of change = (-5(5)^2 + 200) - (-5(2)^2 + 200)) / (5 - 2)

Simplifying further, we get:

Average rate of change = (-125 + 200) - (-20 + 200) / 3

Average rate of change = (75 - 180) / 3 = -35 m/s