Question

A beach ball rolls off a cliff and onto the beach. The height, in feet, of the beach ball can be modeled by the function h(t)=64–16t2

h
(
t
)
=
64
–
16
t
2
, where t
t
represents time, in seconds.

What is the average rate of change in the height, in feet per second, during the first 1.25
1.25
seconds that the beach ball is in the air?

Answer 1

Answer: The average rate of change between 0 seconds and 1.25 seconds is -20 meters.

Explanation:

The height of the ball can be described by the equation

h(t) = 64m - (16m/s^2)t^2

Where we want to find the average rate of change in the first 1.25 seconds that the ball is in the air, we can do this by finding the slope that conects the points h(0s) and h(1.25s).

h(0s) = 64m

h(1.25s) = 64m - (16m/s^2)*(1.25s)^2 = 39m

now, remember that the slope between two points can be written as:

s = (y2 - y1)/(x2 - x1)

where y2 = h(x2) and y1 = h(x1)

so we have that the average rate of change:

s = (39m - 64m)/(1.25 - 0) = -20m

Then we have that the average rate of change betwen 0 seconds and 1.25 seconds is -20m

Answer 2

rate of change = -32 feet/second

Explanation:

h(t) = 64 – 16t2

t = time (seconds)

Look at the start of time where t = 0

h(t) = 64 - 16 * (0) * 2 = 64 feet

Then look at the specified time where t = 1.25 seconds

h(t) = 64 - 16 * (1.25sec) * 2 = 24 feet

rate of change = rate of change y / rate of change x

x = time, y = height

rate of change = (24 - 64) / (1.25 - 0) = -32 feet/second