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, where t represents time, in seconds.
What is the average rate of change in the height, in feet per second, during the first 1.25 seconds that the beach ball is in the air?
I got the answer 39 is that correct?
h(1.25) = 39
h(0) = 64
avg rate of change is totaldisplacement/totaltime = (39-64)ft/1.25s = -20ft/s
you only found the displacement, not the rate of change.
Oh okay thank you but I got the anwser 20 instead of negative -20
Nvm I did the calculations wrong you were right. Thanks again for your help
To find the average rate of change of the height function during the first 1.25 seconds, you need to calculate the difference in height between the starting time and the ending time, and then divide it by the difference in time.
In this case, the starting time is t = 0 and the ending time is t = 1.25.
To find the height at the starting time, substitute t = 0 into the height function:
h(0) = 64 - 16(0)^2
h(0) = 64
To find the height at the ending time, substitute t = 1.25 into the height function:
h(1.25) = 64 - 16(1.25)^2
h(1.25) = 64 - 16(1.5625)
h(1.25) = 64 - 25
h(1.25) = 39
The difference in height between the ending and starting times is h(1.25) - h(0) = 39 - 64 = -25.
The difference in time is 1.25 - 0 = 1.25.
Now, you can calculate the average rate of change by dividing the difference in height by the difference in time:
Average rate of change = (-25) / 1.25 = -20
Therefore, the average rate of change in the height during the first 1.25 seconds is -20 feet per second. Your answer of 39 is incorrect.