Question

An object dropped from rest from the top of a tall building on Planet X falls a distance d (t )equals 8 t squared feet in the first t seconds. Find the average rate of change of distance with respect to time as t changes from t 1equals2 to t 2equals5. This rate is known as the average​ velocity, or speed. The average velocity as t changes from 2 to 5 seconds is nothing StartFraction feet Over sec EndFraction .

Answer 1

The average velocity is

`(128)/(3)\text{ feet per second}`

Explanation:

We are given the following in the question:

`y(t) = 8t^2`

where y(t) is the distance in feet in first t seconds.

Average rate of change =

`(y(t_2)-y(t_1))/(t_2 - t_1)`

a) average rate of change of distance with respect to time from 2 to 5 seconds

Putting values, we get,

Average velocity =

`(y(5) - y(2))/(5-2) = (8(25-9))/(5-2) = (128)/(3)\text{ feet per second}`

Thus, the average velocity is

`(128)/(3)\text{ feet per second}`