Question

An object dropped from rest from the top of a tall building on Planet X falls a distance feet in the first t seconds. Find the average rate of change of distance with respect to time as t changes from to . This rate is known as the average​ velocity, or speed. The average velocity as t changes from to seconds is:________

Answer 1

Complete Question:

An object dropped from rest from the top of a tall building on Planet X falls a distance d(t)=13t^2 feet in the first t seconds. Find the average rate of change of distance with respect to time as t changes from t1=2 to t2=10. This rate is known as the average velocity, or speed.

The average velocity as t changes from 2 to 10 seconds is _______ feet/sec?

Answer:

`v = 156`

Explanation:

Given

`d(t)=13t^2`

Time Interval: (a,b)

`a = 2`

`b = 10`

Required

Determine the average rate of change (v)

This is calculated as thus:

`v = (d(b) - d(a))/(b - a)`

Substitute values for b and a

`v = (d(10) - d(2))/(10 - 2)`

`v = (d(10) - d(2))/(8)`

Calculate d(10): Substitute 10 for t

`d(t)=13t^2`

`d(10) = 13 * 10^2`

`d(10) = 13 * 100`

`d(10) = 1300`

Calculate d(2): Substitute 2 for t

`d(t)=13t^2`

`d(2) = 13 * 2^2`

`d(2) = 13 * 4`

`d(2) = 52`

The equation
`v = (d(10) - d(2))/(8)` becomes

`v = (1300 - 52)/(8)`

`v = (1248)/(8)`

`v = 156`

Hence, the average rate of change is 156ft/s