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:

$v = 156$

Step-by-step 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:

$v = \frac{d(b) - d(a)}{b - a}$

Substitute values for b and a

$v = \frac{d(10) - d(2)}{10 - 2}$

$v = \frac{d(10) - d(2)}{8}$

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 $v = \frac{d(10) - d(2)}{8}$ becomes

$v = \frac{1300 - 52}{8}$

$v = \frac{1248}{8}$

$v = 156$

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