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:
[tex]v = 156[/tex]
Step-by-step explanation:
Given
[tex]d(t)=13t^2[/tex]
Time Interval: (a,b)
[tex]a = 2[/tex]
[tex]b = 10[/tex]
Required
Determine the average rate of change (v)
This is calculated as thus:
[tex]v = \frac{d(b) - d(a)}{b - a}[/tex]
Substitute values for b and a
[tex]v = \frac{d(10) - d(2)}{10 - 2}[/tex]
[tex]v = \frac{d(10) - d(2)}{8}[/tex]
Calculate d(10): Substitute 10 for t
[tex]d(t)=13t^2[/tex]
[tex]d(10) = 13 * 10^2[/tex]
[tex]d(10) = 13 * 100[/tex]
[tex]d(10) = 1300[/tex]
Calculate d(2): Substitute 2 for t
[tex]d(t)=13t^2[/tex]
[tex]d(2) = 13 * 2^2[/tex]
[tex]d(2) = 13 * 4[/tex]
[tex]d(2) = 52[/tex]
The equation [tex]v = \frac{d(10) - d(2)}{8}[/tex] becomes
[tex]v = \frac{1300 - 52}{8}[/tex]
[tex]v = \frac{1248}{8}[/tex]
[tex]v = 156[/tex]
Hence, the average rate of change is 156ft/s
