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Makovka662 [10]

3 years ago

12

The distance traveled, in feet, of a ball dropped from a tall building is modeled by the equation d(t) = 16t2 where d equals the

distance traveled at time t seconds and t equals the time in seconds. What does the average rate of change of d(t) from t = 2 to t = 5 represent?

2 answers:

Elis [28]3 years ago

8 0

Answer

Find out what does the average rate of change of d(t) from t = 2 to t = 5 represent .

To prove

Formula

$Average\ rate\ of\ change = \frac{d(5) - d(2)}{change\ in\ time}$

As given

The distance traveled, in feet, of a ball dropped from a tall building is modeled by the equation d(t) = 16t²

where d equals the distance traveled at time t seconds .

Now

d(5) = 16 × 5 × 5

= 400 feet

d(2) = 16 × 2 × 2

= 64 feet

Now

Change in time = 5 - 2

= 3 second

put all the values in the above formula

$Average\ rate\ of\ change = \frac{400 - 64}{3}$

$Average\ rate\ of\ change = \frac{336}{3}$

Average rate of change is 112 feet/second.

11Alexandr11 [23.1K]3 years ago

8 0

The average rate of change of height is speed.

The average rate of change of d(t) on t ∈ [2, 5] represents the ball's average speed over that interval of time.

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