SOLUTION: An aircraft flying 3,500 feet above the ground at 11:05 A.M. descends to 1,000 feet at 11:10 A.M. What is the aircraft's average rate of change in feet per minute?

Question 1061875: An aircraft flying 3,500 feet above the ground at 11:05 A.M. descends to 1,000 feet at 11:10 A.M. What is the aircraft's average rate of change in feet per minute?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
he starts off at 3500 feet at 11:05 am.
he descends to 1000 feet5 at 11:10 am.

he has descended 2500 feet in 5 minutes.

divide 2500 by 5 and you get an average descent of 500 feet per minute.

the rate of change would be -500 feet per minute.

you can use the general equaion of rate * time = distance.

the time is 5 minutes.

the distance is -2500 feet.

rate * time = distance becomes rate * 5 = -2500

solve for rate to get rate = -2500 / 5 = -500 feet per minute.

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