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? The ball travels an average distance of 112 feet from 2 seconds to 5 seconds. The ball falls down with an average speed of 48 feet per second from 2 seconds to 5 seconds. The ball falls down with an average speed of 112 feet per second from 2 seconds to 5 seconds. The ball travels an average distance of 48 feet
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It represents the average speed during that interval or in other words the distance traveled during that interval/ 3 seconds d(5) = 16 * 25 = 400 d(2) = 16*4 = 64 distance during those three seconds = 400-64 = 336 feet 336 ft/3 s = 112 ft/s = average speed for that interval By the way they mean distance fallen = 16 t^2 the height h = building height - 16 t^2
so would it be A?
yes im pretty sure
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