Asked by Anonymous
The height 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?
Answers
Answered by
Damon
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
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
Answered by
Anonymous
hgy
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