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?

Question

anonymous · Answer

use the difference formula. $$D(t) = d(t2)-d(t1)/(t2-t1)$$

anonymous · Answer

@maggie/tiffanie  you got this ?
you want to compute $$\frac{16\times 5^2-16\times 2^2}{3}$$