Question

The distance traveled, in feet of a ball droped from a tall building is modeled by the equation d(t) = 16t^2 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?

Answer 1

That represents the avg acceleration of the ball from 2-5sec

Answer 2

$\bf slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\ -------------------------------\\\\ d(t)= 16t^2 \qquad \begin{cases} t_1=2\\ t_2=5 \end{cases}\implies \cfrac{d(5)-d(2)}{5-2}\implies \cfrac{400-64}{5-2} \\\\\\ \cfrac{336}{3}\implies 112$

what does 112 ?

well, it means that from the 2nd second to the 5th second, on average the ball was dropping at an average of 112 feet per second.