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?
a. the ball travels an average distance of 112 feet from 2 seconds to 5 seconds.
b. the ball falls down with an average speed of 48 feet per second from 2 seconds to 5 seconds.
c. the ball falls down with an average speed of 112 feet per second from 2 seconds to 5 seconds.
d. the ball travels an average distance of 48 feet from 2 seconds to 5 seconds.

When  t=2, the ball has fallen     d(2) = 16 (2²) = 64 feet .

When  t=5, the ball has fallen     d(5) = 16 (5²) = 400 feet .

Distance fallen from  t=2  until  t=5  is  (400 - 64) = 336 feet.

Time period between  t=2  until  t=5  is  (5 - 2) = 3 seconds.

Average speed of the ball from  t=2  until  t=5  is

                 (distance covered) / (time to cover the distance)

             =            336 feet        /        3 seconds       =  112 feet per second.

That's what choice-C says.        

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.