If a ball is thrown into the air with a velocity of 42 ft/s, its height in feet t seconds later is given by y = 42t − 16t2.
(a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following.

1) 0.5 seconds

if someone could point me into the right direction i'd appreciate it. i need the slope of the secant line, then height/change in time, but I am lost on how to get to that point.

Since it is average velocity over a time period, you don't need to do a derivative. The average velocity is the difference in position aka displacement (y) during the period over the difference in time from start to end of the period (t).

The period of time is given as t=2 to t=2.5. The position at each time (y) can be found by plugging in each of those times into your position equation.

Average velocity = [y(2.5) - y(2)]/[2.5 - 2]

Hard to type out. Sorry not in a place that I can write it.

The slope of the Secant line is the average between two points on a curve. So the value you get as average velocity should be equivalent to that slope.

This post was edited by timmayX on Jun 4 2016 11:10am