# how do I find the average velocity and instantaneous velocity?

This is a question from my Calculus book. I was showed the correct answer but I really need to learn what the process is to achieve the answer. Here is the question:

If a ball is thrown into the air with a velocity of $40 \frac{ft}{s}$, its height in feet $t$ seconds later is given by $y=40t - 16t^2$ . Find the average velocity for the time period beginning when $t = 2$ and lasting

(i) 0.5sec

(ii) 0.1sec

(iii) ...

First you find the height after 2 seconds, we'll denote that as $y_2$. So using the fomula we have:
$$y_2 = 40 \cdot 2 - 16 \cdot 2^2 = 80 - 64 = 16$$
$y_{2.5} = 40\frac 52 - 16 \left(\frac 52\right)^2 =100 - 100 = 0$$So after 2.5 second the height of the ball will be 0\text{ ft}. We know that the average velocity can be calulated using the following formula:$$v = \frac{\Delta s}{\Delta t} = \frac{|y_1 - y_0|}{|t_1 - t_0|}$$So after the substitution we have:$$v = \frac{|0 - 16|}{|2.5 - 2|} = \frac{|-16|}{|0.5|} = \frac{16}{0.5} = 32$$So the average speed in that period will be 32$\frac{ft}{s}$Now you can do the 0.1 second interval by yourself. The average velocity for the time period between$t_1$and$t_2$($t_2>t_1$), is given by$\dfrac{s(t_2)-s(t_1)}{t_2-t_1}$where$s(t)$is the displacement at time$t$. So, (i) is asking for the average velocity between$2\text{ sec}$and$2.5 \text{ sec}$and (ii) is asking for the average velocity between$2 \text{ sec}$and$2.1 \text{ sec}\$.