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:

$$y_2 = 40 \cdot 2 - 16 \cdot 2^2 = 80 - 64 = 16$$

After 0.5 seconds, the time passed after the throw will be 2.5 seconds so for this height we have:

$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 $2.1 \text{ sec}$.


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