What is the average velocity of an object between time intervals? Using the derivative.

Question

Here is my work on this problem so far:

I know to find the average velocity, I should first find the derivative of the height vs time function. $$f(t)=80-4.9t^2$$ $$f'(t)=-9.8t+C$$

Where do I go from here to find average velocity? I know the answers to this problem, but am having a hard time figuring out how they are arrived at.

Any hints or help would be greatly appreciated.

Answer 1

The average velocity is simply the distance, covered by the object, divided by the time it took for the object to cover that distance.

There is no need to calculate $f'(t)$ to answer (a,b,c). You only need to calculate $\frac{f(t_1-f(t_0)}{t_1-t_0}$ where $t_0=1$ and $t_1$ takes three different values.

Answer 2

Notice, we can directly apply the formula for average velocity ($\bar{v}$).

the average velocity of the object is given as $$\bar{v}=\frac{\text{total distance traveled in time interval}\ \Delta t}{\text{total time required}\ (\Delta t)}$$$$=\frac{f(t_1)-f(t_2)}{t_2-t_1}=\frac{f(t_1)-f(t_2)}{\Delta}$$ Since, the height of the object from the ground is decreasing with the time $(t)$

1. from $t_1=1\ sec$ to $t_2=1.1\ sec$ average velocity is $$=\frac{(80-4.9(1)^2)-(80-4.9(1.1)^2)}{1.1-1}$$ $$=\frac{1.029}{0.1}=\color{red}{10.29}$$ I hope you can calculate rest of the values