v = (xf – x0)/(tf – t0)

= (3 – 3)/(4 – 0) = 0/4 = 0

This tells you that the object was in the same place at the end of 4 seconds as it was at the beginning. The average velocity, then, was 0 – the thing didn’t go anywhere.

Does that help?

Edit: The simple formula is the one I quoted, but they may be using different symbols. Average velocity is displacement over time – look for something like that.

As for whether it moved or not, that’s not what averages are about. My uncle used to say “never wade across a river that’s an average of 3 feet deep” – the point being that just because the average is 3 feet doesn’t mean it doesn’t get a whole lot deeper. The average velocity only considers where it started and where it ended – how it got there doesn’t matter. It started and ended at the same place, so it’s just as if it travelled with a constant velocity of 0 the whole time. That’s the whole point of the average velocity.

One way to define it is “the velocity you would have needed to start where you started and end where you ended if you travelled at constant velocity”.

I was intentionally vague, because you indicate that you are really trying to figure this out, not just looking for someone to do your homework. I hope this helps. It’s been a long time since I’ve worked out this type of problem, but I have a degree in civil and environmental engineering, so it’s something I’ve done. Feel free to write me an e-mail if this helps, but you have more questions.

s(t)= 1/2gt2 + v + s

where g is gravity (-9.8 m/s or -32ft/s)

t is time (it’s supposed to be t squared)

v is the initial velocity and s is the height it starts at

then, you take

(s(t) – s(c)) / (t – c) where t and c are the two points

I don’t know the guess method, but what you’re asking sounds like what I just described

-Ignore my overcomplicated explanation.

Even if the graph dipped down, so far as its AVERAGE velocity goes, it didn’t go anywhere

Don’t complicate things unnecessarily.