# Accelerating with gravity equation

1. Jan 4, 2004

### Cr4X

Allright, this isn't exactly a K-12 question, but I am trying to learn something on my own, since I don't think I learn enough at my school. Anyway, I am working with Basic Physics secound edition, and I really need some place to ask for help since I am stuck.

I have come to a place, where I am getting taught how to use an equation to learn how far an object has traveled with gravity:

d = v0t+½at**2

Explanation:
v0 = starting position
d = distance
at = unsure but a = acceleration and t = time
** = It's difficult to explain but: 3**3 = 3*3*3=27

Should I multiply v0 with time or what?

When v0 = 0 it shows me this example for 1 secound of acceleration:
d = ½*9.8*1**2

Here I get that distance traveled = average speed * secound, but I don't get **2

2. Jan 4, 2004

### HallsofIvy

Well, yes, that exactly what "v0t" means!

They are exactly the same (and I'm impressed that you saw the "average speed" connection).

One can show (though it requires calculus) that if acceleration is constant then distance moved will depend on t2 and is the same as averaging initial and final velocity and using that like a constant velocity.

3. Jan 5, 2004

### Cr4X

I understand most of it now, but I still don't get **2...
Let's say we drop a ball from a tall building and it reaches ground after 2 secounds gaining a speed of 19.6m/s

d=½*19.6*2=19.6m

I can't see the problem in this one, but if I now add the **2 equation it would look like this:
d=½*19.6*2**2=39.2m

4. Jan 5, 2004

### Staff: Mentor

Nothing wrong with this, if you know what you are doing. You are using the final velocity (19.6) to find the average velocity 1/2 (19.6). So, you are essentially doing this:

$$d=V_{ave}\times T$$, which becomes..
$$d=\frac{1}{2}(V_{initial} + V_{final})\times T$$
which becomes (since it starts from rest)...
$$d=\frac{1}{2}(V_{final})\times T$$
$$d=\frac{1}{2}(aT)\times T$$

Note that you are already using time once, to caculate Vfinal; if you wish to skip that separate step, you can go directly to distance without calculating Vfinal. In that case you use "a" (acceleration) in the formula, not Vfinal:
$$d=\frac{1}{2}aT^2$$
No. The version of the distance formula that uses T2 (aka T**2) is 1/2 a T2, not 1/2 Vfinal T2! (See the discussion above.)

Does this help a bit?

5. Jan 5, 2004

### Cr4X

I don't really get it, but I as long as I know how to calculate the answer, i'm fine, when I have learnt more I might return and fully understand it.

Thanks a lot for the help.