# $d=\frac{1}{2}at^2$ not giving expected results

Say I have an object that starts falling from rest at 2 m/s2. Its total displacement and velocity at the end of every second should look something like this:

t    d    v
1    2    2
2    6    4
3    12   6
4    20   8


At the end of the first second, it should have accelerated to 2 m/s and moved 2 m. At the end of 4 seconds, it should moved 20 m. But when I use the equation for getting displacement, $d=\frac{1}{2}at^2$, I get 16 m. Shouldn't it be 20 m? What is going wrong here?

-
With $t=1$ and $a=2$, shouldn't $d(1) = 0.5*2*(1*1) = 1$?? – Kyle Kanos Aug 6 '13 at 1:01
@KyleKanos I guess, but even that doesn't make sense. It should have moved 2 meters, according to that it accelerates at 2 m/s^2 – Daniel Pendergast Aug 6 '13 at 1:05
If you took the time to enter $t$ into $d=at^2/2$, you would see that your entire d column is wrong. So the question really is: what are you using to compute $d$? – Kyle Kanos Aug 6 '13 at 1:08
@KyleKanos I didn't realize that for the first second it should have been 1m. I then added what should have corresponded for the next seconds. – Daniel Pendergast Aug 6 '13 at 1:12
I echo what @Kyle said. Without some explanation of why you expect $d$ at 1s to be 2m, I really don't think this is a good question. It comes across as "why don't these numbers I made up fit this formula?" So could you edit the question to explain where you're getting your numbers from? – David Z Aug 6 '13 at 1:27

t.   v     v_avg       d