Is acceleration $a = s/t^2$, or $a = 2s/t^2$, or something third? I'm having trouble understanding some of the stuff regarding movement in my introductory physics class (I never thought I'd say that...)
Acceleration is defined as $ a = \frac{s}{t^2}.$
Distance can be calculated as the area under velocity-time line; given a constant accelation, and an initial velocity of 0, this forms a triangle: $ s = \frac{at^2}{2} $.
The latter, though, gives me the definition $ a = \frac{2s}{t^2}$. Is the former, then, wrong?
Perhaps I'm just better of using integrals?
EDIT (from comment below):
What confuses me is that $a = \frac{\Delta v}{\Delta t}$, and $\Delta v$ when $a$ is constant and $V_0$ is 0 is $v = \frac{\Delta s}{\Delta t}$. Substituting $v$ in the former equation: $a = \frac{\frac{\Delta s}{\Delta t}}{\Delta t}$, or $a = \frac{\Delta s}{\Delta t^2}$. Along with, of course, the SI unit $\frac{s}{m^2}$.
 A: If you would determine $a$ by integration you can see where the factor 2 is coming from:
$a=\frac{\partial^2 s}{\partial t^2}$ so integrating once gives: $a t = \frac{\partial s}{\partial t}$ (ignoring the integration constant). Then integrating a second time you find:
$\frac{a t^2}{2}=s$ or $a=\frac{2s}{t^2}$ (again ignoring the integration constant).
So the factor 2 arises from integrating $t \partial t$.
Now, I  mentioned that I ignored the integration constant. If I would not have done that I would have found that: $s=\frac{a t^2}{2}+v_0 t + s_0$, because the initial velocity or displacement could be non-zero. In this definition the acceleration would be: $a=\frac{s-s_0}{t^2}-\frac{v_0}{t}$, although I think it would be a bit strange to define an acceleration based on this equation.
A: Your error is in stating $\Delta v=\frac{\Delta s}{\Delta t}$. This is not true in your case because that is the expression for the average velocity. Compare it with falling from a height. Your velocity of impact will be larger than your average velocity during the fall.
Thus, $a=\frac{\Delta s}{(\Delta t)^2}$ is incorrect here. Instead, you are correct that $a=\frac{2\Delta s}{(\Delta t)^2}$.
A: You are sloppy with the $\Delta$'s. $\Delta v$ is not equal to $s/t$, even if $a$ is constant; it is equal to $\Delta s/\Delta t$. This is where your factor of 2 trouble lies. For constant acceleration (and $s(0)=v(0)=0$) you have $s=\frac{1}{2}at^2$. Now simply compare the two expressions:
$$
\frac{s}{t} = \frac{\frac{1}{2}at^2}{t} = \frac{1}{2}at \ ,
$$
whereas
$$
\frac{\Delta s}{\Delta t} = \frac{s(t+\Delta t)-s(t)}{\Delta t} = \frac{\frac{1}{2}a(t+\Delta t)^2-\frac{1}{2}at^2}{\Delta t}=\frac{at\Delta t+\frac{1}{2}a\Delta t^2}{\Delta t} \approx at \ ,
$$
where the last "$\approx$" holds for small $\Delta t$.
The second expression is the correct one for the velocity. The first one is wrong (it only holds for constant velocity and $s(0)=0$).
A: You should use differential quotients. Acceleration is defined as the change of velocity with time, $a=\frac{dv}{dt}$ and since velocity is the change of position with time we find that accelaeration is the second derivative of the distance with respect to time:$a=\frac{d^2r}{dt^2}$. If you say velocity is constant, than $v=\frac{s}{t}$. But then you cannot say, that you have an acceleration. If the velocity is not constant due to a non-vanishing a, you have to use the derivative. So the second one is correct for constant a and $v_0=0=r_0$
A: Using the formula $s = ut + \frac{1}{2} at^2$, we can see that if initial velocity is 0, then $s = \frac{1}{2} at^2$, and therefore $a = \frac{2s}{t^2}$.
A: There are several things wrong with what you've posted.
First of all, you state that acceleration is defined as $a = s/t^2$. However, even in the wikipedia article that you link to in your second line, it states that for uniform acceleration,
$s = ut + \frac{1}{2}at^2$
and given that our starting velocity is $u = 0$, we get $s = \frac{1}{2}at^2 \implies a = \frac{2s}{t^2}$. This is the same answer as the one you arrive at using the method of calculating the area under the velocity-time curve.
You may be confused in your first statement from hearing that for uniform acceleration, distance 'goes like' or 'is of the order' $t^2$. That is, if from a standing start, under uniform acceleration, I travel one metre in the first second, then I travel four metres in the first 2 seconds, since $2^2 = 4$.
A: We know 1. s=at²
2. s=vt+1/2(at²)
Above two equation valid for uniform motion.
     Equation two divide by t we get.

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

s/t= Average velocity  or average velocity is also equal to (u+v)/2. Put in above equation.
(v+u)/2 = v + 1/2(at)
Put v =0 above equation become
u=at. —————-[1]
for finding velocity, velocity formula is rate of change in displacement. Mathematically (s2-s1)/t
     Put s2-s1=s Equation become s/t

Where s2= final point
s1= initial point
Put in [1]
  s/t=at

s=at² hence proved
