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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 $\frac{s}{m^2}$.

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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 $s(0)=0$).

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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.

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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 $a=\frac{2\Delta s}{(\Delta t)^2}$.

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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 $v_0=0=r_0$

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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 $a = \frac{2s}{t^2}$.

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  • $\begingroup$ How is this different from michielm's answer? $\endgroup$
    – Bernhard
    Mar 17, 2013 at 17:13
  • $\begingroup$ Thank-you for your comment. I love spot the difference. My answer excludes calculus and thus is easier to understand for those people who haven't studied calculus. The one sentence simplicity of my answer also adds to its value for one to understand the concept with just one quick glance. $\endgroup$
    – Kenshin
    Mar 18, 2013 at 21:59
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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 $2^2 = 4$.

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  • $\begingroup$ What confuses me is that $a = \frac{\Delta v}{\Delta t}$, and $\Delta v$ when $a$ is constant is $\Delta v = \frac{s}{t}$. Substituting $v$ in the former equation results in the incorrect definition. $\endgroup$ Mar 17, 2013 at 13:58
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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

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    $\begingroup$ Hey there, welcome to Physics.SE! Please try to format your answer using MathJax, because it's a little bit difficult to read as it is now. $\endgroup$
    – Urb
    Aug 14, 2020 at 17:48

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