homework and exercises - Calculating Acceleration

I'm having problems calculating acceleration for the following variables. I would have thought it would be extremely straight forward, except I am getting two different answers and do not know which one is correct.

I have the following variables:

$$\begin{align}d &= 229.75\ \mathrm{cm}\\ t &= 1.97\ \mathrm s\end{align}$$

and I need to find acceleration using these variables. I've used both of the following equations, each resulting in a different answer:

$$v=\frac{\Delta d}{\Delta t}\tag{1.1}$$

$$a=\frac{\Delta v}{\Delta t}\tag{1.2}$$

$$a=\frac{2d}{t^2}\tag2$$

Using Equations (1.1) and (1.2), I get the following:

$$\begin{align}v&=\frac{229.75\ \mathrm{cm} - 0\ \mathrm{cm}}{1.97\ \mathrm s - 0\ \mathrm s}\\ &=116.6\ \mathrm{cm/s}\end{align}$$

$$\begin{align}a&=\frac{116.6\ \mathrm{cm/s} - 0\ \mathrm{cm/s}}{1.97\ \mathrm s - 0\ \mathrm s}\\ &=59.2\ \mathrm{cm/s^2}\end{align}$$

Acceleration is $59.2\ \mathrm{cm/s^2}$, according to those equations.

Using Equation (2), I get the following:

$$\begin{align}a&=\frac{2(229.75\ \mathrm{cm})}{(1.97\ \mathrm s)^2}\\ &=118.4\ \mathrm{cm/s^2}\end{align}$$

Which is obviously a different answer than above. It is also double the answer above, which makes complete sense because if I merge Equations (1.1) and (1.2), I get:

$$a=\frac{\frac{\Delta d}{\Delta t}}{\Delta t}$$

or, essentially:

$$a=\frac{\Delta d}{\Delta t^2}$$

and the second equation is the same except it doubles displacement at the top. Therefore, the answer for my second equation is double the answer I got for my first equations.

What I don't understand is which formula I am supposed to be using, and why the formulas result in different answers. I was under the impression that I can use whatever formula I want, as long as I have enough variables to put in and am able to use algebra to solve for the variable I want.

Any ideas as to which I should use?

EDIT: Forgot to mention, acceleration is constant. Initial velocity, initial time and initial displacement are all 0. The information above was measured during a lab in which we timed a cart accelerating down a ramp from rest.

Answer

An accelerating object has a changing velocity. Obviously so since the object starts with zero velocity and the velocity increases with time according to the SUVAT equation:

$$ v = u + at $$

So your equation 1.1 is no use here. It calculates the average velocity. This could actually be used to calculate the acceleration, but the working is a bit involved so I advise not going down that path. Instead you need another SUVAT equation:

$$ s = ut + \tfrac{1}{2}at^2 $$

You know that the cart starts at rest so $u = 0$, and you know $s$ and $t$ so you can calculate $a$.