Two equations for position $x$ but gives different results? It seems that there are two equations for finding $x$:
$$
\begin{align*}
x & =x_0+vt\tag{1} \\
x & =x_0+vt+\frac{1}{2}at^2\tag{2}
\end{align*}
$$
But one can clearly see that the two equations will not give the same answers....
How can I decide which equation to even choose. I have an exam on kinematics but it's all so confusing for me....
 A: tl;dr-  The equation with $a$ is the "real" equation.  The other equation is a simplified version that you can use if there's no acceleration.  The "real" equation always works and you can always use it.
Compare the equations
So you've got two equations:
$$
\begin{align*}
x & =x_0+vt\tag{1} \\
x & =x_0+vt+\frac{1}{2}at^2\tag{2}
\end{align*}
$$
where


*

*$x$ is position;

*$x_0$ is the initial position;


*

*Note that $x=x_0$ when $t=0$; after that, $x$ can change, but $x_0$ stays the same.


*$t$ is time;

*$v$ is velocity;

*$a$ is acceleration.


What happens to Equation (2) if there's no acceleration?
$$
\begin{align*}
x & =x_0+vt+\frac{1}{2}at^2\tag{2} \\
& =x_0+vt+\frac{1}{2}\left(0\right)t^2 \\
& =x_0+vt+0 \\
& =x_0+vt \tag{1}
\end{align*}
$$
Seems like both equations are the same when $a=0$, right?  That's because Equation (2) is the more general form.  Equation (1)'s the simplified version that you can use when there's no acceleration.
Why two equations for the same thing?
Your class probably taught you Equation (1) because it's easier to understand.  But, Equation (2)'s the one that you really need to remember, since it works even when there's acceleration.
If you remember Equation (2) and know to assert $a=0$ whenever there's no acceleration, you can forget Equation (1).
Summary
Equation (1), $x=x_0+vt$:


*

*Easier for new students to remember.

*Works when there's no acceleration.

*Gives you the wrong answer if there is acceleration.


Equation (2), $x=x_0+vt+\frac{1}{2}at^2$:


*

*Harder to remember since it's bigger.

*Always works.

*Remember that if an exam question says that something is moving "with constant velocity" or has "no acceleration", then that means $a=0$.

