Fourth Equation of Motion When I was studying motion, my teacher asked us to derive the equations of motion. I too ended up deriving the fourth equation of motion, but my teacher said this is not an equation. Is this derivation correct?
\begin{align}
v^2-u^2 &= 2ax \\
(v+u)(v-u) &= 2 \left(\frac{v-u}{t}\right)x \\
(v+u)t &= 2x \\
vt+ut &= 2x \\
vt+(v-at)t &= 2x \\
2vt-at^2 &= 2x \\
x &= vt- \frac{at^2}{2}
\end{align}
And why is this wrong to say that this is the fourth equation of motion?
Given 3 equations of motion:-
\begin{align}
v&=u+at \\
x&=ut+ \frac{at^2}{2} \\
v^2-u^2&=2ax
\end{align}
 A: Just to add that when you use these equations it's good to think in a physical way as you go along. So:
$$
v = u + a t
$$
think: "yes the final velocity is the initial velocity plus the change owing to acceleration. This is basically the definition of $a$."
Next
$$
x = u t + \frac{1}{2} a t^2
$$
think: "the distance is travelled is how far it would have gone if it had not accelerated, plus the extra bit owing to the increase in the velocity". This can also be seen as
$$ 
x = \left(\frac{u + v}{2}\right) t
$$
which is a nice way to write it since on the right you have the average velocity multiplied by time, which is easy to remember.
Finally, for the next equation I recommend writing it as
$$
\frac{1}{2} m v^2 = \frac{1}{2} m u^2 + m a x
$$
since then you can understand it as "this is all about energy: the final kinetic energy is equal to the initial kinetic energy plus the work done by the force which is causing the acceleration. The force is $f=ma$ and the work done is $fx$."
The idea is that by understanding the physical meaning of the equations you don't need to think of them as rules presented to you without explanation. Rather you understand the physics and then the equations write themselves. You can then go on to derive from them any further equations you may find useful. As long as the derivations are correct then the further equations are also correct.
A: First of I think it is great that your taking apart equations and putting them back together in a different ways as if they were LEGO pieces. I think most of us (with an interest in physics) did that in high school and it can be very insightful.
So instead of fretting over the characterization of $x = v t - \tfrac{1}{2} a t^2$, debating if it is an equation of motion, a solution of the equations of motion with different boundary conditions, is it just kinematics, or the description of a geodesic path under constant acceleration, do the following.
Examine the expression you have under different situations and see what story it tells you.
I would go one step back in your equations at $(v+u)t = 2x$ and re-arrange it as
$$ \frac{v+u}{2} = \frac{x}{t} $$
How would you interpret this equation?
On the left you have the average velocity between the launch condition and now, and on the right you have displacement over time. So you can state the equation as
$$ x(t) = v_{\rm ave} t $$  where at every instant $v_{\rm ave} = \frac{v(t)+u}{2}$.
This is helpful because it shows that displacement is directly related to the average speed. In fact if you plot speed vs. time you find the above expression as the area under the curve and it looks like a trapezoid. The formula above is the area of trapezoid you learned in geometry. Actually the above is one of the first concepts you are going to learn in calculus. Displacement is the area under the velocity curve and you can use geometry to solve physics problems.
Now specifically for $x = vt- \frac{at^2}{2}$ try to dissect it further but looking at what is known and unknown at specific situations. It results in displacement as a function of time $t$, knowing the acceleration $a$ and the current velocity $v$.
This is different from the standard displacement equation of $x = ut+ \frac{at^2}{2}$ since the current velocity isn't included but instead the velocity $u$ at time zero is included.
Technically speaking your equation is implicit as it depends on future conditions and the standard equation is explicit as it depends on initial conditions only.
Both are correct, but each tells us a different story.
A: The problem of perception as to "What is a new Equation of Motion?" seems to originate with the dogmatic teaching of the Three Equations Of Motion as a Set of Results to be Learned.
They are in fact three results derived from the distillation of Newton's Laws:
$$\mathbf f = \dfrac {\mathrm d} {\mathrm d t} (m \mathbf v)$$
which differential equation is solved, where $\mathbf f$ is set to a constant (and $m$ is taken for granted as being constant also).
This of course can't be done at the elementary level at which the SUVAT equations are initially introduced. So the three convenient "equations of motion" are introduced instead, in a way that the students can get their heads round them.
Whether an equation is given an official Name to Be Remembered is not all that important. What is important is the ability to use them. Working out that fourth equation from the given three is actually a worthy exercise in its own right. Granted it is not a particularly profound equation, as it can be obtained from the other three. But -- get this -- each of the other three has also merely been derived from other equations.
For your teacher to dismiss it as "not an equation" is appalling.
It may be the case that the teacher is teaching from the book, and not from his or her own expertise in the subject. It can be disheartening to be taught by teachers who do not understand the subject they are teaching, but hang on in there, it gets better as you go on in your schooling.
A: Here are the four equations being discussed here:
\begin{align}
v &= u + at \tag{1} \\
x &= ut + \frac{at^2}{2} \tag{2} \\
v^2 - u^2 &= 2ax \tag{3} \\
x &= vt- \frac{at^2}{2} \tag{4}
\end{align}

Is this derivation correct?

The result is always correct. The derivation is correct, except for dividing by $(v - u)$. It is possible that this is zero, ie $v = u$ (equivalent to $t = 0$ or $a = 0$ by equation $(1)$). If $t = 0$, then $x = 0$, so Equation $(4)$ is still correct. And if $a = 0$, then Equation $(4)$ clearly follows from Equation $(2)$, so Equation $(4)$ is still correct.

And why is this wrong to say that this is the fourth equation of motion?

It is not wrong. It is a matter of opinion which equations should be collected together and referred to as standard equations of motion; opinions cannot be right or wrong. And this equation is actually a standard equation of motion, so in that sense, it was actually your teacher who was wrong. We will come back to this.
First, let us look at whether it is useful.
Notice that your derivation only used Equations $(1)$ and $(3)$. Unfortunately, to cover the $a = 0$ case, we were forced to use Equation $(2)$ as well. Equation $(1)$ does not have $x$ at all, and Equation $(3)$ says nothing about $x$ when $a = 0$.
So we need Equation $(2)$. But can we drop Equations $(1)$ or $(3)$, so we are back to using only two equations? Yes:
\begin{align}
x &= ut + \frac{at^2}{2} \\
&= (v - at)t + \frac{at^2}{2} \\
&= vt - at^2 + \frac{at^2}{2} \\
&= vt - \frac{at^2}{2}
\end{align}
We used Equations $(1)$ and $(2)$, but not Equation $(3)$. Actually, this makes sense. Given a constant $a$ and explicit formulas for $x$ and $v$, we should be able to derive all other equations. In fact, we can derive Equation $(3)$ from Equations $(1)$ and $(2)$:
\begin{align}
v^2 - u^2 &= (u + at)^2 - u^2 \\
&= u^2 + 2u(at) + (at)^2 - u^2 \\
&= (2u + at)(at) \\
&= 2a \left( ut + \frac{at^2}{2} \right) \\
&= 2ax
\end{align}
So it would appear that Equation $(4)$ is not that useful – but only if we accept that Equation $(3)$ is not useful either.
But there is another way of looking at it. There are five standard quantities: $t$, $x$, $u$, $v$ and $a$. Equation $(1)$ includes every quantity except $x$. Equation $(2)$ includes every quantity except $v$. Equation $(3)$ includes every quantity except $t$. Equation $(4)$ includes every quantity except $u$. There should be an equation that includes every quantity except $a$:
\begin{align}
x &= ut + a\frac{t^2}{2} \\
&= ut + \frac{v - u}{t} \cdot \frac{t^2}{2} \\
&= ut + (v - u)\frac{t}{2} \\
&= ut + \frac{vt}{2} - \frac{ut}{2} \\
&= \frac{ut}{2} + \frac{vt}{2} \\
&= \frac{(u + v)}{2}t \tag{5}
\end{align}
(And there is that divide-by-zero trap again. But if $t = 0$, then $x = 0$, so Equation $(5)$ is still correct.)
When you look at it this way, you can see the all five equations could be useful. When you replace $x$ with $s$, these become the so-called “SUVAT” equations that some other users have mentioned. All of these are standard equations, as mentioned above.
A: There's nothing wrong with your calculation. But plug in $v=u+at$ and you find:
\begin{equation}
x=(u+at)t-\frac{at^2}{2} = ut + \frac{at^2}{2}
\end{equation}
which is already one of your three equations of motion. So your teacher is probably just saying that it is redundant to call this a "fourth" equation, because it is essentially the same as the second equation.
A: There is no problem with this equation for the time being that the equation is written in a book or an article or discussed amongst people. There is no problem with this equation to be used as a quick approach to a problem.
But the problem arises the moment we try to name it the fourth equation of motion. This is because this equation has two qualities that disqualifies it off the league.

*

*This equation gives no new information : everything we can obtain from this equation is what we already know from the previous ones.


*This equation has almost no practical usage : when you have an equation in $x, v$ and $t$ we call $t$ as a known quantity. We know at what time we want to know the position and speed of the body. The $x$ and $v$ part are unknown to us. If you know the velocity of the body at exactly one time instance, you start your clock accordingly, making it the "initial velocity" $u$. But a linear equation in three variables fails to serve any good cause.
A: One thing that hasn't been raised is that your derivation has a mathematical problem in it.
In the step:
$$ (v+u)(v-u) = 2 \frac{(v-u)}{t}x $$
$$ (v+u)t = 2x $$
you have divided each side by
$$ (v-u) $$
The problem with this step is that this could be the $0$, unless you explicitly require
$$ v \neq u $$
Dividing by $0$ is not a valid operation. It can be the source of many a logically flawed derivation and it's good to keep an eye out for them.
A: When I learned the kinematic equations in high school, this equation was actually presented to us. There is however a reason it's seldom seen.
The main "issue" is that your equation, in order to be valid for all $t$, would be better expressed as $$x(t)=v(t)\cdot t-\dfrac12 at^2$$
This is because when $t$ changes, $v$ changes as well. And if you substitute in $v=at+v_0$, you end up with the regular $x=\dfrac12 at^2+v_0t$ equation.
The use of equations like $x=\dfrac12 at^2+v_0t$ is that $v_0$ is fixed. You can substitute any value of $t$, and the equation will hold.
In the equation you derived, that is not the case. You would have to ensure the value of $v$ is for the correct time $t$. This would confuse students, which is why the equation is not often taught. But, in terms of its derivation, everything is fine.
Also, a teacher should not be dismissing reasonable equation derivations, especially when a student is putting in lots of effort to master the material.
A: Another way to show you are right -:
$x=ut+\frac{1}{2}at^2 = (v-at) t +\frac{1}{2} at^2 = vt -\frac{1}{2} at^2$
A: It is definitely an equation of motion. I think what she meant is that this is no conventional called the 3 equations of motion. Although it seems very strange that can happen. Some people prefer to stick to conventions for simplicity as well as assurance that it is right.
A: Just square $v = u+at$, yields
\begin{eqnarray}
v^2 & = & (u+at)^2\\
& = & u^2 + 2uat+a^2t^2\\
& = & u^2 + 2a(ut+\frac{1}{2}at^2)\\
\Longrightarrow v^2 & = & u^2 + 2as
\end{eqnarray}
Since we have used $s = ut+\frac{1}{2}at^2$ in the derivation, it is not independent equation of motion.
A: Definitely it could be handy for example when we know 3 and wish to know the other from the set x=s,v,a,t.
The simplest and fastest way to derive that equation would seem to be simply consider Time Reversal Symmetry on
x= ut + (1/2) a t^2
and then the “new” equation is kind of obvious with the sign reversals if you take due care.
Many dynamics problems solve very quickly using time reversal though a lot of teachers don’t appreciate it as they want you to practice the hard way. For example if you throw a ball and it goes 10 metres up then how long did it take? Just imagine the fall 10 metres and use x= g(t^2)/2 and time reversal for fast answer. The proper method using an unknown initial u etc is just silly and lacks insight and so is basically dumb and more likely to get a confusion or slip and wrong answer.
Note also there are lots of other useful equations like x=(u+v)t/2 so there will be dispute about which is number 4 or whatever.
