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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}

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – rob
    Mar 5, 2021 at 17:37
  • $\begingroup$ Again, this has nothing to do with Newton’s laws. Why was the tag newtonian-mechanics added? $\endgroup$ Mar 7, 2021 at 12:38
  • $\begingroup$ But my main aim was that the first 2 only are applicable for vectors, rest all are for single straight-line motion! $\endgroup$
    – Prof. Meow
    Mar 14, 2021 at 14:07

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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.

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    $\begingroup$ I've removed a number of comments that were getting too fighty for my taste. Be kind to each other, everybody. $\endgroup$
    – rob
    Mar 5, 2021 at 17:36
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Mar 7, 2021 at 12:47
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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.

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    $\begingroup$ But it is very useful when we are not given the initial velocity (and we need not find it too) and we are to find the position of the particle in the given time. $\endgroup$
    – Prof. Meow
    Mar 3, 2021 at 3:08
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    $\begingroup$ @Prof.Meow Sure, as I said there is nothing wrong with the equation. You could probably justify calling this the "second equation", and getting rid of your current 2nd eq. But keeping both would be as pointless as, say, putting a 100¢ tag on an item that's already been tagged as $1 in a store. $\endgroup$
    – hiccups
    Mar 3, 2021 at 3:19
  • $\begingroup$ Given any two of the equations of motion, you can derive three more. For example saying "distance travelled is average speed times time" $x=\frac{u+v}{2}t$ can be found from any two of the others. So once you have more than two equations, it is difficult to say there are three but not four or five such equations. $\endgroup$
    – Henry
    Mar 3, 2021 at 11:23
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    $\begingroup$ @Prof.Meow Maybe one other big picture comment to make, is that given even one equation, you can derive an infinite number of other, true equations. For example, if we know that $v^2-a^2=2ax$, we also know that $(v^2-a^2)^p=(2ax)^p$ for any $p$. So this is why we insist on having the minimum number of independent equations. In other words, we want to know the smallest set of equations, such that none of the equations in the set can be derived from the others in the set. Each one should provide a new piece of information. Of course, this is not the list of all useful equations. $\endgroup$
    – Andrew
    Mar 3, 2021 at 12:17
  • $\begingroup$ Come to that, the 3rd equation is also redundant, because you can get it by just squaring the 1st one, factoring out the $a$ and substituting for $x$ from the second one. $\endgroup$ Mar 3, 2021 at 13:07
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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.

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    $\begingroup$ Good reminder that there are physical events behind symbols +1 $\endgroup$
    – Winston
    Mar 5, 2021 at 6:43
  • $\begingroup$ There is some good physics reasoning in this post, and I want to upvote it, but I think it is technically not an answer. $\endgroup$ Mar 7, 2021 at 11:34
  • $\begingroup$ @BrianDrake the answer part is at the end where it says "You can then go on ... are also correct." $\endgroup$ Mar 7, 2021 at 12:49
  • $\begingroup$ But you did not actually say that the derivation given in the question is correct, so I still do not see how this is an answer. $\endgroup$ Mar 7, 2021 at 12:54
  • $\begingroup$ @BrianDrake no need to repeat what is already correctly said in other contributions. I am not trying to replace but augment. $\endgroup$ Mar 7, 2021 at 16:02
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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.

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    $\begingroup$ “displacement is directly related to the average speed” Displacement is always related to average velocity, by definition, even if acceleration is not constant. Average speed is a different matter. The next sentence also says “speed”. But the last sentence in that paragraph says “velocity”. Be careful. $\endgroup$ Mar 7, 2021 at 11:30
  • $\begingroup$ I was tempted to say that this is technically not an answer, until I got to the last sentence. The last sentence is an answer, so we do not need to worry about that now. That is good, because aside from the not-an-answer issue, this is one of the best posts here. Perhaps you could rewrite the answer to more clearly address the question? $\endgroup$ Mar 7, 2021 at 11:32
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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.

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  • $\begingroup$ I have massively expanded this post to cover the second part of the question as well. (I did not have time to do this when I first posted.) $\endgroup$ Mar 5, 2021 at 12:44
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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.

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    $\begingroup$ To be fair, $v^2 - u^2 = 2 a x$ also does not give any new information, as it is derived from the other two by straightforward algebra. So what's the difference between the "official" 3 equations and this 4th one? It's just another derived equation, which can be just as useful as the 2nd one if you know the final velocity and not the initial. $\endgroup$ Mar 3, 2021 at 13:05
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    $\begingroup$ I totally agree with your first point that this is not a new equation but I totally disagree that it is useless. It is quite common to have problems where (say) the initial velocity and acceleration are known, as well as the distance travelled, and what we want to know is the final velocity, without needing to know the time it takes to get from the initial to final state. Then $v^2-u^2=2ax$ is a very useful equation. Similarly there are common problems where starting from$x=vt - \frac{1}{2}at^2$ gives the most efficient solution. $\endgroup$
    – Andrew
    Mar 3, 2021 at 13:28
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    $\begingroup$ Also, when you progress from kinematics to dynamics, The equivalent of $v^2 - u^2 = 2ax$ is a very important equation, which expresses the idea of conservation of energy. $\endgroup$
    – alephzero
    Mar 3, 2021 at 13:43
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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.

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    $\begingroup$ Note that OP's equation is in fact still true for $v=u$ (which requires $a=0$ since we are considering constant acceleration). So it's just a matter of re-ordering the algebra to avoid this and not really harmful. $\endgroup$
    – jacob1729
    Mar 3, 2021 at 23:51
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    $\begingroup$ True, but it's a good thing to consider since this is a common way for physics students to mess up proofs and derivations $\endgroup$
    – Dancrumb
    Mar 4, 2021 at 15:18
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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.

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  • $\begingroup$ we don’t normally consider x and u and v as time dependent when we do suvat $\endgroup$
    – blanci
    Mar 5, 2021 at 20:26
  • $\begingroup$ For OP's equation to hold for all $t$, one must hold $v$ as time dependent/displacement-dependent. That's why this equation is not usually taught. $\endgroup$
    – user256872
    Mar 5, 2021 at 21:20
  • $\begingroup$ When using the SUVAT equations, we often treat $x$, $u$ and $v$ as constants – but then we also treat $t$ as a constant. I am not sure what the problem is here. $\endgroup$ Mar 7, 2021 at 11:24
  • $\begingroup$ Go plot $x=\dfrac12 at^2+v_0t$ and $x=vt-\dfrac12at^2$ with $v$ as a constant. They're not the same functions. $\endgroup$
    – user256872
    Mar 7, 2021 at 19:00
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Another way to show you are right -:

$x=ut+\frac{1}{2}at^2 = (v-at) t +\frac{1}{2} at^2 = vt -\frac{1}{2} at^2$

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    $\begingroup$ Oh then this is a new proof of the equation. Means it can be an equation of motion! $\endgroup$
    – Prof. Meow
    Mar 5, 2021 at 6:47
  • $\begingroup$ @Prof.Meow I would say they are equivalent $\endgroup$
    – Shashaank
    Mar 5, 2021 at 7:54
  • $\begingroup$ @Prof.Meow They are not equivalent. Have a look at my expanded answer, discussing which equations are needed to derive others. $\endgroup$ Mar 5, 2021 at 12:48
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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.

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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.

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  • $\begingroup$ I understand your equations, but not your words. Which equation is not independent? Not independent of what? How does this relate to OP’s questions? $\endgroup$ Mar 5, 2021 at 12:55
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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.

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  • $\begingroup$ -1 “a lot of teachers don’t appreciate it as they want you to practice the hard way” When you say the “the hard way”, did you mean “the more rigorous way”? That is what it is, unless you can prove that your symmetry method works. If you really think that the “proper method” is “silly”, you are welcome to edit your answer to add details of alternative methods. $\endgroup$ Mar 7, 2021 at 11:18
  • $\begingroup$ Anyway, it sounds like your way is also hard. “the ‘new’ equation is kind of obvious with the sign reversals if you take due care.” (emphasis added) $\endgroup$ Mar 7, 2021 at 11:20
  • $\begingroup$ Brian -1 ha ha! I am adding information even if you disagree with my off the cuff expletives. Students should intuitively know a rock thrown up takes the same time to go up as to come back down and the parabolic curve is symmetric. The whole point is being pedantic will ruin the enjoyment. $\endgroup$
    – blanci
    Mar 29, 2021 at 20:14

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