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The 5 equations of motion have been chosen such that from the 5 variables of motion: $s$, $u$, $v$, $a$ and $t$; each equation, exclusively omits one of these. This allows us to only ever require the values of THREE variables to work out the FOURTH.

$v = u + at$    -missing $s$     - (1)

$s = ut + \frac12 at^2$   -missing $v$     - (2)

$s=vt - \frac12 at^2$   -missing $u$     - (3)

$s=\frac{(u+v)}{2}t$     -missing $a$    - (4)

$a= \frac{v^2 - u^2}{2s}$     -missing $t$     - (5)

However, I have noticed by looking at these equations, if they are indeed independent equations, then it should be possible to only require TWO variables to deduce all the rest! Yes, I know that physically, this is impossible, BUT mathematically, this appears to be permissible.

So, if we have values for any two random variables (from the five), then there will always be THREE equations where BOTH these variables exist. This means that for this group of THREE equations, there will be THREE unknown quantities but with THREE equations, which means we should be able to solve for those three unknowns. But clearly this is impossible physically. Therein lies the paradox!

For example: If we know $a$ and $s$; let's say $a=6m/s^2$ and $s=18m$, then why can we not use equations (2), (3) and (5) above to solve for the unknown quantities, $u$, $v$ and $t$?

Of course, choose any two random variables to assign values to, and you should be able to solve for the remainder.

Why does the maths contradict physics? It may be that two of the five equations are derived from the other three? I don't know. I suspect these equations are not independent and can be reduced to fewer.

I have played around and discovered some pairings do result in easy results. Pairing $u$ and $v$ (by which I mean, assigning values to $u$ and $v$) leads us to using equations (1), (4) and (5) and doing so leads to a $0 = 0$ scenario where the maths correctly prevents us from being able to calculate values for the others. BUT, with other parings ($a$ and $s$ for example) I get equations (quadratics mixed with square roots making it hard to solve).

Can anyone shed any light on this apparent contradiction?

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    $\begingroup$ What is $u$ and what is $v$? These are not "the" equations of motions. The are just the equations of motion for a particular problem. It seems to me that this is just a constant accelerating body, with $u$ the initial velocity? $\endgroup$
    – Bernhard
    Commented Dec 1, 2021 at 20:08
  • $\begingroup$ These equations are taught in schools around the world (certainly in the UK) as THE equations of motion -to be used, as you quite rightly observed, for bodies only under constant acceleration. u is the initial velocity, v=final velocity. Yes, I am aware there are perhaps more general versions involving calculus, but still, for this particular set of equations, there seems to be a "paradox". $\endgroup$
    – IqbalHamid
    Commented Dec 1, 2021 at 20:16
  • $\begingroup$ "I know that physically, this is impossible, BUT mathematically, this appears to be permissible" - when you have found the impossible to be true, check your assumptions. The equations are not independent. $\endgroup$
    – Nuclear Hoagie
    Commented Dec 1, 2021 at 20:18

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There is no paradox, because your 5 equations are not independent.

  • Fill in (1) in (3) to obtain (2)
  • (2)+(3) = 2*(4)
  • (5) is left as an exercise to the reader.

You also see this in your example. Combining three equations inevitably leads to one equation becoming $0=0$. Proving that the three equations are not independent.

So, you have only 2 equations. One following from $v=x'$ and one from $a=v'$. And these are the two that you should remember.

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  • $\begingroup$ So are you saying there are actually only TWO independent equations? Wow, I have learnt something there. I suspected that at least three of them were independent, but you are suggesting that the two independent ones are actually (1) and (5)? Is that right? $\endgroup$
    – IqbalHamid
    Commented Dec 1, 2021 at 20:21
  • $\begingroup$ It's true - pick any two equations and you can solve any kinematics problem (with constant acceleration). You may have to do it in two steps though, e.g. solving for $v$ first and then $s$. $\endgroup$
    – M. Enns
    Commented Dec 1, 2021 at 20:36
  • $\begingroup$ @IqbalHamid You can choose whichever two equations you want and get the other three. Personally, I would chose (1) and (2), since these are the ones that come directly from integrating the equation of motion. $\endgroup$
    – d_b
    Commented Dec 1, 2021 at 20:38
  • $\begingroup$ @IqbalHamid It doesn't really matter which to you peak. I would just stick to the first two (or better, the two that I propose, so you can derive everything yourself). $\endgroup$
    – Bernhard
    Commented Dec 1, 2021 at 20:38

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