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Let acceleration = -a. I'm was originally trying to find the stopping distance in terms of $v_0$ and $a$

Two relevant equations of linear motion with constant acceleration then become $x = \frac12(-a)t^2 + v_0t$

$0 = v_0 - at$

Now if i substitute in the first equation $t = \frac{v_0}a$, i get

$x = -\frac12a(\frac{v_0}a)^2 + \frac{v_0^2}a$

$x = \frac12\frac{v_0^2}a$

Or, x is inversely proportional to a

Yet if i substitute $v_0 = at$ instead

$x = -\frac12at^2 + (at)t$

$x = \frac12at^2$

Or, x is directly proportional to a

Obviously it can't be both, but i can't seem to put my finger on the error.

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    $\begingroup$ In the expression $x=ay^2/t$, $t$ (the stopping time) is implicitly a function of $a$, meaning that you cannot conclude from that expression that the stopping distance $x$ is proportional to $a$. The first expression is the correct one to use, because $v_0$ is a parameter in your problem, i.e. it is fixed by the initial conditions. That said, both expressions are correct; it just depends on what you're looking at that makes one or the other of the expressions the correct one to look at. $\endgroup$
    – march
    Commented Mar 8, 2016 at 4:52

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Both equations are correct.

What you've hidden in writing the second equation is that $t$, the time it takes to stop, is inversely proportional to $a$. You said it yourself: $t=\frac{v_0}{a}$. So if you double $a$, $t$ will halve, $t^2$ will quarter, and thus $x$ will halve as expected.

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If you wanted to find the stopping distance in terms of $v_0$ and $a$ then you can use this equation :

$v^2 - u_0^2 = 2ax$

Now from your question $v=0$ (as the object finally comes to rest)

so, $-u_0^2 = -2ax$ (acceleration = $-a$)

then by rearranging we get:

$\color{green}{ x = \frac{u_0^2}{2a}}$

PS: This shows that the first method by the poster is more correct, by using another equation or method. But both the methods can lead to suitable answers.

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  • $\begingroup$ I don't think this addresses what the poster was asking. $\endgroup$
    – Jahan Claes
    Commented Mar 8, 2016 at 4:54
  • $\begingroup$ This addresses the correct answer by another method or equation. $\endgroup$
    – hxri
    Commented Mar 8, 2016 at 4:56
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    $\begingroup$ That's fair. I think it's worth noting that both of the poster's methods are correct, and lead to correct answers. $\endgroup$
    – Jahan Claes
    Commented Mar 8, 2016 at 5:03

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