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I wanted to find the relationship of the speed of a falling object in a vacuum at g=9.8 m/s2 compared to its distance. I wanted to get time out of the equation. When I plotted the relationship it looked like it was a square root function.

Then it was easy to find that speed = sqrt(distance x acceleration).

However I'm not sure how to derive that from initial premises like speed = distance x time.

enter image description here

I"ve been interested in trying to imagine a universe where all changes in states are explained by relative motion, and no longer by time which serves as a proxy.

Sorry if this is hyperbasic, sometimes I'm bad at math.

NOTE: The first graph here is incorrect, it overestimates the distance traveled relative to speed. It is a funny (rookie?) mistake, I multiplied velocity by time elapsed at every time step to find the distance traveled. However, one needs to take the integral of velocity with respect to time at every timestep, since the velocity is constantly changing. The correct graph is named distance 2 vs speed.

enter image description here

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  • $\begingroup$ "I"ve been interested in trying to imagine a universe where all changes in states are explained by relative motion, and no longer by time which serves as a proxy" - time serves as a proxy for what? $\endgroup$
    – probably_someone
    Commented Feb 11, 2020 at 22:18
  • $\begingroup$ The title of the question provides a wrong equation. In fact the correct equation is, speed = sqrt(2 x acceleration x distance). (for uniform acceleration) The mistake was due to to my error in calculating distance in the first try. $\endgroup$
    – dlight
    Commented Feb 12, 2020 at 8:10
  • $\begingroup$ Yeah, ok, but that doesn't answer my question. You say time "serves as a proxy"; what does time serve as a proxy for? $\endgroup$
    – probably_someone
    Commented Feb 12, 2020 at 14:53
  • $\begingroup$ @probably_someone Cells emerged and molecules have danced to the tune of life on Earth for 12 galactic rotations. Seems time, as we commonly talk about it, is a convenient standardized way to chop up these relative motions. But I'm not sure we need this "4th dimension" When something happens "after" an initial state, its because it has moved along its trajectory. (even supposedly fixed objects). in a universe where nothing moves, the passage of time becomes impossible to measure, and thus may not exist. (all opinions and not fully fleshed out ideas) $\endgroup$
    – dlight
    Commented Feb 12, 2020 at 17:42
  • $\begingroup$ When comparing two states, how do you tell whether something has moved "along" its trajectory? You look at where the object is "at" on its trajectory in each state, and check if it's "at" the same place. In other words, you need to have a notion of where a particle is "at" on its trajectory. In more technical terms, you have to parametrize the trajectory. And any reasonable parametrization you choose is basically going to be just time by another name (maybe reversed and distorted, but still the same concept). Even more so if you want the Second Law of Thermodynamics to still hold. $\endgroup$
    – probably_someone
    Commented Feb 12, 2020 at 19:21

2 Answers 2

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This follows immediately from the equations of uniform acceleration.

It can be shown that if u, v, a, and s are the initial velocity, final velocity, acceleration and displacement respectively, then the following equation holds:

$$v^2 = u^2 +2as$$

Since $u=0$ it is clear that $v=\sqrt{2as}$.

The equation itself is easy to derive. The definition of acceleration is the rate if change of velocity with respect to time, so for constant acceleration $a=\frac{v-u}{t}$, or equivalently $v=u+at$. By definition, $v=\frac{ds}{dt}$ and both u and a are constants, so integrating gives $s=ut + \frac{1}{2}at^2$, which can be combined with $v=u+at$ to give $v^2=u^2+2as$ with a bit of algebra I'll leave to the reader.

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  • $\begingroup$ Great. integrating the velocity with respect to time in order to find the displacement was a key trick there. thanks $\endgroup$
    – dlight
    Commented Feb 11, 2020 at 23:28
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    $\begingroup$ This is strictly only valid for uniform acceleration. $\endgroup$
    – my2cts
    Commented Feb 11, 2020 at 23:47
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Assume, for simplicity's sake, at $t=0$, $x=0$ and $v=0$, then at $t$: $$v=at \Rightarrow t=\frac{v}{a}$$ and: $$x=\frac12 at^2$$ $$x=\frac12 a \Big(\frac{v}{a}\Big)^2 \Rightarrow x=\frac{1}{2a}v^2$$ $$v=\sqrt{2 a x}$$

A slightly alternative derivation, based on an object falling in the Earth's central gravitational field.

Again, assume $t=0$, $x=0$ and $v=0$.

The object falls a distance $x$ (without drag) and potential energy is converted to kinetic energy, so that:

$$mgx=\frac12mv^2$$ $$\Rightarrow v=\sqrt{2 g x}$$

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