# Is gravity's acceleration rate - squares of all natural numbers? [duplicate]

I've seen some science\history documentary in which they made a replica of Galileo's inclined plane experiment.

They rolled a ball down the plane; and it's progressed in length units each unit of time was - 1, 4, 9, 16 etc. I realized it was squares of natural numbers; but I didn't really understand why this happens. I expected it to be exponential; as in - 1, 2, 4, 8, 16 etc.

I've checked a few more videos since; and now I understand that this progression is "time squared". That makes sense; but I still don't exactly understand why it works this way.

Also in some other video, which is a segment from some science show, they said the ball suppose to roll - 1, 3, 5, 7 etc. Did they just got it wrong?

So why does it work this way - time squared; and what other basic things in physics progress this way; can you give some examples? What about energy needed to accelerate mass?

## marked as duplicate by John Rennie, Kyle Kanos, ACuriousMind♦, Floris, MartinJul 8 '15 at 8:51

• There is no requirement for integer (natural) squares, the video was doing that for the general public who would understand $x\propto t^2$ better with the more commonly known squares: 1, 2, 4, 9, etc. – Kyle Kanos Jul 7 '15 at 14:43
• – John Rennie Jul 7 '15 at 14:44
• @John Rennie - My question is different from the other question. I ask - why is it squares of every natural number and not exponential. The other question is - "Why is it proportional to the square of the time and not just time?" – MrAccident Jul 7 '15 at 15:34
• @Kyle Kanos - Maybe; but my question is - why does it work this way; and specifically - not exponential. – MrAccident Jul 7 '15 at 15:35
• The natural numbers are the times i.e. 1 second, 2 seconds, 3 seconds etc, and since distance is propeortional to $t^2$ the distances are $1^2$, $2^2$, $3^2$, and so on. The question I've linked explains why $s \propto t^2$ rather than $s \propto e^t$. That's why it's a duplicate. – John Rennie Jul 7 '15 at 15:37

$$v = \int \frac{F}{m} dt = a\; t\\ x = \int v\; dt = \int a\; t\; dt = \frac12 a t^2$$