# What does 'seconds squared per meter' mean?

I was wondering about the meaning of seconds^2 / meters (more specifically, period^2 / length - of a pendulum, after the square root graph of period vs length is linearized). I know that acceleration is meters / second^2, but if I have the units 'seconds^2 / meters', what would that mean? Would it be the 'inverse' of acceleration? And is there a term to describe time (in general) divided by distance? I know this sounds like weird questions but any help would be appreciated. Thanks.

• Seconds squared divided by meters is the inverse of acceleration. – Bob D Mar 17 at 18:14
• period vs time Did you mean to write “period vs length”? – G. Smith Mar 17 at 19:28
• There is a famous constant in physics with the units cubic meters per kilogram per square second. Trying to understand what these units of $G$ “mean” is pointless. They are whatever they need to be so that $F=GMm/r^2$ makes sense. You want to understand the equations and the meaning of the symbols in them. Then, if you understand physically the units of $M$, $m$, $r$, and $F$, you “understand” (mathematically) the units of $G$. – G. Smith Mar 17 at 19:44

What do the units of $$1/$$acceleration mean? We can understand it in terms of the units of acceleration. Write $$15~\frac{\textrm{m}}{\textrm{s}^2}$$ as $$\frac{15~\textrm{m}/\textrm{s}}{\textrm{s}},$$ or, in words, "15 (meter per second), per second". Which is to say, the velocity is changing by 15 m/s every second.
Thinking about this in a different way, we can ask the question, how long does it take for the velocity to change by 1 m/s? Assuming a constant acceleration, if it takes 1 second to change by 15 m/s, it should take 1/15th of a second to change by 1 m/s. In other words, 1/acceleration is $$\frac{\textrm{s}}{15~\textrm{m}/\textrm{s}} = \frac{1}{15}\frac{\textrm{s}}{\textrm{m}/\textrm{s}}= \frac{1}{15}\frac{\textrm{s}^2}{\textrm{m}},$$ which is exactly the way you write down that it takes 1/15th of a second for the speed to change by 1 m/s.