Question

If instead of defining velocity as \(v(t) = x'(t)\) we had some other property of motion \(v(x) = t'(x)\) how would that alter things like acceleration and force?

Answer 1

This property of motion would be in units of \(seconds/meter\)

Answer 2

Maybe this would change the paradigm entirely. Just how arbitrary is it that we think about speed as meters/seconds, versus some other property of matter that would use units of seconds/meter.

Answer 3

\[
t = t(x(t))\implies 1 = \frac{dt}{dx}\frac{dx}{dt}\implies t'(x) = \frac{1}{x'(t)}
\]

Answer 4

So\[
\frac{dx}{dt}= \frac{1}{\frac{dt}{dx}}
\]Then \[
x''(t)=\frac{d^2x}{(dt)^2} = \frac{d}{dt}\left( \frac{1}{\frac{dt}{dx}}\right)
\]

Answer 5

\[
\frac{d}{dt}\left(\frac{1}{\frac{dt}{dx}}\right) = -\frac{\frac{d}{dt}\frac{dt}{dx}}{\frac{dt}{dx}\frac{dt}{dx}}
\]

Answer 6

What the heck would \[
\frac{d}{dt}\frac{dt}{dx}
\]be?

Answer 7

\[
\frac{dt}{dx} = t'(x) = t'(x(t))
\]

Answer 8

\[
\frac{d}{dt}\frac{dt}{dx} = \frac{d^2t}{(dx)^2}\frac{dx}{dt}
\]

Answer 9

why seconds/meter?

Answer 10

Why meter/second?

Answer 11

at which something is accelerating or moving

Answer 12

you said, Just how arbitrary is it that we think about speed as meters/seconds, versus some other property of matter that would use units of seconds/meter.

Answer 13

Well, seconds per meter is a more concrete than say \(f(x,t)\), which would be too abstract to example well.

Answer 14

examine

Answer 15

t(x) = x^3 + x

what nonsense... time changes with respect to x ?

Answer 16

laughing out loud doggy

Answer 17

lol how to think it ?

Answer 18

I didn't say it was intuitive.

Answer 19

Are you saying that time doesn't change?

Answer 20

Because if time didn't change, then that would nothing could move.

Answer 21

At the very least we can establish a relationship between time and distance.

Answer 22

so you're redefining position function as well?

Answer 23

\(x\) is position. \(t\) is time.

Answer 24

\(x(t)\) is position as a function of time. \(t(x)=x^{-1}(t)\), since \(t(x)\) is just time as a function of position.

Answer 25

how long was your physics class today? 20 meters

Answer 26

Well, instead of doing a function inverse, you could also consider...\[
\overline{x}(t)=\frac{1}{x(t)}
\]Maybe that would be another way to look at it.

Answer 27

This is a bit less complicated. \[
\overline{v}(t)=\frac{d}{dt}\overline{x}(t)=\frac{d}{dt}\frac{1}{x(t)}=-\frac{x'(t)}{[x(t)]^2}
\]

Answer 28

distance represents physical dimension, while time does not?
20-minute walk away
50-week ant sprint away
5-minute car drive

is this what you're talking about ?

Answer 29

The dimensional analysis would mean that \(\overline{v}(t)\) has units \[
\frac{m/s}{m^2} = \frac{1}{ms}
\]

Answer 30

The question here is...

Suppose instead of thinking about the property velocity \(m/s\) we initially looked at motion differently, specifically as \(s/m\) or \(\Delta time / \Delta distance\). How would this affect other areas of physics?

Answer 31

so ... HOW FAR IS A second?

Answer 32

Would there be a corresponding property to this property of motion similar how acceleration is to speed

Answer 33

We are talking about a moving (but possibly at rest) object. There is a relationship between how it's time changes and it's motion changes.

Answer 34

how its time changes and how its position changes

Answer 35

time can change independent of distance change
distance can remain zero and time still continues to tick tock

Answer 36

dude stop trolling us laughing out loud

Answer 37

I don't see how that matters here.

Answer 38

Motion requires that time or distance change regardless.

Answer 39

Quantum physics has a lot of intuitive stuff, even things being in multiple places at the same time.

Answer 40

unintuitive stuff^

Answer 41

So from a quantum physics point of view, saying that at any given time something has to have one position is somewhat absurd as well.