inverse speed

[michel123456]

And what would happen with acceleration?

 

If V can be written as t/d {reversed to the common way) and Acceleration as a rate of V/t, then from A=V/t you would obtain an acceleration rate described by (seconds/meters)/seconds The sec. cancelling each other you get units of 1/meters for acceleration.

 

Or once V is described as t/d (reversed) then you MUST describe acceleration as a rate of A=t/V (also reversed) and why would that be an obligation?

[Janus]

And what would happen with acceleration?

 

If V can be written as t/d {reversed to the common way) and Acceleration as a rate of V/t, then from A=V/t you would obtain an acceleration rate described by (seconds/meters)/seconds The sec. cancelling each other you get units of 1/meters for acceleration.

 

Or once V is described as t/d (reversed) then you MUST describe acceleration as a rate of A=t/V (also reversed) and why would that be an obligation?

If inverse speed is t/d, then the equivalent concept to acceleration would t/d/d or t/d^2 in the MKS system this is s/m^2 or the rate at which the time it takes to travel one meter changes per meter traveled.

 

If we call call inverse speed "rapidness"® then q = t/d. And if we call the equivalent to acceleration "quickness"(q), then we can say that r = t/d and q=r/d

Thus r= qd, and the anti-derivative of r with respect to d is qd^2/2, or qd^2/2=t which gives us the time to travel a given distance with at a constant quickness.

[michel123456]

If inverse speed is t/d, then the equivalent concept to acceleration would t/d/d or t/d^2 in the MKS system this is s/m^2 or the rate at which the time it takes to travel one meter changes per meter traveled.

 

If we call call inverse speed "rapidness"® then q = t/d. And if we call the equivalent to acceleration "quickness"(q), then we can say that r = t/d and q=r/d

Thus r= qd, and the anti-derivative of r with respect to d is qd^2/2, or qd^2/2=t which gives us the time to travel a given distance with at a constant quickness.

Typo after rapidness. You ment r=t/d

_--------------------------

And i suspect you wrote all the above because youn knew what would be consistent answer. But truly, what is the logic behind?

Why t/d^2 and not (t/d)/t?

(edit)

Why

the rate at which the time it takes to travel one meter changes per meter traveled.

 

And not

the rate at which the time it takes to travel one meter changes per second traveled.

??

Edited April 18, 2015 by michel123456

[Endy0816]

Provided you remain consistent, everything is good.

 

s²/m = s * s/m

 

You can even go further and do:

 

(s²/m) / kg = N^-1

[michel123456]

How do you know when you remain consistent and when not?

 

here we have very simple things: speed, acceleration, time, distance.

When do we know what is the correct way? And why some other way is "inconsistent". What is the law that we have to follow?

Imagine the same things for processes that remain purely theoretical?

Provided you remain consistent, everything is good.

 

s²/m = s * s/m

 

You can even go further and do:

 

(s²/m) / kg = N^-1

Now i am really confused

Your s^2/m is different from the s/m^2 from Janus

[Endy0816]

How do you know when you remain consistent and when not?

 

For consistency, everything on one side of the equals sign needs to equal everything on the other side of the equals sign.

 

That is the only hard rule though.

 

Now i am really confused

Your s^2/m is different from the s/m^2 from Janus

 

they are describing different things. The first is 1/acceleration.