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Posted
45 minutes ago, ImplicitDemands said:

Acceleration is velocity over time. Velocity is distance over time. So, acceleration is distance over time^2

Those are the units, but not the equation. Is there a point here?

Posted (edited)
5 hours ago, swansont said:

Those are the units, but not the equation. Is there a point here?

It is the equation. a(t) = V(t)/s(t)^2 or miles/hours^2 aka mph^2

Position=S(t)->(distance over time)

Velocity=v(t)->s(t)/time

Acceleration=a(t)->v(t)/S(t)->distance/time^2

11 hours ago, geordief said:

any  change in the perceived passage of time

Time squared, yes it is covering the same distance over more time. Hence the outside world is experiencing a greater passage of time. Whether or not the observer is experiencing more time as well is a matter of debate. We really can't go fast enough to tell. 

Edited by ImplicitDemands
Posted
8 hours ago, ImplicitDemands said:

It is the equation. a(t) = V(t)/s(t)^2 or miles/hours^2 aka mph^2

Position=S(t)->(distance over time)

Velocity=v(t)->s(t)/time

Acceleration=a(t)->v(t)/S(t)->distance/time^2

 

a = dv/dt  

It will be v/t only if acceleration is constant and you start from rest.

a = 2d/t^2 for constant acceleration, starting from rest. Doing the proper math gives you the factor of 2, which doesn’t come from unit analysis 

This would seem to be irrelevant to the twin paradox discussion.

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