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Physics

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  • $\begingroup$ thanks a lot! That is (?): Acceleration is defined using inifinitesimally small intervals and a vector that is perpendicular to the instantaneous velocity vector. The confusion about the existence of an acceleration vector having a length when speed is constant disappears if you consider the definition of acceleration that refers to infinitely small intervals (with infinitesimally small vector lengths) $\endgroup$
    – Sylvia
    Commented Feb 28 at 16:14
  • $\begingroup$ The acceleration vector is not, in general, dependent on the magnitude of velocity. Geometrically, one can think of it as related to the curvature of the path on which a particle moves; even if the magnitude of velocity is small, a sharp turn indicates a large acceleration. See, for example, discussions on the "TNB" frame. $\endgroup$
    – Albertus Magnus
    Commented Feb 28 at 16:25
  • $\begingroup$ ok. @Albertus Magnus, I don't understand whether the acceleration vector is perpendicular to the v1 (v at t1), or to v2 (v at t2)... How can you tell? $\endgroup$
    – Sylvia
    Commented Feb 28 at 18:58
  • $\begingroup$ @Sylvia In the limit that $\Delta t$ is small, then the acceleration vector is perpendicular to both of the vectors. $\endgroup$
    – Albertus Magnus
    Commented Feb 28 at 22:53
  • 1
    $\begingroup$ thanks a lot for your answers! $\endgroup$
    – Sylvia
    Commented Mar 5 at 9:32