Timeline for Magnitude of Acceleration Vector when Speed is Constant
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 5 at 12:22 | comment | added | Albertus Magnus | No problem, glad I could help! | |
| Mar 5 at 9:32 | comment | added | Sylvia | thanks a lot for your answers! | |
| Feb 28 at 22:53 | comment | added | Albertus Magnus | @Sylvia In the limit that $\Delta t$ is small, then the acceleration vector is perpendicular to both of the vectors. | |
| Feb 28 at 18:58 | comment | added | Sylvia | 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? | |
| Feb 28 at 16:25 | comment | added | Albertus Magnus | 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. | |
| Feb 28 at 16:14 | comment | added | Sylvia | 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) | |
| Feb 28 at 15:52 | vote | accept | Sylvia | ||
| Feb 28 at 15:27 | history | edited | Albertus Magnus | CC BY-SA 4.0 |
edited body
|
| Feb 28 at 15:00 | history | answered | Albertus Magnus | CC BY-SA 4.0 |