Timeline for Finding an elegant proof of rectilinear motion
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 22, 2022 at 17:44 | comment | added | Christophe AGUETTAZ | Amazing stuff, I wish I had the reputation to upvote your answer. | |
| Oct 22, 2022 at 17:30 | comment | added | Tomek Czajka | @ChristopheAGUETTAZ I added a derivation that it is true for $r \ne 0$. | |
| Oct 22, 2022 at 17:29 | history | edited | Tomek Czajka | CC BY-SA 4.0 |
added 909 characters in body
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| Oct 22, 2022 at 17:06 | vote | accept | Christophe AGUETTAZ | ||
| Oct 22, 2022 at 17:06 | comment | added | Christophe AGUETTAZ | Ok so just to rephrase the insight of your answer - the crux is that when c=0, those cross product constraints are easily satisfied by coming to rest at point 0, and canceling the acceleration as well. At which point you're free to switch directions without violating the constraints. The flaw in my "proof" is that it breaks down when r(t) = 0 since anything is "collinear" with 0. Setting c ≠ 0 prevents such direction-changing, passing-through-0 shenanigans. Nice, thanks a lot! | |
| Oct 22, 2022 at 16:56 | comment | added | Christophe AGUETTAZ | Oh yes sorry my bad. | |
| Oct 22, 2022 at 16:55 | comment | added | Tomek Czajka | The function is differentiable twice everywhere. $r'(1) = r'(2) = 0$ and $r''(1) = r''(2) = 0$. | |
| Oct 22, 2022 at 16:53 | comment | added | Christophe AGUETTAZ | Nice answer, thanks! But is it correct? You're creating a function that's not differentiable at points 1 and 2, which means r(t)×r′(t)=0 and r(t)×r′′(t)=0 is not true at these points. Since the initial assumptions don't hold anymore for the whole function, the conclusion doesn't either. | |
| Oct 22, 2022 at 16:36 | history | edited | Tomek Czajka | CC BY-SA 4.0 |
added 38 characters in body
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| S Oct 22, 2022 at 16:31 | review | First answers | |||
| Oct 22, 2022 at 19:19 | |||||
| S Oct 22, 2022 at 16:31 | history | answered | Tomek Czajka | CC BY-SA 4.0 |