Timeline for What conditions are required for the derivative of kinetic energy to be F.v?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2017 at 18:53 | vote | accept | Daniel Underwood | ||
| Aug 2, 2017 at 7:28 | comment | added | Farcher | @danielunderwood I have updated my answer to try and explain better your error. | |
| Aug 2, 2017 at 7:28 | history | edited | Farcher | CC BY-SA 3.0 |
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| Aug 2, 2017 at 2:45 | comment | added | Daniel Underwood | So from what you're saying, it looks like I messed up saying $\frac{d}{dt}v^2 = 2v\dot{v}$? I understand that the vector time derivative needs a time derivative of its basis vectors, but I don't seem to understand why this would be needed to handle the time derivative of $v$ since it's a scalar function. Does the magnitude of a vector not work like a scalar function? I feel like I'm missing something rather obvious, but I'm not sure what it is. | |
| Jul 31, 2017 at 10:49 | history | answered | Farcher | CC BY-SA 3.0 |