Timeline for If I apply a constant force to an object until I've reversed its starting velocity, does its final position remain unchanged?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 30, 2020 at 9:48 | history | edited | Qmechanic♦ |
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| Apr 30, 2020 at 9:09 | answer | added | Max | timeline score: 1 | |
| Apr 30, 2020 at 8:50 | comment | added | Marius Ladegård Meyer | Your quantity $s(t)$ is the coordinate of the object at time $t$, i.e. the distance from the origin of the implicit reference frame. It should not be confused with the distance travelled. | |
| Apr 30, 2020 at 8:41 | comment | added | Max | @bemjanim Then $s(t) = \frac{v_0^2 - v(t)^2}{2 F/m}$, and at $t = 2t'$, $v(t) = -v_0$ and $s(2t')=0$, as above. | |
| Apr 30, 2020 at 8:12 | answer | added | Anton Baranikov | timeline score: 3 | |
| Apr 30, 2020 at 8:11 | comment | added | bemjanim | Why not use $v^2=v_0^2+2as$ instead? | |
| Apr 30, 2020 at 8:09 | comment | added | Max | I could argue for either, but the textbook answer looks to me like the equivalent of 50m! | |
| Apr 30, 2020 at 8:03 | comment | added | Dr Chuck | I think it comes down to this. If you walk 100m in the x direction and then walk directly back to where you started from, is the distance yoou have travelled 200m or 0m? | |
| Apr 30, 2020 at 7:52 | history | edited | Max | CC BY-SA 4.0 |
added 365 characters in body clarifying variable t vs parameter t'
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| Apr 30, 2020 at 7:48 | review | First posts | |||
| Apr 30, 2020 at 8:15 | |||||
| Apr 30, 2020 at 7:38 | history | asked | Max | CC BY-SA 4.0 |