Timeline for Instantaneous Velocity at a Sharp Point
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 4, 2020 at 16:03 | history | edited | CommunityBot |
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| Oct 30, 2019 at 20:17 | vote | accept | j18w | ||
| Oct 30, 2019 at 20:15 | comment | added | jacob1729 | Interesting. I would have interpreted "at $t_1$ the person stops pushing the block" to mean $t_1$ is when the force becomes zero, i.e. a little to the right on your graph than where you have drawn it. That restores the 'textbook' solution to being correct. | |
| Oct 30, 2019 at 20:00 | comment | added | j18w | @SV You are right. The graph should continue the line of the parabola in x but as a straight line. My bad, the point was just to differentiate it from a parabolic increase. But you are correct. | |
| Oct 30, 2019 at 19:24 | history | edited | Qmechanic♦ |
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| Oct 30, 2019 at 19:22 | answer | added | Bob D | timeline score: 0 | |
| Oct 30, 2019 at 19:03 | comment | added | Bob D | @dmckee Why don't you post that as an answer rather than a comment? | |
| Oct 30, 2019 at 18:53 | answer | added | JMac | timeline score: 1 | |
| Oct 30, 2019 at 18:48 | comment | added | dmckee --- ex-moderator kitten | Instantaneous changes of velcoity like that have the same significance as frictionless surfaces; massless pulleys; semi-infinite, rigid rods; infinite square wells; and all the other idealizations of physics teaching. That is, they are good for teaching and good for some kinds of approximating but you must understand them as idealizations. Real system can only approximate instantaneous acceleartion, so the chenge is, in reality, smooth. | |
| Oct 30, 2019 at 18:35 | comment | added | user137661 | But also in a realistic case there are no frictionless surfaces and instantaneously applied forces. There is also air resistance. For that matter there are no constant forces either. Even if you release the force gradually the problem is still as idealised as it can get... For an ideal problem the ideal condition seems to be most appropriate. | |
| Oct 30, 2019 at 18:31 | comment | added | pyroscepter | You are correct, in the realistic case, there can't be a discontinuity in the acceleration. Hence, it must smoothly go down. So the velocity must be less than the final velocity. | |
| Oct 30, 2019 at 18:28 | comment | added | user137661 | Your position graph is wrong. The velocity gives you the slope of the position, when $v$ becomes constant then you just continue the line of the parabola in $x$ but now as a straight line (you stitch together a parabola with the line tangent to it). | |
| Oct 30, 2019 at 18:25 | review | First posts | |||
| Oct 30, 2019 at 20:54 | |||||
| Oct 30, 2019 at 18:20 | history | asked | j18w | CC BY-SA 4.0 |