Timeline for The instant an accelerating object has zero speed, is it speeding up, slowing down, or neither?
Current License: CC BY-SA 4.0
21 events
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| Jun 17, 2019 at 15:35 | comment | added | jarhill0 | Let us continue this discussion in chat. | |
| Jun 17, 2019 at 11:47 | comment | added | Michael Grant | Everyone is focusing so much on the calculus they are missing the commonsensical view of the physics. Math's job is to explain the physics not the other way around. | |
| Jun 17, 2019 at 11:43 | comment | added | Michael Grant | Speed cannot get more negative if it is zero. It has a nonzero change in speed, we know this, and a zero speed. There is one only way to go. It is like being at the south pole; any change takes you north. | |
| Jun 17, 2019 at 4:53 | comment | added | jarhill0 | I think you're also getting caught up in the velocity vs. speed issue, which is the crux of this question. Velocity is getting more negative, but at the specified point, speed is getting both more negative and more positive (or neither). | |
| Jun 17, 2019 at 1:39 | comment | added | Michael Grant | But even if, for the sake of argument, we were measuring the instantaneous change in speed, in effect taking the left-derivative we would discover that it was negative—and the actual speed itself zero. It would be reasonable to conclude that sign change in acceleration is about to occur. | |
| Jun 17, 2019 at 1:37 | comment | added | Michael Grant | Why would I be more concerned about what we observe about what is occurring as opposed to what actually is occurring? But to answer your question: it would not make a difference if we are measuring velocity instead of speed. Speed is an artificial function of the natural dynamic variable, velocity. | |
| Jun 17, 2019 at 0:13 | comment | added | jarhill0 | As a counterpoint, imagine observing a particle for some amount of time until its speed reaches zero. In that instant, if you were to compute the change in speed of the particle, the data you would have would suggest that the particle is slowing down. At any given instant, we have no information about how the particle will behave in the future. This is because of the directionality of time. By your logic, then, shouldn't we only look at the left-sided derivative? | |
| Jun 16, 2019 at 17:25 | comment | added | Michael Grant | One more argument. Suppose you began observing the particle only at the instant it had zero velocity. After observing the slope of the speed curve from that point forward, it would be entirely sensible to conclude it had nonzero acceleration. What happed immediately prior to your observation is irrelevant. | |
| Jun 16, 2019 at 17:20 | comment | added | Michael Grant | There is an infinite family of speed-acceleration functions that are consistent with the cusped speed plot. They are identical of course at every instant of time except the instant where the cusp occurs. So then we have a choice to make: which of those acceleration functions should we select? I argue the right-continuous version is the one that aligns best with physical reality. Yes, I'm ignoring calculus, because calculus is in service to physics, not vice versa. | |
| Jun 16, 2019 at 17:11 | comment | added | Michael Grant | The fallacy here is that people are constructing the speed plot and then looking at the derivative to obtain acceleration, as if that's the only way you can think about it. But in fact, in the velocity space, the acceleration is a continuous function, and we know it is nonzero at that crossing point. A speed acceleration that is right-continuous is entirely consistent with the math, and with the physical reality. | |
| Jun 16, 2019 at 17:08 | comment | added | Michael Grant | Because time evolves in the positive direction, that's why. | |
| Jun 16, 2019 at 17:08 | comment | added | jarhill0 | @MichaelGrant In this case, the one sided derivatives do not agree (one is negative and one is positive). How do you justify choosing one over the other? | |
| Jun 16, 2019 at 14:39 | comment | added | Michael Grant | My issue with these answers based on the derivative is that the standard derivative assumes no directionality on the independent variable. But time does have a direction. For this reason, it is best to consider a one-sided derivative (see, for example, this Wikipedia article). And in this context, there is an unambiguous answer. | |
| Jun 15, 2019 at 11:01 | vote | accept | CommunityBot | ||
| Jun 15, 2019 at 11:01 | |||||
| Jun 14, 2019 at 18:41 | comment | added | Nuclear Hoagie | @Rad80 "Not true" and "false" are identical for statements with well-defined truth values. By applying the "less than" sign in a situation where it is not applicable, we have crafted a logically self-inconsistent statement that is neither true nor false. The resolution, as in the Liar Paradox, is that we cannot meaningfully assign truth values of true (not false) or false (not true) to the original statement | |
| Jun 14, 2019 at 17:45 | comment | added | jarhill0 | @NuclearWang you raise a good point. Perhaps to be more precise we define "speeding up" as "slope exists and is positive." Then, we have no issue with undefined comparison. This is similar to definitions in calculus, for example the limit definition of continuity, which requires a limit to exist and be equal to a certain value. | |
| Jun 14, 2019 at 17:41 | history | edited | jarhill0 | CC BY-SA 4.0 |
fix typo (meant -> mean)
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| Jun 14, 2019 at 15:07 | comment | added | Rad80 | Rather he is saying the statements are not true. Which is correct. This is the best answer IMHO. | |
| Jun 14, 2019 at 13:00 | comment | added | Nuclear Hoagie | "Slope > 0" means speeding up, "Slope < 0" means slowing down. Since slope is undefined at the point in question, we wind up with the comparisons of "Undefined > 0" and "Undefined < 0". Your answer suggests both of these statements are false, but really they cannot be evaluated. It's like asking "is this sandwich positive, negative, or neither?". Saying an undefined value is nonpositive is like saying a sandwich is nonpositive - it's a meaningless statement because you're using qualifiers that simply don't apply. I agree "neither" is probably the correct answer, but it's a terrible question. | |
| Jun 13, 2019 at 23:15 | review | First posts | |||
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| Jun 13, 2019 at 23:14 | history | answered | jarhill0 | CC BY-SA 4.0 |