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In everyday life, we experience the universe in a non-relativistic classical way. We are familiar with the concept of time and space. Defining the velocity as the ratio between a distance traveled in a given time interval is a much more natural choice instead of defining the rapidity. If we all lived at relativistic speeds, or close to the event horizon ...

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Your mistake is in assuming that within those one second intervals the velocity is constant. The velocity as a function of time $v(t)$ for motion under constant acceleration $a$ is given by $$v(t)=at+v_0$$ where $v_0$ is the velocity at $t=0$. So, you are right that $v(0)=20\,\mathrm{m/s}$, $v(1)=12\,\mathrm{m/s}$, and $v(2)=4\,\mathrm{m/s}$. But, for ...

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I think that a good way to highlight the error in your "concept of acceleration" answer is to draw a velocity against time graph and to remember that the area under such a graph is the displacement. You will see that in the first second the velocity is changing and to get the displacement you need to use the average velocity during that time interval ...

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It is not particularly easy to add rapidities pointing in different directions. e.g. Suppose B moves with rapidity $\rho_{1}$ with respect to A in the common $x$-direction of both of them. And suppose that C moves relative to B with rapidity $\rho_{2}$ in the common $y$-direction of B and C (we are taking it that their axes are aligned in this way). Now ...

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Suppose you are travelling at a uniform velocity and you cover 1 meter in 1 second. Your average velocity is $$\frac{1\ {\rm m}}{1\ {\rm s}} = 1 \frac{\rm m}{\rm s}.$$ If you consider a 1 millisecond interval within that 1 second, you cover 1 millimeter. Your average velocity in that milisecond is $$\frac{1\ {\rm mm}}{1\ {\rm ms}} = 1 \frac{\rm m}{\rm s}... 39$$v_\text{average}=\frac{\Delta s}{\Delta t}v_\text{instantaneous}=\lim_{\Delta t\to0}\frac{\Delta s}{\Delta t}$$If the time interval gets infinitesimally small \Delta t\to 0, then you are dividing with something very, very tiny - so the number should become very big:$$\frac{\cdots}{\Delta t}\to \infty \quad\text{ when } \quad\Delta t\to0$$If the ... 31 The contradiction is in the setup. It is impossible that those three relative velocities can simultaneously be true. You can't just add relative velocities together directly in special relativity. If an observer moving at velocity v measures, in their own frame, a velocity v' of some passing object, then a stationary observer would measure the velocity ... 28 In English, it seems that: Position is a vector. Distance/length is a name of its magnitude. Velocity is a vector. Speed is a name of its magnitude. Acceleration is a name of a vector and its magnitude. Force is a name of a vector and its magnitude. Momentum is a name of a vector and its magnitude. ... Velocity/speed as well as position/length seem to be ... 27 People understand what velocity means, and how it is used. Anyone who needs to use relativistic models knows enough to distinguish between which model they are using. No additional words were needed when it was created, and trying to invent new words now would just cause more confusion, not less. 25 After the invention of modern calculus and notions like continuity and differentiability, the answer is quite trivial in Newton's formulation of mechanics assuming the body is moving along a line. The second derivative of the position should be always defined as it equals the total force acting on the body. Therefore the first derivative must be ... 24 While this varies by country, it's fair to say that kids have a qualitative understanding of speed before kindergarten. By grade 8, they're talking about speed, acceleration, and related issues of energy quantitatively. Vectors are introduced during high school. Most students won't get a quantitative discussion of Special Relativity until freshman physics.... 23 You, referring to yourself, always have a 4-velocity:$$ u_{\mu} = (c, \vec 0) $$which has:$$ u_0 = c $$Hence, moving through the time-direction at the speed of light. You can't move through space (otherwise, there would be a preferred frame in which you were not moving through space). A moving observer that sees you moving through his definition of ... 22 As well as issues of practicality, it doesn't answer the question of why there's a speed limit. (It can't as it's just a mathematical transformation.). The question becomes 'in the formula for w, why do you take c=3 \times 10^8 m/s'? 17 In the real world, a displacement time graph can never be discontinuous. The only not-so-physical meaning that it has is that the body teleported from one place to another such that its displacement changed instantaneously/discontinuously. And as you can see, it's impossible to define the velocity of this teleportation. 16 Given velocity v(t), the distance moved after a certain time t is not v(t)t - this formula works at constant velocity, but when the velocity is changing, the correct expression is \int^{t_f}_{t_0} v(t) dt. Therefore your friend's third line is incorrect. 16 For clarity of thought in the physics of distance, time and motion, one should be careful with words such as "moving", especially when people speak of "moving through time". The trouble is that this creates a confusion of two meanings of the word "moving". There is physical motion, when bodies have relative motion and two different worldlines have different ... 16 This is a simple but important point. While the speed of light is invariant among all inertial observers, the direction of light is not. Let's show this explicitly. Let's say that light is propagating in the negative-y direction w.r.t. observer \mathcal{O}_1 who is standing still. Now, consider an observer \mathcal{O}_2 running at a uniform speed v ... 16 In general velocity is a vector. In 1D motion we can get away with saying the velocity is positive or negative by associating each sign with a direction. However, once you move in more dimensions you can't say the velocity is positive or negative. You just have the vector (in Cartesian coordinates)$$\mathbf v=v_x\,\hat x+v_y\,\hat y+v_z\,\hat z$$Of ... 14 This is roughly the simplest you can get it:$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-v/c} \sqrt{1+v/c}} \approx \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1-v/c}}.$$In other words, if \Delta v is how far the speed is below c, then$$\gamma \approx \sqrt{\frac{c}{2 \Delta v}}.$$13 For ultra-relativistic particles, c-v basically stops being an experimentally accessible observable. Unless you are extremely careful about timing, you assume that the beam is traveling at c and measure the Lorenz factor by comparing the kinetic energy per particle to the particle mass, \gamma = E/mc^2. (If you care about the difference between total ... 12 I don't like these definitions of momentum. They may try to give an intuitive explanation but some will still find it hard to understand why these definitions correspond specifically to m\vec{v} in Newtonian mechanics. The truth is that momentum is just a vector that we discovered that it is conserved when there is no external force on the system. It is ... 12 In retrospect, having seen special relativity, what you suggest is reasonable. In fact, one can unify the geometries of Euclid, Minkowski, and Galilean-relativity after one distinguishes that "Galilean rapidity" is different from "[Minkowskian] rapidity". So, I think we should use at least introduce it (or aspects of it) if one is going to discuss special ... 12 I think the main reason, as already captured in part by other answers (e.g. Davide Dal Bosco's), is the following: velocity is a physical quantity, it tells us how far something goes in a given time. Rapidity may be mathematically convenient due to its relativistic addition properties, but what does it tell us? As an example, the rapidity of light is w = \... 12 There are really 2 questions here, so let's answer them separately. Why don't we use rapidity in daily life? Simple, most people don't even know what relativity is don't understand its implications or how to reason about them. They are not capable of using relativistic quantities. If you are suggesting they simply don't worry about relativity and just ... 11 Aaron Stevens is right, but maybe I can clarify his explanation. Consider the first second. At its beginning, the velocity is indeed 20 m/s, and at its end it is 12m/s, as you correctly wrote. But what distance was covered during this first second ? Since the acceleration was constant the mean velocity was the average, (20+12)/2 m/s, namely 16m/s. So the ... 11 As Lewis Carroll Epstein explains in Chapter 5 “The Myth” of his excellent book Relativity Visualized: There is afoot an errorneous idea. It is that in physics the ultimate reality is a mathematical prescription, an equation. In fact, the ultimate reality is a little story or myth. Then he divided myths into 2 categories: good and bad ones. Good ... 10 I assume you are talking about an object that reverses direction through a collision. Take the case of a ball bouncing off the ground. As it hits the ground, the ball deforms and the ground compresses a tiny amount. The resistance of the ball to deformation and the resistance of the ground to compression decelerate the ball to the point at which its centre ... 10 There are several answers to this question. Rapidity is not taught from the outset in introductory physics classes in part because it would unnecessarily confuse students, and besides, you only need to worry about it when you are dealing with relativistic speeds (or precise-enough measurements that can detect relativistic effects). But there are other ... 9 The first definition transforms as a four-vector: \dfrac{dx^{'\mu}}{d \tau} = \Lambda^{\mu}{}_{\nu} \dfrac{dx^{\nu}}{d \tau}. The second definition transforms not quite as a four-vector: \dfrac{dx^{'\mu}}{d t'} = \dfrac{dt}{dt'} \Lambda^{\mu}{}_{\nu} \dfrac{dx^{\nu}}{d t}. This makes sense, since in the first definition you divide the differentials of ... 9 The other answers are correct in showing why what is being proposed in your text is correct, but I want to show you where your error is and how to easily fix it directly. Your issue is that you are plugging velocities in for t instead of v_x(t) i.e. you have done$$\int_{v_0}^{v}\mathrm{d}v_x(t) = \bigl[v_x(t)\bigr]_{t=v_0}^{t=v} = v_x(v)-v_x(v_0) ...

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