105

A notable property of frictional forces is that they resist motion (as opposed to other types of forces, which might resist displacement, for example, which is how a spring behaves). As a result, the brakes on your car slow down the motion of your wheels that produces forward movement of your car—but they also slow down motion that produces reverse movement....


102

Ah, what a tricky mistake you've made there. The problem is that you've simply confused some notions in multivariable calculus. Don't feel bad though-- this is generally very poorly explained. Both steps 1 and 3 above are incorrect. Rest assured, the Euler-Lagrange equation is not trivial. Let's first take a step back. The Lagrangian for a particle moving ...


69

The easy explanation is that the tennis ball is hollow. When you merely drop the objects, they are subjected to the same acceleration - the aceleration due to gravity - and nothing else. Conservation of energy then says that their gravitational potential energy should be completely transformed into kinetic energy at the ground: $$mg\Delta h=\frac{1}{2}mv^2\...


65

Before telling you why an observer in free fall does not feel any force acting on him, there are a couple of results that should be introduced to you. Newton's second law is only valid in inertial frames of reference: To measure quantities like the position, velocity, and acceleration of an object, you need a coordinate system $(x,y,z,t)$. Now the ...


64

When you say "why aren't things being destroyed", you presumably mean "why aren't the chemical bonds that hold objects together being broken". Now, we can determine the energy it takes to break a bond - that's called the "bond energy". Let's take, for example, a carbon-carbon bond, since it's a common one in our bodies. The bond energy of a carbon-carbon ...


63

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 ...


61

Velocity is a vector. Speed is its magnitude. Position is a vector. Length (or distance) is its magnitude. A vector points in a direction in space. A negative vector (or more precisely "the negative of a vector") simply points the opposite way. If I drive from home to work (defining my positive direction), then my velocity is positive if I go to ...


60

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 ...


58

You are right that, without a force acting on it, water falling from a tap could not follow a spiralled path. The tap, however, creates an illusion - the water appears to be spiralling, but it isn't - it's falling straight down. The illusion is created by the "turbine" inside the nozzle, which rotates the ring of spouts that the water falls through. The ...


57

If you assume rigid bodies (no deformation) then actually the coin never needs to have zero velocity since it can instantaneously change directly. In reality though, the centre of mass must pass through zero velocity (in your frame of reference). However, this does not mean that the train will stop! It will slightly decrease its speed due to change of the ...


54

Within the context of Newtonian mechanics, there's a simple answer: velocities are not absolute, but differences in velocities are. So you can state that acceleration occurs unambiguously. In special relativity, this is a bit more complicated because of relativistic velocity addition, but all observers can unambiguously compute a "proper" acceleration for ...


46

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 ...


43

People already answered your question from a usefulness standpoint, but I just want to add that your reasoning isn't correct: Speed is usually defined as the magnitude of (instantaneous) velocity. So one could assume that average speed would be defined as the magnitude of average velocity. That's not how it works. If we have [speed] = [magnitude of ...


43

Escape velocity is the velocity an object needs to escape the gravitational influence of a body if it is in free fall, i.e. no force other than gravity acts on it. Your rocket is not in free fall since it is using its thruster to maintain a constant velocity so the notion of "escape velocity" does not apply to it.


43

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 ...


41

What does change the velocity of an object? Force of course. (Acceleration is the change of velocity). The question is how much time does it take to make that change? A long time? Infinitely small time? Momentarily? Well, it depends on the acceleration itself! If you accelerate in your car, it'd take a significant amount of time to go from $0$ to $100\ \...


40

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 ...


38

Your question is legitimate and I don't understand why it got downvoted. The confusion arises in the difference between average and instantaneous velocity. Consider this example: a car moves at 10 m/s for 5 seconds, then stops at a light for another five seconds. What is the velocity of the car after 7 seconds? According to your calculation, it would be $\...


37

The tricky part of this question is that you are given a graph of velocity but asked a question about speed. Several others have said essentially the same thing, but what really makes this clear for me is a graph of speed: The above is the graph of $$ y = \left \lvert 4 - \left ( \frac{x - 2}{2} \right ) ^2 \right \rvert \text{,}$$ which is just the ...


36

In non-relativistic mechanics, time $t$ is a (universal) parameter and a particle's coordinates (in some inertial coordinate system) can be expressed as three functions, $x(t),y(t),z(t)$ of this universal parameter $t$. The velocity of the particle (in these coordinates) is then the derivative of the position with respect to the parameter $t$: $$\mathbf{v} ...


35

The complete answer to that question is an open problem in fluid mechanics, as exact closed form solutions to the irrotational surface gravity water wave equations are unknown. However, under certain asymptotic approximations, we can estimate the speed of these waves. Irrotational inviscid surface waves are governed by Laplace's equation, i.e. $$\nabla^2 ...


35

It is incorrect to link the feeling of being accelerated to being accelerated itself. You can be under constant velocity or be continuously accelerated, yet you need not feel anything at all. Let me explain. The reason you feel compressed or stretched when you are accelerated in a lift is because of the presence of the normal force from the ground on you. ...


35

Cars move because the wheels are spinning in a certain direction. Brakes work by making the wheels not spin, not by making them spin in the opposite direction. If instead of slamming the brakes you "brake" a car by having some other kind of force pushing it backwards, like a super huge fan in front of it, then yes, it might begin moving backwards.


33

Let us consider that you are at rest and the car, which emits at frequency $f_0$, approaches you with speed $v$. The frequency you receive increases to $$f_1=f_0\frac{c}{c-v},$$ where $c$ is the speed of sound. When the car get passed you the perceived frequency is reduced to $$f_2=f_0\frac{c}{c+v}.$$ The ratio is $$\frac{f_1}{f_2}=\frac{c+v}{c-v}.$$ Now ...


33

There is a theory of rainbows due to elliptical droplets. It was started by Willy Möbius [1] but is perhaps more clearly described in the modern papers [2,3]. Unfortunately the math is messy. While this may change the appearance of the rainbow, the biggest effect is on the angles between the different order rainbows: by measuring them one can presumably work ...


32

The definition of speed (please, let me call it velocity hereinafter) is not random at all. It seems you understand that it must depend on the distance $d$ and the time $t$, so I'll skip to the next stage. Evidently (for a constant $t$) velocity increases if $d$ does; and (for a constant space) $v$ decreases if $t$ rises. That constrains the ways we can ...


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 ...


30

Acceleration is the general term for a changing velocity. Deceleration is a kind of acceleration in which the magnitude of the velocity is decreasing. The reason this might be confusing is because the word 'acceleration' is sometimes used to mean that the magnitude of the velocity is increasing, to contrast it with deceleration. One cannot go wrong, however, ...


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 ...


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