New answers tagged

3

I have noticed this effect often in cars and sometimes in trains. This is the reason I think it happens though I can't claim to have done any research. The car stops because the breaks are applied, the wheels stop turning and there is a force of static friction between the road and the tires. In the car frame of reference I experience this backwards ...


0

Take it one step at a time Time to reach $v_1$ under acceleration $a_1$ $$t_1 = \frac{v_1}{a_1}$$ Distance traveled to reach $v_1$ $$x_1 = \frac{1}{2} a_1 t_1^2$$ Total distance traveled during cruise (time $t_1$ to $t_2$) $$ x_2 = x_1 + v_1 (t_2-t_1) $$ Total time to stop from $v_1$ under deceleration $a_3$ $$ t_3 = t_2 + \frac{v_1}{a_3}$$ Total distance ...


4

If a body is moving, this doesn't mean that a net force certainly must be exerted on it. It can move without any net force (First Law of Newton). You might say "How that body has started its motion without any net fore?" The answer is: "Equations of motion are moment equations, i.e. they are stated for moments not for a time interval ($F(t_0)=ma(t_0)$). So ...


3

You are correct in your definition of force. A car, not accelerating, has zero net force associated with it. However, if the car were to hit something--let's say it's me standing in the middle of the street--it would exert a net force on me, and by Newton's Third Law experience a net force equal in magnitude and opposite in direction. So how can an object ...


6

You are correct. The car has no net force on its environment, and the environment has no net force on the car. This is true of any object traveling with a constant velocity. This is even true in the vertical direction. There is a force of gravity pulling down on the car, and there is a force caused by the road pushing up on the car. If the car is not ...


1

The brief answer is 'yes’. Here is a thought experiment which I think makes it easy to see that the answer must be yes. Consider the standard twin 'paradox': twin a hangs around in free-fall; twin b zooms off on their spaceship at some enormous speed with respect to twin a, turns around (in some smooth way, undergoing acceleration), and returns. Well, ...


1

Consider what you stated: $$a=v\frac{dv}{dx}$$ Now rewrite it: $$\frac{dv}{dx}=\frac{a}{v}$$ If $v$ is tiny, then you know that $\frac{dv}{dx}$ must be enormous in order to produce the acceleration that you darn well know exists. Accleration at the top of that trajectory is surely $9.8$ $m/s^2$. As $v$ gets smaller and smaller as it gets to the apex, $\...


0

I am not versed in general relativity but I see a problem with your scheme. Let us take the ensemble of distant stars as our inertial frame. Whenever a body has acceleration $\textbf{a}$ w.r.t. inertial frame it experiences an inertial force equal to $\textbf{F}_{inertial}=-m\textbf{a}$, where $m$ is body's mass. Now suppose that the large mass $M$ in your ...


1

I think your right. Due to equivalence principle, to the free-falling body it will seem like it has no acceleration, as opposed to a body standing on ground which is equivalent to the ground pushing the body so that it will accelerate upward, and hence experience the normal force from the ground. Because the acceleration due to gravity will depend only on ...


3

Primary question: This is very similar to the old question of what would happen if you fell into a black hole. You are correct that you wouldn't feel the acceleration due to gravity per se, but you'd still need to worry about tidal forces. These have complicated geometric dependence - they're negligible near the center of a planar mass $M$, and for a ...


0

The question tells you that the speed is decreasing. It is NOT constant. For the speed to decrease, there has to be a force, and hence an acceleration, acting opposite to the initial direction of motion. This decrease in speed required a -ve acceleration. The question also says that the train turns round a circular bend. The centripetal force required to ...


0

Problem statement says "the train slows down at constant rate", not "at constant velocity" - in fact, if velocity were constant the train wouldn't be slowing. What the statement means is that the train is slowing with a constant acceleration - at least, the tangential component of acceleration is constant. Since velocity is decreasing, the sign of the ...


0

The problem doesn't say the train "is slowing down at a constant velocity." That doesn't even make sense --- either it's slowing down, and the velocity is not constant, or it's not slowing down, and the velocity is constant. It can't be both accelerating and at a constant velocity, since the very definition of acceleration is change in velocity with respect ...


0

The magnetic force on a particle is given by the Lorentz Force: $F=q(\vec{v}\times \vec{B})$, where $\vec{v}$ is the velocity of the particle and $\vec{B}$ is the magnetic field Using Newton's second law, $m\vec{a}=q(\vec{v}\times \vec{B})$ where $m$ is the mass of the particle This is equivalent to: $ma=q(vBsin\theta)$ where $\theta$ is the angle ...


0

The charged accelerometer will register a non vanishing acceleration. The reason why the setup you are proposing (interaction via electric charge) gives a physically distinct result from the setup in the answer you linked (interaction via gravity), even though they can be described by the same mathematical force, is because the Equivalence Principle applies ...


6

There is indeed a nett force on the body owing to the electrostatic attraction / repulsion. Therefore, there is nonzero four acceleration, and the body will have a different orbit from the ones defined by the spacetime geodesics for the metric describing the massive body's neighborhood. From the standpoint of an observer stationary with respect to the ...


13

Your error is simply that you are assuming that $v(x)$ is differentiable with respect to $x$ at $v=0$. The chain rule needs that all derivatives involved exist before you can apply it. In the case of just letting go of something, the function $v(x) = \sqrt{2gx}$ is not differentiable at $x=0$, which is where $v=0$, so you are not allowed to apply the chain ...


3

When an object starts at rest, the change in velocity when it has made an infinitesimal displacement is infinite - in other words, $\frac{dv}{dx}$ is undefined. You can see this most easily by plotting the curve of $v$ as a function of $x$ for an object starting at rest: $$x = \frac12 a t^2\\ v = at\\ x = \frac12 a \left(\frac{v}{a}\right)^2\\ v = \sqrt{...


-1

This equation is best understood in integral form $$ \left. {\rm d}x = \frac{v}{a}\,{\rm d}v \right\} x_2-x_1 = \int \limits_{v_1}^{v_2} \frac{v}{a}\,{\rm d}v $$ It gives you the distance traveled by a varying acceleration between two speeds. "A car accelerating from 0 to 60 mph needs X distance". By stating that $v=0$ always not only it implies that $a=0$...


0

The acceleration along the axis in this case depend upon the choice of axis. If you have chosen vertical as y axis then there will be only acceleration along y axis as it cannot have any component along x axis. I think you have problems with why it only affect the motion along the vertical. It is because tho orthogonal vectors doesn't affect each other as ...


3

As a general rule, whenever you are setting sudden discontinuities (as in your example with the Heaviside function) it is not a surprise that these may reflect in discontinuous distributions, or derivatives, or infinities here or there. Do keep in mind that it is just a result of the mathematical simplifications that we are introducing (although ...


1

The kinematic equations are derived from differential equations. This means that you can start with the acceleration, then move to velocity, and then position given that the necessary information is available (like initial velocity or position). I find that understanding this helps me visualize the connection between the three. You can not rely on the ...


0

You're measuring the wrong thing. Measure the momentum of a punch, not the "force". The force will be different if you hit something soft or hard, so you can't depend on that. The momentum is just the mass of hand+glove+forearm times velocity of punch. How to measure that? Hit the punching bag and see how far it swings.


-1

They are, as Einstein pointed out, equivalent. So why distinguish between the two? Well the only real difference is that they are measured differently. To measure inertial mass, we exert a given force to something with an unknown mass. To measure gravitational mass we compare the force of gravity from an object with an unknown mass to the gravitational force ...


0

Gravity causes acceleration, but acceleration can happen from a lot of other things as well, for example, on electromagnetic effects. In most cases, the acceleration depends on some charge-like quantity. For example, a body with a mass of 1kg and with a charge of 1C will accelerate faster in the same electric field, as a body with 2kg of mass and with the ...


1

"Comfortable ride" is a tricky thing to quantify. Jerk is not the right metric to use. The reason it works for roller coaster design is the fact that in a roller coaster, you brace yourself against the rather large low-frequency acceleration. If you make a sharp turn to the left, you will want to lean left. If you then suddenly make a large turn to the right,...


1

If velocity at bottom of slope is all you want then energy method in Gert's and Dr Xorille's answers is an elegant and easy way to get at it. However you seem to be tangled up with resolving vectors. In working with vectors you should set up a coordinate system and stick to it until the end. If you have to set up more than one, and worse, have to switch ...


-1

Potentials do not have retarded time effects. If electron is removed from the sun its effect on the potential on esrth is immediate . So the electronics will be likely tip top. The limit of speed nowadays in space is set by the rocket gas engines that they use. Basically they use the chemical bonds energy to accelerate and push gas outside.


11

The power radiated by a charged particle moving at non relativistic speeds, whose acceleration is $a$ and charge is $q$, is given by the Larmor Formula $$P =\frac{2q^2a^2K}{3c^3}.$$ The product above is nowhere near $c^3$ for planetary motion and hence the radiation emitted is negligibly small.


1

To formalize the comments (now in chat here): Associated jerk is probably what you want to calculate, as it is the measure of how violently something is shaken.$^1$ Jerk is the derivative of the acceleration with respect to time. To properly calculate this, you would use the formula $\left| a \right| = \sqrt{a_x^2+a_y^2 +a_z^2}$ (from the Pythagorean ...


1

Have a read through the answers to How can you accelerate without moving? and If $F=ma$, how do can we experience both gravity and a normal force even though we are not accelerating as these explain in some details exactly what is meant by acceleration in relativity. If you are falling freely then by definition you are weightless i.e. your proper ...


0

No. Acceleration and velocity have different units, so their magnitudes cannot be compared. Whether one is larger than the other numerically depends on what units are used. Acceleration and velocity can have the same direction, but this is not necessary - eg a ball thrown upwards has upward velocity but downward acceleration, until it reaches maximum ...


3

I think they mean the acceleration due to gravity, on the surface of the moon. The moon equivalent of "g" (9.8 m/s²), in other words. The second value (1.232 * 10^-5) could be the centripetal acceleration on the moon surface due to the rotation on it's own axis.


1

Gravity acceleration is... acceleration, measured in $\mathrm{m/s^2}$. It is the rate of change of the velocity.. Velocity is measured in $\rm m/s$. It is the rate of change of the position. The vertical velocity and the acceleration due to gravity of a body are collinear but they can have different magnitudes as well different orientations. Think about a ...


-1

I believe I understand what you are asking and the answer is that the observer does not see an accurate representation of what happens near the event horizon. In the first graph the line would curve up to infinity at the end because as the object fell into the hole the light would be slowed to a stop at the event horizon. As the object begins it's approach ...


1

Dispense with most of the math and think of this question in more practical terms and in an easier to understand way: You have 3 theoretically "perfectly identical", "perfectly symmetrical" cars "A", "B" and "C". If car "A" traveling at "exactly 50kp/h" hits a theoretically "immovable wall" "head on", (that is to say, exactly perpendicular to the plane of ...


2

I would choose as notation conventions: $\Delta X$ for a constant increment, ie if $X_{n}-X_{n-1}=\Delta X$ for all n and with underscript otherwise: $\Delta X_n=X_{n}-X_{n-1}$ I prefer the increment to have the index of the value it produces (adding increment $\Delta X_n$ gives value $X_{n}$) but that's a matter of taste. It's not clear to me what the A ...


2

I would name the period at $n$ as $P_n$, and to the change in period as $\Delta P_n$. In such a case you have the rate of change of the period, that is the change in period per unit of time is $\Delta P/t_A=\left(\frac{1}{f_{n+1}}-\frac{1}{f_n}\right)/t_A$. However, to define the angular acceleration, $\alpha$, you do not need the period, the definition is ...


2

You typically cannot write subscripts or superscripts (easily) as comments in code, so I'll assume you are rather interested in writing this into a specification or some kind of documentation that does support such notation. When notation becomes cumbersome, here's what I'll generally do: 1) for discrete increment always use $(n), or (k)$ or the like ...


0

Yes and no. Coordinate acceleration doesn't need to be relative, but proper acceleration is always invariant.



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