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0

You are actually completely right, and then at the last minute you look the wrong way!!! Lets examine the last bit. If you were to continue applying a net force on an object for an infinite time it would indeed accelerate infinitely. There is nothing wrong here. The key point is the NET force, that is, the net amount of force, and its direction, that is ...


0

This is an extended comment on Valter's answer, so please upvote his answer not this one. In Relativity (General and Special) there is no unique way to divide spacetime into space and time. Different observers, using different coordinate systems, will disagree about whether a four vector is just a displacement in time or just a displacement in space. So to ...


2

In a certain sense (regime) acceleration is caused by the curvature of time more than the curvature of space. Actually, the curvature is of the spacetime so that, making rigid distinctions has no much sense. However, if you consider the motion of a particle free falling in a region of spacetime, the equation of its story is the geodesical one: ...


2

I'm hardly a GR expert, so if you want a more technical analysis I'm sure others will be able to give you one. However, the answer to your apparent questions is fairly straight forward. It is not the curvature of space or the curvature of time that causes accelerations, it is the curvature of space-time. We live in a four dimensional universe (ignoring ...


0

The second law states that the net force on an object (F) is equal to the rate of change (that is, the derivative with respect to time) of its linear momentum p (p = m*v, where v is the velocity) in an inertial reference frame. When F is zero, due to the above equality, dp/dt is also zero. This means that d(mv)/dt = 0 --> mv = const --> v = const. ...


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Here's what you do: You know $v_0$, $v_f$, and $x$. This means you can use the equation $$v_f^2=v_0^2+2ax$$ Re-arrange it to get $$a=\frac{v_f^2-v_0^2}{2x}$$ You also know the mass, $m$. You can use Newton' second law, $F=ma$, here: $$F=m\frac{v_f^2-v_0^2}{2x}$$ This should lead you to the answer.


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Here's a simple demonstration: Consider flat space (i.e. Minkowski), viewed in a rotating frame (in e.g. cylindrical coordinates one just replaces $\phi$ by $\phi'=\phi+\omega t$). One can calculate (without too much trouble) that, in these coordinates, a spatial line element can be expressed in terms of the canonical cylindrical coordinates as $$ ...


3

But what about the case where the velocity only changes its direction keeping its magnitude constant? The first equation you give is a vector equation which, in this case, means there are three equations: $$\vec a = \frac{d\vec v}{dt}\Rightarrow a_x = \frac{dv_x}{dt}; a_y = \frac{dv_y}{dt}; a_z = \frac{dv_z}{dt}$$ If we have the additional ...


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There’s a slight issue with some of your wording. Measuring and calculating a quantity are two very distinct tasks - one is experimental in nature, and one lies upon mathematical formalism and theory. That is not to say that they are entirely separate, but it’s worth noting from a pedagogical standpoint. Fundamentally, the answer to your question relies on ...


1

The change in velocity can be calculated by vector subtraction. ($d\vec{v} = \vec{v_f} - \vec{v_i}$). Divide by the time between the two velocities to generate an acceleration. The direction of the acceleration will be the same as the direction of the difference vector. The magnitude of the acceleration will be the same as the magnitude of the difference ...


2

Since both velocity and acceleration are vectors, it is possible as in the case of circular motion for the magnitude of the velocity vector to remain constant but for the direction of the vector to change. Since acceleration is the rate of change of the velocity vector, acceleration would be non-zero even if the velocity vector magnitude is constant, but ...


0

Terminal Velocity depends on two things- surface area and speed. These are inversely proportionate. Terminal velocity depends on two things: Drag force and gravity. These very much are not inversely proportional to one another. Terminal velocity is reached when the drag force is equal but opposite to gravitational force. To a bacterium, drag force is ...


2

The only thing that the device "knows" when it is hit, is the force with which it gets hit, and the duration of that hit. Transfer of momentum $m\Delta v = F\Delta t$. So what matters is the momentum of the hammer's head - or more specifically, the momentum that you are able to transfer. Ultimately it comes down to giving the most momentum to the head of the ...


1

One way to think of this conceptually is that the inertial mass of the system is in fact different from the gravitational mass. That is, while the net force on the systems are the same, the two systems have a different amount of mass, and so they resist changes in velocity to different degrees. Except that in this case, we are considering a compound ...


-1

A while back, I generated illustrations for a treatment of this problem in The Physics FAQ. You can read my updated illustrated version in full on my mirror here. Here is a paraphrased conclusion: How did the ships get farther apart, if they maintained the same constant acceleration at all times? As shown in the following drawing, for any chosen P ...


0

You should integrate the velocity to get the distance $$s(4)=x_0-x(4)=\int_{x(0)}^{x(4)}dx=\int_0 ^4vdt=\int_0 ^4( 2 + 3t^2 )dt=2t+t^3|_0^4=2*4+4^3 = 72 \,\text{m}$$ The initial position doesn't matter since de distance $s$ is the difference between the initial position and the position at a given time (in your case, $t=4$). I don't know what is that ...


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Generally true because during the acceleration phase fuel that will later be burned is accelerated along with the ship, whereas upon deceleration the overall mass of the ship will be smaller, requiring less fuel to slow it.


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This post is a continuation of Ben's answer. I will use $\alpha = a'$ to avoid notational clutter. As Ben showed, we can write for 1-dimensional motion $$ \ddot{x} = \gamma\alpha, $$ where the dots are derivatives wrt proper time. The problem is that $\gamma$ contains $v = dx/dt$, so it is a function of coordinate time $t$ instead of proper time. We can ...


2

This is an answer for motion in 1+1 dimensions. Let a dot stand for differentiation with respect to the rocket's proper time $t'$. The rocket's four-velocity is normalized, so $$\dot{t}^2-\dot{x}^2=1\quad.\qquad (1)$$ Since the norm of the acceleration four-vector is invariant, we have $$ \ddot{t}^2-\ddot{x}^2=-a'^2 \quad . \qquad (2)$$ Implicit ...


1

How to describe arbitrary accelerations in special relativity In trying to address this (first) question, I recommend the following coordinate-free, and invariant (frame independent) way of describing the acceleration of a participant ("object", "rocket", ...) $A$: Given the trajectory of $A$ (coordinate-free, and invariantly) as the (ordered) set $\{ ...


4

There does not need to be a force on an object for it to move, only for it to accelerate, as can be seen from Newton's second law: $$F=m \cdot a$$ I think your confusion arises from forgetting to take into account frictional forces. In practice, a moving object will slow down because of friction: the net force is not zero! Therefore you need to apply an ...


0

That is because when you apply force on any object momentum is transferred which means velocity is transferred to an object at rest. now we know rate of change of velocity is proportional to force applied so if velocity does not change, so rate of change of momentum is zero and so force applied is also zero and hence the object will continue to move. however ...


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I've read your question a number of times in order to try and understand precisely what is puzzling you about this well known 'paradox'. It seems to me that this is it: Well the two ends of the rope are moving at the same velocity, Let's stipulate that, in the inertial frame of reference in which the two spacecraft are initially at rest, the two ends ...


2

Bell's thought experiment is set up in such a way that the distance between the ships, call it $d$, remains the same in the stationary frame; after all, both ships have the same velocity $v$ at the same time $t$, so their distance never changes. Let's use $(x,t)$ as coordinates in the stationary frame and $(x',t')$ in the space ships' frame, we have $\Delta ...


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I see in the comments section this question became something else entirely, but I'll try to sum up all points in my answer -- forgive me if I repeat what others said. First, why should the proton not accelerate forever? It will surely accelerate as long as the force acts upon it. I think what you mean is akin to the Zeno's Paradoxes, in which basically it ...


0

When a force is applied to a body initially at rest for a finite amount of time, the velocity of that body is given by conservation of momentum: $$m\Delta v = F\Delta t$$ When initial velocity is zero, $\Delta v = v$ - the final velocity is equal to the change in velocity. So we can write $$v = \frac{F\Delta t}{m}$$ From this it can be seen that ...


3

Summarizing some of the comments (since they have a tendency to disappear over time): The force lasts as long as you apply it. Force is not a property of material - momentum is. As you know, momentum of a particle initially at rest will be $F\Delta t$ after you apply a force $F$ for a period of time $\Delta t$. More generally, the change in momentum ...


0

The force on the car (due to gravity) is proportional to the mass. The inertia of the car is also proportional to the mass. If you set up your solution properly you should see the two terms cancel out. You never need to know the force - you are asked about the acceleration (which is force divided by mass). Put differently: the acceleration due to gravity ...


-3

Joshua, there is no paradox here. For a start, notice that the paradox is based on contradictory assumptions. In the link you provide it says: The distance between the spaceships does not undergo Lorentz contraction with respect to the distance at the start, because in S it is effectively defined to remain the same, due to the equal and simultaneous ...


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"To an observer in the original rest frame, the spaceships stay the same distance, d, apart.". But why do they stay the same distance apart to an observer in the original rest frame? The spaceships move with constant mutual distance in the original rest frame, since their corresponding parts have the same velocity function of time. The description of ...


3

You understand wrongly I'm afraid. It's commonly said (by non-relativists) that special relativity doesn't describe acceleration but this is quite incorrect. Accelerating frames can be described perfectly well by special relativity. As an example look at my Q/A How long would it take me to travel to a distant star?. This analyses the motion of an ...


1

General relativity describes measurement and observations for a general observer. Theoretical problems with it lie mainly in its interaction with quantum mechanics and statistical mechanics, and there are no experiments that disagree with its predictions.


2

babou, this is really an extended comment to your own answer. I think your answer is pretty much spot on, but I would simply the reasoning a bit. What kills you is when the distance between different parts of your body changes. You give the example of the separation between your head and feet changing by more than 5% (something exploited by hangmen over the ...


2

It really is the stress that kills you. Velocity, acceleration & jerk are all fine as long as they are spatially uniform. It is a postulate of general relativity that you can not even detect acceleration due to a uniform gravitational field, no matter how intense. However if the there are spatially non uniform forces applied to your body, then there will ...


2

If you fly head first into a wall, at the moment of impact the top of your head will accelerate very rapidly while the est of your body continues to travel at a constant speed until they make contact with the wall. Clearly having different parts of your body rapidly accelerate in different direction will lead to some very large forces on various parts of ...


5

Acceleration does not kill us any more than speed. If your head and feet do not move at the same velocity long enough, whatever the cause, you are in trouble. Velocity does not kill us when the whole body has the same velocity. Similarly, I doubt acceleration kills us when all parts of the body accelerate, but without having to transmit forces. It is said ...


2

The problem is, there isn't just one way in Newtonian mechanics to kill someone. You can cause as little or as much acceleration as you want. A few things worth analyzing are: Whiplash. If you're under constant acceleration and you reach a steady state (and aren't dead yet), a change in acceleration (jerk) could cause a whip effect. The Earth-Sun system. ...


0

That's because the equation of motion is a 2nd order differential equation. F=ma. If you integrate it to get r(t), you get two arbitrary integration constants. So you have two degrees of freedom, making the absolute r(t) and the absolute v(t) invariant when adding a constant. This holds with relativistic equations of motion, and even relativistic QM. As ...


0

Acceleration under constant power is $$ a(v) = \frac{P}{m v}$$ Direct integration has $$ t = \int \frac{1}{a(v)}\,{\rm d} v = \int \frac{m v}{P} \,{\rm d}v = \frac{1}{2} \frac{m}{P} \left( v^2 - v_0^2 \right) $$ $$ \boxed{ v(t) = \sqrt{v_0^2 + \dfrac{2 P t}{m}} }$$


0

A certain version of the 3rd option is correct. Let's analyze the case for a person that just stands on a step of the escalator. The person comes walking at a certain speed, presummably the same as the horizontal speed of the escalator's speed, so when the step into the platform is taken no acceleration occurs. But shortly after, the platform changes ...


1

$P=d(mv^2/2)/dt=m/2 * d(v^2 )/dt=m/2 * d(v^2 )/dv * dv/dt=m/2 * 2v * dv/dt=mva $ $dP/dt=d/dt * (mva)=m[dv/dt * a+v * da/dt]$ $dv/dt * a+v * da/dt=0$ $dv/dt * a=-v * da/dt$ $dv/(-v)=da/a$ $ln⁡(-v)+C=ln⁡(a)$ $C=ln⁡(a/v)$ $a/v=e^C$ $v=a/e^C $ $v=dv/dt * k$ , where $k=1/e^C$ $dv/v=1/k * dt$ $ln⁡(v/v_0)=1/k * t$ $v=v_0 e^{1/k * t}$ $v=C_1 e^{C_2t}$ ...


2

I got the answer. There are two separate cases. CASE 1 If $m_1$ is exerting a force on $m_2$, which is in turn causing constant acceleration in both the bodies, then $$a_{net}=\frac{F}{m_{1}+m_{2}} ;~~~~~~\therefore F_{12}=\frac{Fm_2}{m_{1}+m_{2}}$$ CASE 2 If the acceleration on both the bodies is caused by an external force separately, as in the case of ...


2

Q1 does not allow for a conclusive answer. Without knowing whether the force pushes $m_1$ or pulls $m_2$, no statement regarding the contact force between the two masses can be made. The bodies may even be in free fall, then there would be no contact force. Assuming that the force acts on $m_1$, the contact force between $m_1$ and $m_2$ is compressive. You ...


1

The original question asks about the force on $m_2$. The system is explicitly defined to be object 2. Thus, the contact force is not an internal force. You are given the acceleration, so the force of $m_1$ on $m_2$ is $m_2a$. By the way, in your answer you say $a_1=F/m_1$. That is not correct as it does not take into account the force that 1 applies ...


0

@imakesmalltalk: You are right, except for $a_1$. It is equal to $a_{\rm{net}}$.


2

How you approach this depends on the time and resources you have available to you. If a problem like this cropped up as part of a scientist's work the best approach would be to try and understand the underlying physics. You would develop a mathematical model for the system then fit the data to a function derived from that model. In this case the model would ...


2

I'll go with an Equivalence Principle argument. For a model system, consider a test particle in a highly elliptical orbit around a neutron star; the particle will pass through regions of greatly different field strength. But it feels no force as it "falls" around the star. Per the Equivalence Principle, at each point there is a locally inertial ...


0

At each point you have $F=ma$. The instantaneous acceleration is the slope of the speed-time curve. It looks like acceleration may be a linear function of time here. You could pick three points and calculate the acceleration and speed as a function of time and see if they match the curve. It will be hard, as errors will depend on the error of the speed ...



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