# Tag Info

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What if you had a proton travelling at .99999c towards a heavy object? Would it have to keep accelerating or would the acceleration of the proton slow down to zero and only it's mass would increase? It would keep accelerating, but there's a problem. See what Einstein said in the second paragraph about the speed of light being spatially variable: The ...

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Good question. Although the speed of gravity is approx. 3x10^8 meters per second, the acceleration at the Earth's surface is as you describe ~9.8m/s^2. If you choose the Earth as the heavy mass your proton in question is approaching, and forgo magnetic influence, then you would realize that the increase in acceleration is negligible as compared to the ...

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In order for that proton to reach the speed of light, it would need an infinite amount of energy. Perhaps if the proton reached the center of a black hole, converging with the singularity, it would cease to accelerate, but it would not travel neither faster or at the speed of light.

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You should look at it like an asymptote. Yes the proton would accelerate but it would probably accelerate to .999991c or more likely less due to the massive energy required to accelerate something so fast already. Therefore you could always keep accelerating your particle but it would never cross the Light-Barrier.

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Ok, so here are is my solution. I'll be happy is someone can provide something simpler. $a_1, v_1, t_1 - acceleratin, terminal\ velocity,\ time\ to\ reach\ it\ (for\ Tiger)$ $a_2, v_2, t_2 - acceleratin, terminal\ velocity,\ time\ to\ reach\ it\ (for\ You)$ $s_0 - starting\ distance$ Not let's consider three cases: 1) $a_1 < a_2 \wedge v_1 < v_2$ ...

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Having read John Rennie's answer above, I'm going to give an answer that's hopefully of the same sense, which hopefully makes sense, but which hopefully brings out an issue. 1. Is the free fall acceleration the same as the coordinate acceleration for a hypothetical observer at rest on the star surface? Yes and no. Yes because the falling body falls ...

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I'm guessing your questions all amount to whether general relativistic effects become important at the surface of a neutron star. To answer this we can compare the flat space metric (in polar coordinates): $$ds^2 = -c^2dt^2 + dr^2 + r^2 d\Omega^2 \tag{1}$$ with the Schwarzschild metric that describes the geometry outside a spherically symmetric mass: $$... 0 Try applying conservation of energy, E_i = E_f, as the problem suggests. To start, you know that the ball initially has kinetic energy and gravitational potential energy but ends with only kinetic energy. 1 Acceleration is any change in velocity, whether it's the direction of the velocity or its magnitude. In the case of a perfect orbit it is the direction, the tangential velocity of the satellite should keep a constant magnitude and just change direction as the satellite orbits. When the velocity is slightly less than what is needed for a perfect orbit, it ... 16 For simplicity, consider a perfectly circular orbit; the gravitational acceleration is always at a right angle to the velocity vector. This means that the speed cannot change despite the fact that there is constant acceleration. Note that for the speed to change, there must be a non-zero component of acceleration parallel (or anti-parallel) to the velocity ... 0 I was told forces can depend on time, location and velocity, but never on acceleration. That depends on what one means by "force". The concepts of "force" as used in Newton's second law and in Newton's third law are distinct but related concepts. That force and acceleration are intimately connected with one another entire the point of Newton's second ... 1 F = ma when mass is constant: I think that's a common misconception. It doesn't really represent the law correctly (unless mass is constant). Newton's second law of motion states, F = d/dt(mv) the external forces acting on an object in an inertial frame is equal to the change in linear momentum (the measure of motion) I can see why he made that ... 1 An actual example in which there is a non-zero change in acceleration, that is, jerk, occurs is a spring. A spring's motion is described by a sinusoidal function. The derivative of a sinusoidal function is just another sinusoidal function. As a result, you can differentiate such a function infinitely many times, and will never have a derivative that's 0/a ... 2 An example of a force that depends on position (of the particle) is the force due to a spring:$$F_x = -kx $$An example of a force that depends on velocity (of the particle) is the force due to a dashpot$$F_v = -c\dot x$$Now, consider a hypothetical force that depended only on the acceleration of a particle:$$F_a = -d \ddot x$$The differential ... 0 To clarify the law should be stated$$\sum F(t,x,v) = m\,a $$which is used to solve problems by re-arranging into$$a(t,x,v)=\frac{1}{m} \sum F(t,x,v)$$The point is that various forces depending on time, position and velocity are used to define accelerations. So yes, you are correct. A less obvious case is in problems of nverse dynamics. Here all ... 12 http://wordpress.mrreid.org/2013/12/11/jerk-jounce-snap-crackle-and-pop/ Speaking derivatives to time: First position x, then velocity v=x'=\frac{dx}{dt}, then acceleration a=x''=\frac{d^2x}{dt^2}, then jerk x'''=\frac{d^3x}{dt^3}, then jounce/snap x''''=\frac{d^4x}{dt^4}, then crackle x'''''=\frac{d^5x}{dt^5}, then pop ... 0 Unless you are talking only about classical mechanics, I think that it is not true. For example, the Abraham-Lorentz force depends on jerk (time derivative of acceleration). 7 Yes. Rate of change of acceleration is called jerk. Yes its dimensional formula is [M^0, L^1, T^{-3}]. Similarly one could also define higher time derivatives of acceleration if required for a particular problem. 3 Yes. usually we name them a' . and there can be even a speed of my a' that I can call that a'' and it goes on like that. it is only used it real life calculation that the calculation should be very precise like rocket science. and the equation of displacement (with a constant a') will be :$$x =\frac16 a't^3 + \frac12 a_0t^2 + v_0t+x_0$$(EDIT: ... 2 For most of the trials the racket head reached peak speed just at the time of impact. The racket head showed great acceleration just before impact where the racket head speed went from around 10 m/s to its peak value [35 m/s] in less than 0.1 seconds Probably tou have misinterpreted the text, it nowhere says that the acceleration gives more power, ... 1 Several things here. First - how efficiently you transfer momentum depends in part on whether you hit the "sweet spot". The location of the sweet spot is a function of the instantaneous center of rotation - an almighty swing from the shoulder puts the center of rotation further back so the same velocity of the racket might be less efficient. That depends on ... 1 They are the same thing. Parts of the racket though produce more force than others, suppose I completely missed the strings and hit the metal part (that's why without technique is so important). Usually, when you play you hit in what is called the sweet spot which is a vibration node and although it may not give you the biggest pop or bounce it gives you ... 0 Calculate the MI (moment of inertia) of the bowling ball around the axis of rotation , i.e - shoulder. Assuming the bowling ball to be a solid sphere (it is actually a shell?), since the shell thickness isn't given, this would be- Icm = (2/5)MR^2 (MI about it's center of mass) Hence MI around shoulder, using parallel axis theorem - Ishoulder = Icm + Md^2 ( ... 0 After further research, I found that the moment of inertia of a system consisting of multiple objects, like the arm-bowling ball system in the problem, can be found simply adding the moments of inertia of each object 1, though use of the horizontal axis theorem may be necessary if the object is rotating around an axis parallel to its typical axis of rotation ... 1 A nice way to compare both is to invoke the definitions:$${\vec a}_{\rm avg} = \frac{\Delta {\vec v}}{\Delta t}$$and$${\vec a}_{\rm inst} = \lim_{\Delta t \rightarrow 0}\frac{\Delta {\vec v}}{\Delta t} = \frac{d{\vec{v}}}{dt}$$Graphically, and if you consider change over an infinitesimal time period \Delta t \rightarrow 0, the same definition ... 0 The instantaneous acceleration is the time derivative of the velocity vector:$$ \vec{a} = \frac{d\vec{v}}{dt} $$If the velocity is changing then the acceleration will be non-zero. 1 If only the forces of gravity are present, all objects fall at the same rate. This is what one calls equivalence principle. In classical mechanics it shows up in the force law for two particles of gravitating mass m_G and M_G, where M_G shall denote the earth's mass.$$ m_i \cdot \vec{a} = -G \cdot \frac{m_G \cdot M_G}{|\vec{r} - \vec{r} '|^2 } \cdot ...

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Here is an extremely simple explanation: Force = Mass x Acceleration Force / Mass = Acceleration Mass x Acceleration due to Gravity / Mass = Acceleration Acceleration due to Gravity = Acceleration For further intuition, consider this: The greater the mass, the greater the inertia. The greater the inertia, the greater the difficulty to accelerate the ...

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Without air resistance all objects are accelerated with gravity $$g = 9.81 \frac{m}{s^2}$$ Only the air causes a "slower" acceleration. This effect depends from the density and shape.

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"So the idea, is that performing experiments himself, inside the train it is possible to detect the value of the acceleration?" You're spot on. Couldn't write it better myself. BTW in GR we have the same situation, as long as you qualify what you mean. All observers will agree on the reading on an accelerometer fixed in a certain frame, and on the readings ...

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Speed has no direction while velocity does. For example, if I say that I'm running at 10 mph, I have given you my speed. If I say that I'm running 10 mph north, then I have given you my velocity. Acceleration is the rate of change in velocity. Imagine this: I am in my car and you look at me before I even press the gas pedal. You close your eyes then open ...

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The effects of inertial acceleration is best understood from the physics of Newtonian mechanics. A good site to support your understanding of this physics: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html I recommend studying the trajectory calculator as it applies to your problem.

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I do not know where you read that, but "net acceleration of the body is = acc.due to gravity - acc. of the body" does not much sense to me. Usually, you calculate the net acceleration of a body once you know what forces are acting on it. For a body under free fall, the acceleration will be "g", the acceleration of gravity because it is the only force that ...

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The acceleration of the plane will cause an offset in the motion of the mass/spring system; if it was centered on x=0 when the plane was stationary, its motion will now be centered around $x = -\frac{m\cdot a}{k}$ where $k$ is the spring constant, and $F = m\cdot a$ is the force needed for the mass to accelerate with the plane. If the plane is accelerating ...

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There are various ways of calculating acceleration and jerk from your data set. THe value of $dt$ you want is the time step generally, in your case $100$ ms. The method that you suggest $$a=(v2-v1)/dt$$ will give you an answer and is reasonnable, but it will not give you the best value I expect. It might be better to use something like Savitsky-Golay ...

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The total force acting on a raindrop equals $g$ minus air resistance which increases with velocity. In other words, as the raindrop speeds up, air resistance increases which decreases the acceleration (until eventually the acceleration equals zero and terminal velocity has been reached).

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Who says all derivatives have to change continuosly? With a suitable setup, acceleration can jump from zero to something. Location and velocity can not change discontinuously, but acceleration can. No, there are no "infinitely many derivatives of velocity". In your question, you write "acceleration has to increase from zero and therefore the third ...

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Have a look here: https://en.wikipedia.org/wiki/Non-analytic_smooth_function There are functions which are identically zero for negative arguments, non-zero for positive arguments and smooth everywhere.

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Yes, if an object is stationary, then starts moving in the positive direction, all derivatives of the position can become positive when the object starts moving. But so what? It's not a paradox. It's just a mathematical statement. All the derivatives are positive. No big deal. It's hard to tell why you find this confusing. You might be thinking that there ...

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You can achieve this in two distinct ways: 1) use a reference (I'll give an example below) 2) measure other quantities that can be converted into an acceleration. Here are some possibilities: 1a) Gravity. At rest, the accelerometer should measure the local g. 2a) Put it on a rotating platform. Knowing the rotational speed and distance to the pivot, you ...

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force = acceleration * mass, hence acceleration will be $a=50N/22kg \approx 2.27m/s^2$ Distance it moves might be found by integration: $\int_0^{1.2}v(t)dt=\int_0^{1.2}atdt$, since speed $v(t)=at$ Answer to (B) then is 1/2*1.2*(2.27*1.2)=1.63, which seems pretty close to what you have got

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Yes you can! For example: when you are sitted, doing nothing. From the reference system of the earth you are firm, with no velocity BUT there's the gravity acceleration. We can say that we have everytime a potential to have a velocity if we have a force. For the reason that says that for every force there's an other force with same value but with a contrary ...

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