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26

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 ...


20

It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles. So assuming the spins of the electron-positron pair is averaged, and that you don't care about the photon ...


17

You have scrambled across one of Zeno's many paradoxes - the so-called Dichotomy paradox. The resolution lies in the fact that sum of terms in an infinite series do not necessarily add up to produce an infinity, so the basic premise is flawed. In particular, the series $$S = \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = \frac{1}{2} + \frac{1}{4} + ...


14

We do. The LHC accelerates two protons, each with 3.5 TeV of energy, giving a total of 7 TeV in the CoM frame (The energies are from the initial phase of the previous LHC run. Later in the run this was increased to 8 TeV and the combination of the two dataset was what discovered the Higgs boson. The energies are roughly doubling now for Run II, to 13 TeV). ...


13

Have a look at the article by Phil Gibbs on the relativistic rocket. This describes the motion of a rocket that is accelerating with a constant acceleration. In this context constant acceleration means the crew of the rocket feel a constant acceleration. Technically the rocket has a constant four-acceleration. Anyhow, the velocity of the rocket as observed ...


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 ...


8

The train does accelerate as it goes around a curve. Velocity is a vector, with magnitude and direction. Speed is the magnitude. The train changes direction. Acceleration is caused by a force. If the force causes a change in direction with no change in speed, it must be perpendicular to the direction of velocity. For example, A planet in a circular orbit ...


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.


6

Is there some other formula ... which ... does not allow the speed ... to surpass the speed of light? That would be the equations of special relativity mentioned by sahin in a comment. Image from Loodog? Another factor you have to take into account with classical mechanics is to work out how a constant force can be applied to your object over 11 ...


5

Many modern particle accelerators do accelerate both particles towards each other. LEP accelerated electrons and positrons in opposite directions in the same chamber, and the Tevatron did the same for protons and antiprotons. The LHC is a proton-proton collider, and so it has two stacked rings that accelerate protons in different directions. For the BaBar ...


4

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 ...


4

If the object was released from rest ($u=0\,{\rm m/s}$), what is its speed after $1\,{\rm s}$ if $a=10\,{\rm m/s}^2$? Using: $$v=u+at$$ you will find that the object was not released from rest...


4

Well, if you have $2{\pi}r$ and you know the time in which you traveled it, then you could find speed. Nothing prevents this.


3

The only way to purely rotate a rigid body about its center of mass is to apply a pure torque (no net force). If the net force applied is zero then the center of mass is not accelerating. However and combination of translation and rotation of the center of mass can be viewed as a pure rotation about the instant center of rotation. So to effectively answer ...


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: ...


3

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.


3

On an intuitive level, the initial speed $v_0$ can be considered to have two effects: one on the horizontal velocity, and one on the vertical; the former affects the range in a direct sense, and the latter increases the time the projectile is in the air. The combination of both of these gives an overall $v_0^2$ contribution. If you were to increase the ...


3

First of all, you need to be careful when saying the train does not accelerate. In Physics, acceleration is the rate of change of velocity. As velocity is vector, if the velocity of the train changes direction, as it does on the curved portion of the track, then the velocity is changing over time (changing in direction). Therefore, there must be an ...


2

You must solve for the objects initial velocity first: $$ v(t)=u+at\\ v(0)=u\\ v(1)=u+10\text{m/s}\\ =20\text{m/s}\\ u=10\text{m/s} $$ With this adjustment you should find the correct answer.


2

I will try to explain this in simple english. Since the number of steps are infinite, then the distance you travel at each step is infinitly short. As such, the time teaken in each step is close to zero but not zero. Therefore the total time in taking this infinite number of steps does not necessary equal to infinite. Mathematically the total time take is ...


2

It's a very strange question. In theory a ball is thrown in the shape of a parabola. You can get the graph of the thrown ball with this equation: $$y=\frac{-gx^2}{2V^2 \cos^2 (a)} + x\tan a$$ where $a$ is the angle at which the ball is thrown, $g$ is the gravitational acceleration, and $V$ is the initial velocity. If you want to calculate the initial ...


1

As the wheels try to roll they are prevented from rolling by the frictional force acting in the opp. direction.As the traction force exceeds the limiting frictional force the wheel starts rolling forward w.r.t rails. The force tries to induce relative motion between the wheels and rails.As the rails cannot move backwards (due to friction) the wheels have to ...


1

No, the baseplate will not move if the two motors are applying torques in such a way that the rotors remain at rest. To analyse this problem, it is best to consider the forces on the three parts (base disk and two rotors) individually, using free body diagrams. There are two motors, each connecting a rotor to the base plate, and the effect that each motor ...


1

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 ...


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 ...


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 ...


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 ...


1

Your question is "Is there an infinite series of higher derivatives of position for this to work?" Answer: No. Acceleration can jump from zero to something. When it does, its derivative is not defined, so the series of position derivatives stops after the second one. From the question: "A change is velocity is acceleration, so the value of the ...


1

Of course they both give informations about the motion of bodies. The kinematic equations tell us simply what are the valors of the variables of the specific motion , that because kinematic studies only the variables of the motion and their changing. kinematic equations give us indications about : Velocities (the most simple and known equation of kinematic ...



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