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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 ...


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 ...


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, ...


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 ...


2

It's worth drawing a diagram: The equations of motion are: $$x_1 = x_0 + \frac12 a_1 t^2$$ for $t_1 < \frac{v_1}{ a_1}$ And $$x_1 = x_0 + \frac{v_1^2}{2 a_1} + v_1 (t - \frac{v_1}{a_1})$$ for the steady state. The same equations, with different suffixes, hold for the tiger. To solve, you start by considering all different orders of $t_1, t_2, ...


2

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) $a_1 < a_2 \wedge v_1 < v_2$ ...


1

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.


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

$F = ma$ 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

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 ...


1

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: $$ ...


1

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 ...


1

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.


1

"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 ...


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

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 ...


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 ...



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