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4

\begin{equation} \mathbf{v}\left(t\right)\equiv \dfrac{d\mathbf{r}}{dt}= \dot{\mathbf{r}} \tag{01} \end{equation} We'll use one upper dot for the 1st derivative with respect to $\:t\:$, for example \begin{equation} \dot{\mathbf{r}}\equiv \dfrac{d\mathbf{r}}{dt}\;, \quad \dot{\theta}\equiv \dfrac{d\theta}{dt} \tag{02} \end{equation} Now, let a system of ...


4

I think such a function may only exist in the Maxwell-Boltzmann limit. Here's why: For simplicity let us parametrize everything in terms of $\beta = 1/T$ and denote $Z(\beta) = \int{d^3p\; f_{eq}(p, \beta)}$. Rewrite the latter as $$ Z(\beta) = 4\pi \int_0^\infty{dp\;\frac{p^2}{e^{\beta E_p}\pm 1}} = 4\pi \int_m^\infty{dE\;\frac{E\sqrt{E^2-m^2}}{e^{\beta ...


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Observe that by chain rule, we have that $$a=\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}=\frac{dv}{dx}\cdot v=v\frac{dv}{dx}$$


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These are a couple of classical equations of motion derived from Newton's laws, dealing with the motion of a body with initial velocity of 50 m/s at an angle of $\theta$ degrees (or radians) with respect to the horizontal/ground/$x$-axis, assuming that the coordinate system is flat. The object is subject to a downwards acceleration, which we infer is due to ...


3

What an interesting question! It depends. In modern calculus, $\frac{df}{dt}$ is just a symbol for the derivative $$\lim_{h \to 0} \frac{f(t+h)-f(t)}{h}$$ As a matter of fact, mathematicians prefer different notations for the derivative of a function $f$, as $D f$ or $f'$. But the above definition of derivative became rigorous only when the concept of ...


3

In physics there are no infinitesimals, so dx is always treated as a "small but finite interval" during discussions, and only when the actual calculation is being done do we switch to mathematical mode, and "take the limit." During the 17th and 18th centuries, mathematicians and physicists both did this all the time. As they say in sports "no harm, no ...


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The question is inconsistent. At least one of the numbers (mass, force, stopping distance, or stopping time) is wrong. Your calculation of the acceleration from force and mass is correct, but an acceleration of $24$ m/s$^2$ for $2$ seconds means that the toy car was initially traveling at $48$ m/s. This is over $100$ mph ($160$ kph) and there is no way the ...


2

Differentiate $v$ with respect to time. On the left hand side you will get the acceleration $a$. On the right hand side you will get an expression which includes $s$ and $\frac {ds}{dt}$ which is $v$ which is ?


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During the acceleration phase the object's movement can be modeled with the quadratic curve $$x=x_0 + v_0t+\frac{1}{2}at^2 \qquad\text{where } x_0 \text{ is the initial position, and }v_0 \text{ is the initial velocity}$$ During the constant velocity phase, the object's movement can be modeled with the linear equation $$x=x_1 + v_1(t-t_1)$$ where $x_1$, ...


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The first equation holds good for average acceleration, but the second is the equation for uniform acceleration. The value obtained using option 1 is correct. In the time interval from 0 to 6 s, the acceleration changes (a constant value from 0 to 3 s and another constant value from 3 to 6 s). Then you cannot apply the uniform acceleration equation as ...


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If you draw a velocity-time graph you will see that you do not have enough information to find the initial and final velocity. The gradient of the graph is fixed because it is the acceleration. The distance is the area under the velocity-time graph. As you will see from the graph you can draw an infinite number of trapeziums (or triangles) which satisfy ...


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(Source of image: Mohsin Khan, http://cslearners.blogspot.com/2009/08/equation-of-motion.html) Here they are! All the formulas. Sorry to say, you cannot find anything if you have only acceleration and distance. Think like this, Say you have an object that has an acceleration of 2 m/s^2. and if i say that if travels a distance of 2 meters. It can travel ...


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Yes. It just means that the velocity is in a direction opposite the direction of your reference frame. If you make "down = positive" then $g$ would be positive, and so would the velocity.


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Accelerations being equal doesn't necessarily mean that the velocities are equal, or vice versa. For example, your two cars could have the same acceleration, but if one starts before the other, the one that got going earlier wlil obviously be moving faster. An even simpler example, if one car is standing still and the other one is moving at constant speed, ...


2

$$ a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} v $$ The physics notation makes the truth of the derivation above clear because $dx/dx$ cancels. In mathematicians-style presentation of the rule, it's the chain rule for the derivative of a composite function, $v'(t(x)) =\dots$


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shouldn't we use the parallel axis theorem ... to compute the moment of inertia? You could...if you already had the moment of inertia of the object about its center of mass. Since you don't, it's far easier to simply sum the moments of inertia about the $z$ axis.


1

Consider the picture below. In Cartesian coordinates $$\hat r=\cos\theta\hat i+\sin\theta\hat j,$$ and $$\hat \theta=-\sin\theta\hat i+\cos\theta\hat j.$$ Therefore $$\frac{d\hat r}{d\theta}=\hat\theta$$


1

From your diagram you have $x = y \tan \theta$. Taking the time derivative, and using the speed of the plane is $\dot{x} = v$ (taken as constant) you have $v = y \, \sec^2 \theta \, \dot{\theta}$ assuming that the height of the plane, $y$, is constant. So, for constant height (and taking the speed constant) you have $\dot{\theta} = C \cos^2 \theta$ where ...


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I'm too lazy to do the detailed math, but it's clear that, for constant velocity and constant rate of climb, it is possible to distinguish between a level path and a rising or falling path. Take point A as the intersection point of a level path and a rising path. At some time the aircraft both occupied A. Point B is any aircraft location on the level path ...


1

Mandelstam $t = (p_1 -p_1')^2$ corresponds to the square of the momentum transfered between the two scattering particles, in an elastic scattering process. If you look at the scattering in the centre of mass frame, like you suggest, then clearly they cannot transfer any energy between them, since that would vilolate conservation of momentum.


1

The equations look like the displacement of an object subject to zero acceleration (first equation) or the acceleration of gravity (second equation). If we assume that the first equation represents horizontal motion and the second, vertical motion (suggested by the 9.80 m/s$^2$), we can try to come up with a scenario. By inspection of equation (1) we see ...


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You have made some errors in you calculation of distances. Let $$ \mathbf{r}^{\prime}_{k}=\mathbf{r}_{k}-\mathbf{r}_{_{CM}} $$ Then \begin{aligned} r^{\prime\;2}_1 &= \Big(\frac{24}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 = \frac{1872}{121}\quad \text{(upper left particle)}\\ r^{\prime\;2}_2 &= \Big(\frac{20}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 ...


1

Suppose you take any quadratic equation: $$ y(t) = At^2 + Bt + C $$ for constant $A$, $B$ and $C$. The velocity is $dy/dt$ giving: $$ v = \frac{dy}{dt} = 2At + B $$ and the acceleration is given by differentiating again: $$ a = \frac{d^2y}{dt^2} = 2A $$ and since $A$ is a constant that means the acceleration has the constant value $a=2A$ i.e. any ...


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No. When the other object is far away from the observer the bearing will change very little; when the object is closest to the observer the bearing will change the most. The 'bearing rate' will only be constant (and equal to zero) when the object and observer are on a parallel course moving in the same direction or when they are moving on the same course ...


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The approach really fails, because the proposed motion is kind of broken. Including the integration constant, the distance traveled as a function of time becomes: $s(t) = \frac{1600}{(t+C)^2}$ Since we want $s=0$ when $t=0$, the constant C becomes infinity. This means that, the motion can not really start from rest -- it takes infinite time to do so. This ...


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I think that what you are asking is the position, velocity and acceleration of a space probe relative to the ever changing positions of the components which make up the Solar system? In the article cited below it describes the coordinate system as follows "Calculation of the trajectory of a space probe requires the use of an inertial coordinate system as ...


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If we assume that $g$ (acceleration due to gravity) is constant and parallel to our $y$ axis (but in the opposite direction), then we have: (Initial velocity=$v_0$, Launch angle=$\theta$, Initial position=$(x_0,y_0)$) $$a_y=-g\qquad (1)$$ $$\Longrightarrow \; v_y=-gt+v_0\sin \theta\qquad (2)$$ $$\Longrightarrow \; y=-\frac 12 gt^2+v_0 \sin \theta\; ...



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