# Tag Info

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It's because of the product rule of derivatives, that states d(fg(t))/dt=f(t)(dg/dt)+g(t)(df/dt). In this case, let's call this part: (-isinθ+jcosθ)=g(t), and solve it: d(rωg)/dt=r(dωg/dt)=r(ω*(dg/dt)+g(dω/dt)). If you substitute g for (-i*sinθ+jcosθ), you get the equation for acceleration your book presents.

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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, ...

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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 ...

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there are ways to find the velocity- you can look at the distance and time graph of the motion of the ball and the slope at a point of time will give ya the velocity. If you are looking at the acceleration, well the initial velocity must be zero I guess because it starts at rest...

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I think what confused here is the difference between derivative and partial derivative. For example, $$\frac{d f(x)}{d t} = \frac{\partial f}{\partial x} \frac{d x}{d t}$$ But, $$\frac{\partial f(x)}{\partial t} = 0$$ So, in your question, you just do derivative on both side and notice that $\frac{d x}{d t} = v_x$ and $\frac{d v_x}{d t} = a$. Then, you ...

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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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Your equation for the $\mathbf{v_{AB}}$, the velocity of $\mathcal{A}$ relative to $\mathcal{B}$ with the following formula is correct and general for all 'real' values of $v_{X_{i}}$ where $X=A, B$ and $i=x, y, z$. ...

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Calm down. aren't there an infinite number of distances. This Statement is necessarily false. Distance is an scalar and is unique with a magnitude. It is true that that the possible distances are infinite but at a particular time (say t = 10 min) there will be an unique value of distance. The answer to what term do we use to refer to the rate of ...

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You are overthinking this. Speed is rate of change of distance. Velocity is rate of change of displacement. Because distance and displacement are not necessarily the same, speed and distance are not necessarily the same either. Yes, you are correct - many different distances could correspond to the same displacement, so one particular velocity could ...

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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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Assuming constant acceleration You have to know the time also. If you know the distance traveled $s$ after time $t$ then you can write $$s = v_0 t + \frac{1}{2} a t^2$$ and solve for the initial velocity $$v_0 = \frac{s}{t} - \frac{a t}{2}$$ Once the initial velocity is known, then the final velocity is $$v_1 = v_0 + a t$$

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No, that's not possible. Unless the body is starting or ending at rest in which case $v_f$ or $v_i$ would be zero and you could substitute in $0$ in the eq: $v_f^2 = v_i^2+2aS$ Intuitively this makes sense. If the acceleration and distance for which the body accelerates are known, you can only determine $v_f^2- v_i^2$ You don't know whether $V_i = 0m/s$ and ...

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Under constant acceleration, the relationships between velocity, acceleration, distance and time are: \begin{align}v_t &= v_0 + a_t\\ x_t &= x_0 + v_0 t + \frac12 a t^2\end{align} When a and d ($x_t$) are given, you are left with two equations and three unknowns: $v_0$, $v_t$ and $t$. This means you cannot come up with a numerical (unique) ...

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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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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=$(x_0,y_0)$) $$a_y=-g\qquad (1)$$ $$\Longrightarrow \; v_y=-gt+v_0\sin \theta\qquad (2)$$ $$\Longrightarrow \; y=-\frac 12 gt^2+v_0 \sin \theta\; ... 0 Average acceleration is defined the same way as average velocity : Average velocity is change in displacement / change in time. Average acceleration is change in velocity / change in time. Your 1st calculation gives the constant acceleration which would give the same change in velocity in the same time. This is correct. Your 2nd calculation gives the ... 0 If I understand correctly, the speed of the vehicle as it passes a particular point must be one of the values in a list. The list is different for each point. The overall time is minimized by choosing the maximum speeds possible at each "checkpoint", subject to the constraints. I think what you might be missing is that, if the starting speed on a section ... 1 Mandelstam t = (p_1 -p_1')^2 corresponds to the square of the momentum transfered between the two scattering particles, in an elastic scattering process. If you look at the scattering in the centre of mass frame, like you suggest, then clearly they cannot transfer any energy between them, since that would vilolate conservation of momentum. 2 During the acceleration phase the object's movement can be modeled with the quadratic curve$$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, ... 0 The average value of acceleration should just depend on the initial and final velocity and the time interval between them. Since the average value of a function over the interval a to b is the integral of the function from a to b divided by (b-a), and since the integral of acceleration gives you velocity then if the limits are t_1 and t_2 the average ... 2 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 ... 0 I assuming that you wish to find the distance from the thrower as a function of time. Suppose the thrower is at the origin, and the boomerang moves anti-clockwise in a circle of radius a with centre at (0,a) with angular speed \omega. Then the position of the boomerang at time t is (x,y) where x = asin(\omega t) y = a(1-cos(\omega t)). The distance ... 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 ... 3 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 ... 0 I'm getting a different formula. Please someone point out my error if you see it. So the host vehicle is to the left of the target vehicle. In order to avoid collision, in the worst case scenario, by the time the host vehicle decelerates to the target vehicle's velocity, the target vehicle is still to the right of the host vehicle. We're assuming positive ... -1 if we assume the initial velocity of host car to be u and initial velocity of other car as i relative velocity of other car with respect to te first one would be (u+i). Let initial separation be d applying equation of kinematics v^2 = u^2 + 2as 0 = (u+i)^2 + 2ad so, a=-(u+i)^2/2d Now, required deceleration can be calculated! Moreover, if the same ... 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}$$As a matter of fact, mathematicians prefer different notations for the derivative of a function f, as D f or f'. But the above definition of derivative became rigorous only when the concept of ... 0 Using the word fraction is a more elementary way of naming the set of rational numbers : numbers that are represented by a ratio of integers. In physics when we model physical systems with differential equations we generally work in the domain of real numbers and sometimes complex numbers since we tend to think of real physical systems as existing in a ... 0 I'd say the premise is the other way around: We assume \Delta A to be small, as we intend to make it infinitesimal later on, and so \Delta \theta is also small. Also, when you look at \Delta A\Delta \theta, this is even smaller, or "quadratically small" so to say. When doing math with small deltas or infinitesimals, you usually only have to take the ... 3 In physics there are no infinitesimals, so dx is always treated as a "small but finite interval" during discussions, and only when the actual calculation is being done do we switch to mathematical mode, and "take the limit." During the 17th and 18th centuries, mathematicians and physicists both did this all the time. As they say in sports "no harm, no ... 4 I think such a function may only exist in the Maxwell-Boltzmann limit. Here's why: For simplicity let us parametrize everything in terms of \beta = 1/T and denote Z(\beta) = \int{d^3p\; f_{eq}(p, \beta)}. Rewrite the latter as$$ 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 ...

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 ...

0

This is my guess for 1D motion with air resistance (drag) + wind blowing in negative-$x$ direction and projectile moving in the positive-$x$ direction: $$\ddot x(t) = - D (v+v_w)^2$$ $$\dot x(t) = v$$ where $v_w$ is the wind velocity and $D$ is a constant. The point is that in my opinion you cannot take wind into account simply with a constant force ...

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The formula is $$v = \varphi\cdot r/t$$ Where v is the velocity, φ the traversed angle in radians, r the distance and t the time, if the object is moving in a transversal direction. If it is approaching or receding you have to split the radial and transversal components with Pythagoras. In your example I get around 1140 ft/sec.

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This is the same as the problem of a projectile launched on an inclined plane. The link below shows you how to solve it to find range and time of flight. You will need to adapt your problem to fit that described. O is the point of launch (A) and P is the target (B). OP is the range. In this scheme acceleration is $g=A$ along the -y axis. I think ...

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The displacement in x direction will be a straight line all the time including the travel on the incline plane. There is no reason it will change its value. The displacement in y direction will be zero on the horizontal plane and change in parabolic because of gravity.

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The straight line should be followed by a continuous, downward parabolic curve. There is a gradual change between the straight line and the parabola. Because while the block is on the plane, the x coordinate is still increasing with time but now at a decreasing rate because of the acceleration $gsin(\theta)$ along the plane, opposite to the velocity of the ...

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Your first calculation is correct. Nothing wrong with it. The mystery for me is why you go to so much trouble trying to disprove it!

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At every point the tangential direction is the unit vector of the velocity vector. If you have the velocity components $\boldsymbol{v} = (\dot{x}, \dot{y})$ at every instant, the you decompose this into a magnitude (speed $v$) and direction $\hat{\boldsymbol{e}}$ \begin{align} v & = \sqrt{\dot{x}^2+\dot{y}^2} \\ \hat{\boldsymbol{e}} & = ... 0 The easy way of doing this is to parametrice the trajectory. We have the cartesian definition, so let  \textbf{r} : \mathbb{R} \rightarrow {\mathbb{R}}^{2}  be: \textbf{r}(t)=(t,f(t)) \quad\quad t\in (-\infty,\infty) $$So we got the vector position as function of "time". Then, the velocity and acceleration vectors are defined by:$$ ...

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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 ...

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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.

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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 ...

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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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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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To calculate the minimum energy needed for the reaction the products are assumed to be stationary, i.e. the momentums are zero. With $\pmb p^2 = E^2 - \vec p^2 = E^2 = m^2$ follows: ({p_1^\mu}' + {p_2^\mu}' + {p_3^\mu}' + {\bar p_4^\mu}')^2 = (E_1 + E_2 + E_3 + E_4)^2 - (\vec p_1 + \vec p_2 + \vec p_3 + \vec p_4)^2 = (E_1 + E_2 + E_3 + E_4)^2 = (m_p + ...

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This is a standard projectile motion problem with the complication that the projectile collides elastically with vertical walls. Assuming that the only force exerted on the projectile by the walls is in a direction which is normal to the walls then: the vertical velocity does not change, and the horizontal velocity reverses direction but the magnitude ...

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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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