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In special relativity the transition from one frame to another is given by the Lorentz boosts. This is not quite the same as an acceleration, but a transformation that relates observations on one frame with another. We can think of an acceleration as being a succession of infinitesimal Lorentz boosts that map one frame to another. The infinitesimal distance ...

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Everything is a function of the angle $\theta$ and its derivatives $\dot{\theta}$ and $\ddot{\theta}$. From there use the chain rule of differentiation. \begin{align} x & = \ell \sin \theta & y & = \ell (1-\cos \theta) \\ \dot{x} & = \ell \dot{\theta} \cos \theta & y & = \ell \dot{\theta} \sin \theta \\ \ddot{x} & = ... 1 As you noticed, if we use Euler-Lagrange equation on L= \frac 1 2 (\dot x^2 + \dot y^2) -mgy we get\ddot x=0\ddot y = -g$$Something is clearly missing: gravity is not the only force acting on our mass: we have to take into account the tension of the rod/string. But why doesn't it come out from the equations? The point is that system only has ... 3 It would be easier to answer your question clearly with a drawing. In the following, the angle coordinate of the pendulum is the angle it makes with the vertical line. When the pendulum swings right(left), the angle will be positive(negative). With this setting, I get the exact same answer as you by working out the equations of motion. However, there ... 0 Ok, lets' say that at a distance from the earth, 3 spaceships were constructed. There were two small ships, and one long mother ship. Each ship was programmed to accelerate in the same manner, such that a specific G-force curve is followed as motion begins, and this is continued onward up until the intended target velocity is achieved. Each ship is sent ... 0 Derivative is defined at a point not for an infinitesimal interval.$$\text {If $y=f(x)$} \Longrightarrow \; y’(x)|_{x=a}=\frac {dy}{dx}|_{x=a}=f’(a)=\lim_{\Delta x\to 0}\large{\frac {f(a+\Delta x)-f(a)}{\Delta x}}$$\large{\frac {d\vec \Omega}{dt}} doesn’t represent a fraction. That represents derivative of the \vec \Omega relative to t at a time ... 2 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, ... 0 To generate more power, an engine needs to be able to burn more fuel; laws of physics, chemistry and thermodynamics dictate this requires a larger displacement (bigger volume in which you burn the fuel). Larger volume = larger area over which you generate friction, more more importantly, more air being pulled through per stroke. Some engines with a lot of ... 0 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... 3 Tl;DR: It is mainly do to the skier's positioning on the ski and how being thrown off-balance by the avalanche affects it. Concerning the first part of the first question: 1) If all free falling objects accelerate at the same rate (this was on a fairly steep mountain section), why did we get "trapped" into the avalanche, when our acceleration already ... 0 You are right Jeff. Using a static number for torque, and no velocity dependent frictional term will give you an unreasonably increasing acceleration. As your intuition predicted, electric motors start out with high torque at rest and decrease to zero torque as speed increases to max speed (Good discussion here). According to your spec sheet, your motor has ... 0 I recommend using an accelerometer to find the experimental acceleration of the robot. Probably the best quality would be to use a LabQuest and/or LoggerPro paired with the Vernier Motion Sensor, but you could also use SparkFun accelerometers alongside an arduino/RPi. Another option, which very well may be the best given constraints for your time and ... 1 For the first part of your question, you have to realize that your net velocity (the one that you plug into the expression for centripetal force) is the vector sum of the surface velocity and your velocity relative to the surface. If you were running West as fast as the earth turns East, you would "stay in place" and the sun would appear to stop moving in ... 1 Any object with mass that accelerates (is it linear or angular acceleration) produces gravitational waves, though in most occasions those will be much too small to be detected. As @CuriousOne pointed out, same happens with electromagnetic waves and accelerating charges. The gravitational waves that can be detected usually come from very massive objects (such ... 1 One promising way of implementing arbitrary physics simulations, is by programming in terms of 'constraints'. I highly recommend reading this article that covers constraints very well: http://gamedevelopment.tutsplus.com/tutorials/simulate-tearable-cloth-and-ragdolls-with-simple-verlet-integration--gamedev-519 I was very surprised when I first saw this ... 2 The acceleration should be$$a = \frac{G\cdot M}{r^2 \cdot \sqrt{1-r_s/r}}$$with r as the height above the center of mass and the Schwarzschildradius$$r_s = \frac{2\cdot G\cdot M}{c^2}$$The force to hold the ball at rest is$$F=m\cdot a$$As one can see it now takes an infinite force and energy to keep a body at a fixed height when r=r_s. 0 Average acceleration is defined the same way as average velocity : Average velocity is change in displacement / change in time. Average acceleration is change in velocity / change in time. Your 1st calculation gives the constant acceleration which would give the same change in velocity in the same time. This is correct. Your 2nd calculation gives the ... 0 If I understand correctly, the speed of the vehicle as it passes a particular point must be one of the values in a list. The list is different for each point. The overall time is minimized by choosing the maximum speeds possible at each "checkpoint", subject to the constraints. I think what you might be missing is that, if the starting speed on a section ... 2 During the acceleration phase the object's movement can be modeled with the quadratic curve$$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'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 ... 2 I think you are confusing the escape velocity with terminal velocity. While for the local velocity the limit is only the speed of light, terminal velocity is achieved much sooner because of the air resistance. The equation can be found here and depends on the shape, size and density of the asteroid. For particles travelling near the speed of light you have ... 2 For the sake of simplicity (at the expense of real-life accuracy), let's assume that an asteroid is already travelling at some speed v_0 directly toward the Earth, and it never deviates away from that direct line path (despite the revolution of the Earth around the Sun). Let's also assume the distance from Earth is very large compared to the radius of ... -5 No, you won't speed up and the fastest speed the earth can accelerate you to is at escape velocity. 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 $$and since A is a constant that means the acceleration has the constant value a=2A i.e. any ... 0 To understand the acceleration, you need to know the parametric equation of position on a circular uniform motion. r(t)=rcos(ωt)ux + rsin(ωt)uy (parametric equation of position on c.u.m.) The velocity is a derivate of position: v(t)=-rωsin(ωt)ux + rωcos(ωt)uy The acceleration is the derivate of velocity: a(t)=-rω^2cos(ωt)ux - rω^2sin(ωt)uy In the ... 0 This is the same as the problem of a projectile launched on an inclined plane. The link below shows you how to solve it to find range and time of flight. You will need to adapt your problem to fit that described. O is the point of launch (A) and P is the target (B). OP is the range. In this scheme acceleration is  g=A  along the -y axis. I think ... 0 At every point the tangential direction is the unit vector of the velocity vector. If you have the velocity components \boldsymbol{v} = (\dot{x}, \dot{y}) at every instant, the you decompose this into a magnitude (speed v) and direction \hat{\boldsymbol{e}}$$ \begin{align} v & = \sqrt{\dot{x}^2+\dot{y}^2} \\ \hat{\boldsymbol{e}} & = ...

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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: $$... 0 It turns out you cannot use an accelerometer to determine the attitude of a rocket at any time other than when it is experiencing the normal force from the earth. This is explained in some detail in this article: Thinking About Accelerometers and Gravity. The key point from the article is this line: "An accelerometer never senses gravitational acceleration ... 0 If you have found that the initial angular acceleration of the disc is \alpha = \frac {2g}{3R} that must mean that the initial centre of mass acceleration a = \frac {2g}{3} because a = R \alpha. If the force acting on the pivot is X upwards then applying Newton's second law for the centre of mass motion gives mg-X = ma, so X = \frac {mg}{3}. ... 1 I would think that the difficulty is related to gravity. Without gravity, water will go everywhere in all directions. 1 There's a physical reason why the block slides down along the slope of the inclined plane. If you perform this experiment, and if frictional force is small enough, you will observe that the block slides down along the slope of the plane. It will never move in a direction perpendicular to the slope. For it to never move in the perpendicular direction, we ... 1 The normal force is not playing a role in this case because the force is perpendicular to the moving direction of the box. By "not playing a role" I mean, during this motion, \ F_N is only be used to balance out \ mgcos\theta  so that the object won't be able to move "into the inclined plane". To find the force causing the block to accelerate down, we ... 1 Since acceleration is a vector you can decompose it in the coordinate system you find convenient. If you define a cartesian coordinate system whose axis are along the normal to the plane and the plane itself you see there is a component of the acceleration g\sin\theta along the plane. This is why the block accelerate in this direction. Notice that along ... 1 Astronauts do not feel weightless on the Moon where the gravitational field strength is one sixth of that on the Earth. They feel "lighter" in spite of the heavy suits that they have to wear. A spring balance would give a reading for the weight of an astronaut on the Moon which was one sixth of that on the Earth. An astronaut in orbit feels weightless and ... 0 Gravitational acceleration is a vector and thus has a direction. But, at the center of a spherically symmetric mass distribution, there is no preferred direction while, if there were a non-zero gravitational acceleration vector, it would point in a direction. Now, if one rotated the mass distribution through some angle, the acceleration vector should ... -1 Intuitively, of course, it makes sense that the force in all directions at the center of a massive body would be equal. There are a few ways of formalizing this. The most rigorous would be an integral:$$G\rho\int_{0}^{r}\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\vec{r}}{|r|^3}|r|^2sin(\phi)d\theta d\phi dr where $\rho$ is the mass density. You'll need to ...

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If you measure a force (weight) for a given acceleration (gravity) in order to determine the mass of an object and you haven't started measuring then the mass is undefined. As soon as you apply an acceleration $a>0$ and you measure corresponding force $F>0$ you can determine the mass. Equations are useful only when they can be used to measure things, ...

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