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Acceleration
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== Special cases == ===Uniform acceleration=== {{See also|Torricelli's equation}} [[File:Strecke und konstante Beschleunigung.png|thumb|Calculation of the speed difference for a uniform acceleration]] ''Uniform'' or ''constant'' acceleration is a type of motion in which the [[velocity]] of an object changes by an equal amount in every equal time period. A frequently cited example of uniform acceleration is that of an object in [[free fall]] in a uniform gravitational field. The acceleration of a falling body in the absence of resistances to motion is dependent only on the [[gravitational field]] strength [[standard gravity|{{math|g}}]] (also called ''acceleration due to gravity''). By [[Newton's second law]] the [[force]] <math> \mathbf{F_g}</math> acting on a body is given by: <math display="block"> \mathbf{F_g} = m \mathbf{g}.</math> Because of the simple analytic properties of the case of constant acceleration, there are simple formulas relating the [[Displacement (vector)|displacement]], initial and time-dependent [[velocity|velocities]], and acceleration to the [[time in physics|time elapsed]]:<ref>{{cite book |title=Physics for you: revised national curriculum edition for GCSE |author =Keith Johnson |publisher=Nelson Thornes |year=2001 |edition=4th |page=135 |url=https://books.google.com/books?id=D4nrQDzq1jkC&q=suvat&pg=PA135 |isbn=978-0-7487-6236-1}}</ref> <math display="block">\begin{align} \mathbf{s}(t) &= \mathbf{s}_0 + \mathbf{v}_0 t + \tfrac{1}{2} \mathbf{a}t^2 = \mathbf{s}_0 + \tfrac{1}{2} \left(\mathbf{v}_0 + \mathbf{v}(t)\right) t \\ \mathbf{v}(t) &= \mathbf{v}_0 + \mathbf{a} t \\ {v^2}(t) &= {v_0}^2 + 2\mathbf{a \cdot}[\mathbf{s}(t)-\mathbf{s}_0], \end{align}</math> where * <math>t</math> is the elapsed time, * <math>\mathbf{s}_0</math> is the initial displacement from the origin, * <math>\mathbf{s}(t)</math> is the displacement from the origin at time <math>t</math>, * <math>\mathbf{v}_0</math> is the initial velocity, * <math>\mathbf{v}(t)</math> is the velocity at time <math>t</math>, and * <math>\mathbf{a}</math> is the uniform rate of acceleration. In particular, the motion can be resolved into two orthogonal parts, one of constant velocity and the other according to the above equations. As [[Galileo]] showed, the net result is parabolic motion, which describes, e.g., the trajectory of a projectile in vacuum near the surface of Earth.<ref>{{cite book |title=Understanding physics |author1=David C. Cassidy |author2=Gerald James Holton |author3=F. James Rutherford |publisher=Birkhäuser |year=2002 |isbn=978-0-387-98756-9 |page=146 |url=https://books.google.com/books?id=iPsKvL_ATygC&q=parabolic+arc+uniform-acceleration+galileo&pg=PA146}}</ref> ===Circular motion=== {{multiple image |align = vertical |width1 = 100 |image1 = Position vector plane polar coords.svg |caption1 = Position vector '''r''', always points radially from the origin. |width2 = 150 |image2 = Velocity vector plane polar coords.svg |caption2 = Velocity vector '''v''', always tangent to the path of motion. |width3 = 200 |image3 = Acceleration vector plane polar coords.svg |caption3 = Acceleration vector '''a''', not parallel to the radial motion but offset by the angular and Coriolis accelerations, nor tangent to the path but offset by the centripetal and radial accelerations. |footer = Kinematic vectors in plane [[polar coordinates]]. Notice the setup is not restricted to 2d space, but may represent the [[osculating plane]] plane in a point of an arbitrary curve in any higher dimension.}} In uniform [[circular motion]], that is moving with constant ''speed'' along a circular path, a particle experiences an acceleration resulting from the change of the direction of the velocity vector, while its magnitude remains constant. The derivative of the location of a point on a curve with respect to time, i.e. its velocity, turns out to be always exactly tangential to the curve, respectively orthogonal to the radius in this point. Since in uniform motion the velocity in the tangential direction does not change, the acceleration must be in radial direction, pointing to the center of the circle. This acceleration constantly changes the direction of the velocity to be tangent in the neighbouring point, thereby rotating the velocity vector along the circle. * For a given speed <math>v</math>, the magnitude of this geometrically caused acceleration (centripetal acceleration) is inversely proportional to the radius <math>r</math> of the circle, and increases as the square of this speed: <math qid=Q2248131 display="block"> a_c = \frac {v^2} {r}\,.</math> * For a given [[angular velocity]] <math>\omega</math>, the centripetal acceleration is directly proportional to radius <math>r</math>. This is due to the dependence of velocity <math>v</math> on the radius <math>r</math>. <math display="block"> v = \omega r.</math> Expressing centripetal acceleration vector in polar components, where <math>\mathbf{r} </math> is a vector from the centre of the circle to the particle with magnitude equal to this distance, and considering the orientation of the acceleration towards the center, yields <math display="block"> \mathbf {a_c}= -\frac{v^2}{|\mathbf {r}|}\cdot \frac{\mathbf {r}}{|\mathbf {r}|}\,. </math> As usual in rotations, the speed <math>v</math> of a particle may be expressed as an [[angular velocity|''angular speed'']] with respect to a point at the distance <math>r</math> as <math display="block" qid=Q161635>\omega = \frac {v}{r}.</math> Thus <math> \mathbf {a_c}= -\omega^2 \mathbf {r}\,. </math> This acceleration and the mass of the particle determine the necessary [[centripetal force]], directed ''toward'' the centre of the circle, as the net force acting on this particle to keep it in this uniform circular motion. The so-called '[[centrifugal force]]', appearing to act outward on the body, is a so-called [[pseudo force]] experienced in the [[frame of reference]] of the body in circular motion, due to the body's [[linear momentum]], a vector tangent to the circle of motion. In a nonuniform circular motion, i.e., the speed along the curved path is changing, the acceleration has a non-zero component tangential to the curve, and is not confined to the [[Principal normal vector|principal normal]], which directs to the center of the osculating circle, that determines the radius <math>r</math> for the centripetal acceleration. The tangential component is given by the angular acceleration <math>\alpha</math>, i.e., the rate of change <math>\alpha = \dot\omega</math> of the angular speed <math>\omega</math> times the radius <math>r</math>. That is, <math display="block"> a_t = r \alpha.</math> The sign of the tangential component of the acceleration is determined by the sign of the [[angular acceleration]] (<math>\alpha</math>), and the tangent is always directed at right angles to the radius vector.
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