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From Wikipedia

Acceleration

In physics, acceleration is the rate of change of velocity over time. In one dimension, acceleration is the rate at which something speeds up or slows down. However, since velocity is a vector, acceleration describes the rate of change of both the magnitude and the direction of velocity. Acceleration has the dimensionsL T−2. In SI units, acceleration is measured in meters per second per second (m/s2).

Proper acceleration, the acceleration of a body relative to a free-fall condition, is measured by an instrument called an accelerometer.

In common speech, the term acceleration is used for an increase in speed (the magnitude of velocity); a decrease in speed is called deceleration. In physics, a change in the direction of velocity also is an acceleration: for rotary motion, the change in direction of velocity results in centripetal (toward the center) acceleration; where as the rate of change of speed is a tangential acceleration.

In classical mechanics, for a body with constant mass, the acceleration of the body is proportional to the net force acting on it (Newton's second law):

\mathbf{F} = m\mathbf{a} \quad \to \quad \mathbf{a} = \mathbf{F}/m

where F is the resultant force acting on the body, m is the mass of the body, and a is its acceleration.

Average and instantaneous acceleration

Average acceleration is the change in velocity (Δ'v) divided by the change in time (Δt). Instantaneous acceleration is the acceleration at a specific point in time which is for a very short interval of time as Δt approaches zero.

The velocity of a particle moving on a curved path as a function of time can be written as:

\mathbf{v} (t) =v(t) \frac {\mathbf{v}(t)}{v(t)} = v(t) \mathbf{u}_\mathrm{t}(t) ,

with v(t) equal to the speed of travel along the path, and

\mathbf{u}_\mathrm{t} = \frac {\mathbf{v}(t)}{v(t)} \ ,

a unit vector tangent to the path pointing in the direction of motion at the chosen moment in time. Taking into account both the changing speed v(t) and the changing direction of ut, the acceleration of a particle moving on a curved path on a planar surface can be written using thechain rule of differentiation and the derivative of the product of two functions of time as:

\begin{alignat}{3}

\mathbf{a} & = \frac{d \mathbf{v}}{dt} \\ & = \frac{\mathrm{d}v }{\mathrm{d}t} \mathbf{u}_\mathrm{t} +v(t)\frac{d \mathbf{u}_\mathrm{t}}{dt} \\ & = \frac{\mathrm{d}v }{\mathrm{d}t} \mathbf{u}_\mathrm{t}+ \frac{v^2}{R}\mathbf{u}_\mathrm{n}\ , \\ \end{alignat}

where un is the unit (inward) normal vector to the particle's trajectory, and R is its instantaneous radius of curvature based upon the osculating circle at time t. These components are called the tangential accelerationand the radial acceleration or centripetal acceleration (see alsocircular motion and centripetal force).

Extension of this approach to three-dimensional space curves that cannot be contained on a planar surface leads to the Frenet-Serret formulas.

Special cases

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 g (also called acceleration due to gravity). By Newton's Second Law the force, F, acting on a body is given by:

\mathbf {F} = m \mathbf {g}

Due to the simple algebraic properties of constant acceleration in the one-dimensional case (that is, the case of acceleration aligned with the initial velocity), there are simple formulae that relate the following quantities: displacement, initial velocity, final velocity, acceleration, and time:

\mathbf {v}= \mathbf {u} + \mathbf {a} t
\mathbf {s}= \mathbf {u} t+ \over {2}} \mathbf {a}t^2 = \over {2}}

where

\mathbf{s} = displacement
\mathbf{u} = initial velocity
\mathbf{v} = final velocity
\mathbf{a} = uniform acceleration
t = time.

In the case of uniform acceleration of an object that is initially moving in a direction not aligned with the acceleration, 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, as in the trajectory of a cannonball, neglecting air resistance.

Circular motion

An example of a body experiencing acceleration of a uniform magnitude but changing direction is uniform <


From Encyclopedia

acceleration

acceleration change in the velocity of a body with respect to time. Since velocity is a vector quantity, involving both magnitude and direction, acceleration is also a vector. In order to produce an acceleration, a force must be applied to the body. The magnitude of the force F must be directly proportional to both the mass of the body m and the desired acceleration a, according to Newton's second law of motion, F = ma. The exact nature of the acceleration produced depends on the relative directions of the original velocity and the force. A force acting in the same direction as the velocity changes only the speed of the body. An appropriate force acting always at right angles to the velocity changes the direction of the velocity but not the speed. An example of such an accelerating force is the gravitational force exerted by a planet on a satellite moving in a circular orbit. A force may also act in the opposite direction from the original velocity. In this case the speed of the body is decreased. Such an acceleration is often referred to as a deceleration. If the acceleration is constant, as for a body falling near the earth, the following formulas may be used to compute the acceleration a of a body from knowledge of the elapsed time t, the distance s through which the body moves in that time, the initial velocity vi , and the final velocity vf :


From Yahoo Answers

Question:another question,..lol's......anybody can help me for this question?this is my assignment in physics.

Answers:Instantaneous acceleration is acceleration (change of velocity/time) at any given time. In layman's terms, it is the acceleration of a certain body (or particle) at any particular given or chosen instant. Hope this helps.

Question:Is centripetal acceleration an instantaneous acceleration, an average acceleration, both, or neither? Explain? This one has got me stumped because when I first consider it I would assume it is both because when dealing with an object in a circular path, the acceleration is always the same value and it's direction is always towards the centre of the circle meaning that centripetal acceleration would represent both instantaneous and average acceleration. Could anyone give me a hand?

Answers:Only if an object is traveling in uniform circular motion is centripetal acceleration both an average acceleration and an instantaneous acceleration. Uniform circular motion means constant speed, and a circular path of constant curvature. If an object travels in a non-circular path, the curvature changes with position, and so does the centripetal acceleration magnitude. If an object changes speed, centripetal acceleration magnitude changes. ALSO, even if an object travels in uniform circular motion, the direction of acceleration continuously changes. SO, using the full definition of acceleration, I'd say that centripetal acceleration ALWAYS is an instantaneous acceleration, EVEN in uniform circular motion.

Question:An object starts from rest at x=0 when t=0 and moves in the positive x direction. When the clock reads 6 sec the object is at x=+30m with an instantaneous velocity of 14m/s. Is acceleration constant? How do we know? I calculated (Sf-Si)/(Tf-Ti) = (30-0)/(6-0) = 5 m/s but what about the instantaneous velocity?

Answers:Let's try this again. All you've done is calculate the average velocity, which isn't helpful here. We know that velocity = Vo + acceleration x time, so let's figure out the acceleration assuming that it's constant: acceleration = velocity / time = 14 m/s / 6 second = 2.333 m/ss. Now, let's use one of the other formulas we know--say, the one that'll tell us the distance the object should have gone if indeed it had accelerated at that rate for six seconds. S = Vo time + 1/2 a time time = 1/2 x 2.333 x 6 x 6 = 41.999 m. Which is not the same as 30 meters. This means that (1) either our formulas don't work or (2) we've violated one of the assumptions under which we derived them. And so we have: the acceleration is not constant. Had it been, the formulas would have been consistent. There are other ways to do the same check. I'd suggest you check my work, because I won't.

Question:I am confused! I know instantaneous acceleration is tangent to acceleration graph. But then does that mean instantaneous velocity equal to acceleration since acceleration is tangent of velocity?

Answers:Instantaneous acceleration is the tangent of the **velocity** graph, not the tangent of the acceleration graph. That's probably why you're confused. If you have a graph of acceleration versus time, and you want to know the instantaneous acceleration at any particular time, just read the graph at that time - you don't need to measure the tangent or calculate the slope. You only need to use the tangent or slope if you need to figure out the acceleration from the graph of velocity versus time.

From Youtube

AP Physics B: Instantaneous Acceleration :A free lecture from our AP Physics B series at www.educator.com Other subjects include Algebra, Geometry, Pre-Calculus, Pre-Algebra, Trigonometry, Calculus, Biology, Chemistry, Statistics, and Computer Science. -All lectures are broken down by individual topics -No more wasted time -Just search and jump directly to the answer

3.006 Displacement, Average Velocity, Instantaneous Rate of Change and Acceleration :