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# example problem of law of acceleration

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&nbsp;Tâˆ’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):

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 (Î”'v) divided by the change in time (Î”t). Instantaneous acceleration is the acceleration at a specific point in time which is for a very short interval of time as Î”t 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 :

Question:I have got a physics problem which is described below. There is a ring which radius is R. A little ball moves inside this ring. Ring's flat is normal to the surface of ground. When a ball is moving inside a ring (ring is in quiet), the ball reaches a height which is equal to R/2. The ring starts to move plumb with a fixed acceleration. What is the value of the fixed ring acceleration, if the ball inside the ring reaches the top of the ring? (I know the answer of this problem - acceleration is equal to 4g/5 and a ring moves down with this acceleration. How to achieve this answer?) Thanks in advance. Sorry for my bad English :(. "Ring's flat" is Ring's plane. So "Ring's plane is perpendicular to the surface of ground" "The ring starts to move plumb" means that the ring starts to move upright. Hope this will help to understand my question.

Answers:Under these conditions, the acceleration w/r/t the ring is g/5 downward. What's not clear from the question is where is the ball when the ring acceleration starts? It makes a big difference. For example, if the ball is at R/2, it has no kinetic energy w/r/t the ring. Since acclereation of the ring creates a relative gravit of g/5, it would still osccilate within the ring at R/2. If the ball is at the bottom of the ring, then there is kinetic energy when it starts of m*g*R/2 (I am using this versus .5*m*v^2). This will get converted to potential energy at a rate of m*(g-a)*h, or m*g/5*h. Since the translational distance within the frame is 2*R, then m*g*R/2=m*(g-a)*2*R simplify g/2=2*g-2*a I get a= 3*g/4 If the problem is also to keep the ball in the ring so it doesn't fall, then it has to have kinetic energy at the top to counter act the force of gravity relative to the ring. j

Question:hi folks! need sample problems with sol'n about uniformly accelerated motion with 2 object as given. tnx

Answers:consult a physics book.. they have a lot of problems there. =D

Question:I am very confused about these word problems for Physical Science. In these word problems they ask me to find the velocity or time but only give me a small part of info. Ex) What is the velocity of a quarter dropped from a tower after 10 seconds? Well I know Velocity= Distance/Time , but they only give me time. My teacher said if something is dropped or falling, then use 9.8 meter/second/seconds. I'm not sure what she meant. Can you please explain how to solve the example problem and where I would plug in 9.8 M/S/S. That helped with that particular question but the net one says, If a block of wood dropped from a tall building has attained a velocity of 78.4 m/s, how long has it been falling? Now I know T=D/V but if I don't have the time then how can I figure out the distance?

Answers:Under accelerated motion (in this case free fall without an initial velocity) the distance an object falls is 1/2 x g x time squared (where g = 9.80 m/s^2) So distance = 1/2 x 9.80 x 10^2 = 490 m.

Question:I noticed that there are two equations for speed and three for acceleration: Speed: 1. Speed= distance/time 2.Average Speed= distance covered/time interval How do I know which ones to use when solving a speed problem? Acceleration: 1. A=change in velocity/time interval 2. A=change in speed/time interval 3. A=net force/mass When solving an acceleration problem, how would I know which equation to use? Please keep the explanations simple and add examples in necessary. Thank you!!

Answers:It sound like you're very new to doing physics problems. What you need is a paradigm for working problems. Here's what I do when answering questions in this forum. First, and foremost, identify the physics involved. In HS physics there are just a few options: conservation of momentum, conserations of energy, and Newton's laws of gravity and motion. So learn to recognize what physics you are dealing with first. [In this case, you are dealing with Newton's laws of motion.] Second, put down everything you know about the problem. For example, if it's asking for distance (S), write down all the distance bearing equations you can think of. S = vt and v^2 = u^2 + 2aS, for instance, are two popular distance carryiing equations. They are some of the so-called SUVAT equations you should be intimately familiar with in HS physics. S is distance, U is initial velocity, V is final velocity, A is acceleration, and T is time. Finally, look at the equations to identify what's missing in value and, of course, which factor is the variable you are trying to solve (like acceleration a). For example, you are looking for acceleration, set up the last equation as a = (v^2 - u^2)/2S and now you know you need to have the two velocities and a distance to solve for a. If one or more of these factors are missing, you must come up with additional relationships or equations with them in them. Once you've found the v, u, and S, just plug them in to solve for a. Bottom line: ID the physics first. Write all the equations you can about that line of physics. ID the variable you are looking for and rewrite the equations so the variable (like a) is on the LHS of the = sign. ID what missing from the RHS of the = sign; and solve for those missing values using additional equation as needed.