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I am a little confused as to what the magnitude of acceleration is and what it means.

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  • $\begingroup$ it would be clear if you know calculus. $\endgroup$ – Shing Oct 26 '16 at 15:59

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Your question is kind of vague but I will try to respond. Acceleration is defined as the time rate of change of velocity. Since velocity has both magnitude and direction, so does acceleration. In other words, acceleration is a vector. The length of the vector is its magnitude. Its direction is the direction of the vector. So the magnitude of acceleration is the magnitude of the acceleration vector while the direction of the acceleration is the direction of the acceleration vector. This is, of course, true of all physical quantities defined as having a magnitude and a direction. As an example, if a car is traveling north and accelerating at a rate of 10 feet per second per second, then the magnitude of the acceleration is 10 feet per second per second and the direction of the acceleration is north. If the car was traveling south but accelerating at the same rate, then the magnitude of its acceleration vector would be the same but its direction would be south.

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Acceleration is simply a rate of change of velocity.

So the magnitude tells you, how quickly velocity changes.

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If you are talking about linear motion, then the magnitude of acceleration is simply a measurement of change in speed per unit time. As an example, say you are in a car starting from rest and you begin to speed up. Say that you reach a speed of $20 {m \over s}$ in $2$ seconds. This means the magnitude of your acceleration is: $$ a = {20 {m \over s} \over 2s} = 10 {m \over s^2}$$ That is, your speed changed by $20 {m \over s}$ every $2$ seconds, or $10 {m \over s}$ every second. Thus, when we talk about the magnitude of acceleration, we are talking about how quickly your speed changes in a given unit of time.

It is important to note that this is only the magnitude of acceleration. Acceleration is a vector, meaning it has both magnitude and direction. Therefore, the magnitude only describes part of any accelerated motion. Also, as is pointed out in a comment below a more precise definition of acceleration is needed when talking about nonlinear motion.

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  • $\begingroup$ I don't agree with this, or at least not in the general case of non-linear motion. Magnitude of acceleration is not change in speed per unit time, it is the absolute value of the change in velocity per unit time. These quantities are different, for example, in uniform circular motion, where $\left | \vec a \right | = v^2/r,$ but $\left | d|\vec v|/dt \right |=0$. $\endgroup$ – BMS Jul 24 '14 at 15:35
  • $\begingroup$ @BMS thank you for pointing that out. I've edited my answer to emphasize that my answer is not precise enough for nonlinear motion $\endgroup$ – wgrenard Jul 26 '14 at 16:39
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In the context of linear motion (as BMS correctly points out in a comment of a different answer), the magnitude of acceleration is a measure of how much speed you are gaining per second.

The difference with the acceleration vector is that the vector form also encapsulates the direction in which this gain in speed is happening.

So as an example an acceleration magnitude of $2$ $m/s^2$ means that every second your speed is $2$ $m/s$ higher. Therefore, if my starting speed is $0$ $m/s$, my speed after 1 second is $2$ $m/s$, $4$ $m/s$ after 2 seconds, $6$ $m/s$ after 3 seconds and so on...

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When a particle moves along a prescribed path, with tangent vector $\hat{e}(t)$ and normal vector $\hat{n}(t)$ then the velocity and acceleration vectors are decomposed as such:

$$ \vec{v} = v(t) \hat{e}(t) $$ $$ \vec{a} = \dot{v}(t) \hat{e}(t) + \frac{v(t)^2}{\rho(t)} \hat{n}(t) $$

which is interpreted as

  1. The magnitude of the velocity vector is the speed along the path.
  2. The direction of the velocity vector is tangent to the path.
  3. The magnitude of the acceleration vector along the path is the time rate of change of speed.
  4. The magnitude of the acceleration vector normal to the path is the centripetal acceleration as it goes around the instantaneous radius of curvature $\rho(t)$.
  5. The combined magnitude is the combination of the above and does not have a direct interpretation.

See https://physics.stackexchange.com/a/99570/392 for more details.

Note that item 3 forms a screw vector field, but 4 does not.

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A. Putting it physically

The magnitude of acceleration tells you how much the rest of the world is affecting a particle's state of motion.

One curious characteristic of our universe is that it has a natural state of motion. If a particle is left alone, such particle will:

  1. Move along a straight line.
  2. It’s linear velocity won’t change, i.e., the rate of change of its velocity along that straight line will remain 0 (zero).

This is known as Newton’s first law or Galileo's law of inertia.

So, every time the rest of the world messes (actual word is interact) with a particle, it will cause any or both of these conditions to change. Now, any time these conditions change the magnitude of acceleration will change as well, because

Magnitude of acceleration = Rate of change of in the magnitude of velocity + Rate of changing the direction of motion

The rate of change of in the magnitude of velocity is known as linear acceleration (let it be $alinear$), and the rate of changing the direction of motion is known as centripetal acceleration (let it be $acurve$).

B. Putting it mathematically:

$$alinear = \frac{v2-v1}{t2-t1}$$ $$acurve = \frac{ \theta 2 - \theta 1}{t2-t1}$$

where,

$v$ is the magnitude of the velocity

$t$ is the time

and $\theta$ is the angle between the first and the second direction.

quantities with subscript 2 represent the final states, and those with subscript 1 represent initial states.

if one knows calculus these equations should be put as $$alinear = \frac{dv}{dt}$$ $$acurve = \frac{d \theta}{dt}$$

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When you are asked to "find" or "use" the "magnitude of the acceleration," what it means is that you need not be concerned with its direction, just its value! For example: a boat on a river is traveling at 4 m/s, and there is a cross wind pushing it at 3 m/s. What is the resultant magnitude of the velocity of the boat? Answer: V = $(4^2 + 3^2)^{1/2}$ = 5 m/s. The same applies to acceleration.

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Magnitude refers to size or quantity alone. When it comes to movement, magnitude refers to the speed at which an object is traveling or its size.

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In physics, magnitude is the size of a phusical object, a property by which the object can be compared as larger or smaller than other objects of the same kind. More formally, an object's magnitude is an ordering (or ranking) of the class of objects to which it belongs.

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Acceleration is basically : (final velocity - initial velocity) / change in time.

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  • 2
    $\begingroup$ Hi and welcome to Physics.SE. While this does answer the question, it's not clear to me what it adds above the already-existing answers. $\endgroup$ – rob Oct 26 '16 at 18:47

protected by rob Oct 26 '16 at 18:48

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