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If acceleration of something $= - 10 \text{ m s}^{-2}$

And forwards is define as north.

Does that mean the object is getting slower (decelerating) or accelerating in the reverse direction (south)

How can you tell the difference?

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This is a very fundamental question that I hear a lot from beginning students, and I don't really know where ti fits into our convention on not answering basic exercises. Input from the commentariet would be appreciated. – dmckee♦ Sep 9 '11 at 15:34
@dmckee, that convention is quite annoying, I'm trying to learn more and get to the harder stuff but its quite hard when moderators decided to close questions based on their level. Stackoverflow has many basic questions that are answered, and still retains it's experts. – Jonathan. Sep 9 '11 at 16:20
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@dmckee: if you're talking about the convention of not giving away complete answers to homework questions, I don't think that applies here as this is not a homework question. Jonathan: we never close questions based on level (at least, I don't, though I suppose I can't speak for the other mods); we close questions for being poorly formulated. See meta.physics.stackexchange.com/q/76, meta.physics.stackexchange.com/q/209, meta.physics.stackexchange.com/q/1, and others – David Zaslavsky♦ Sep 9 '11 at 18:13
I guess I misunderstood the question, I deleted my answer and comment. What is the question? If you have an acceleration vector, it tells you how the velocity changes. If you have a velocity it tells you how the position changes. It is the exact same relation. Are you mystified by how a particle which is at -10 can be moving both in a positive and negative direction, which increases the distance from the origin in one case, and decreases it in another? – Ron Maimon Sep 9 '11 at 20:08

2 Answers

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Does that mean the object is getting slower (decelerating) or accelerating in the reverse direction (south)

It really doesn't matter. Basic kinematic formulas are designed to work just as well in either case, which is why physicists don't generally use the word "decelerating." It's just another kind of acceleration.

That being said, if you want to determine whether the object's speed is increasing or decreasing (which correspond to the popular meanings of "accelerating" and "decelerating" respectively), you can just look at the orientation of the acceleration with respect to the velocity. If the acceleration is parallel to the velocity, the object will be speeding up. If it's antiparallel, the object will be slowing down. You can see this mathematically by taking the derivative of the kinetic energy:

$$\frac{\mathrm{d}}{\mathrm{d}t}\biggl[\frac{1}{2}mv^2\biggr] = m\vec{v}\cdot\frac{\mathrm{d}\vec{v}}{\mathrm{d}t} = m\vec{v}\cdot\vec{a}$$

So the sign of the dot product $\vec{v}\cdot\vec{a}$ tells you whether the speed is increasing or decreasing.

Do note that velocity is reference frame-dependent. So two different inertial observers looking at the same object at the same time could have differing conclusions as to whether it is speeding up or slowing down. That's one big reason why the distinction is not important in physics.

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so you must be given velocity for it to mean anything? – Jonathan. Sep 9 '11 at 19:01
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@Jonathan. Not at all. Acceleration always has meaning, but if you don't know the velocity you don't know if the acceleration will result in a speed increase, a speed decrease, or a change of direction. For example, the Earth is constantly experiencing acceleration towards the sun (so its velocity is always changing), but its speed remains constant (or close to constant anyway, since it does not have a perfectly circular orbit, and other bodies have some impact too). – Kevin Cathcart Sep 9 '11 at 19:30
I'm sorry for such stupid questions, but what use is acceleration if you don't know if it's a speed increase, decrease or change of direction? – Jonathan. Sep 9 '11 at 19:39
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It lets you calculate force on a constant-mass object, for one thing. But the main reason acceleration is useful is that, in combination with initial velocity, knowing acceleration enables you to calculate the entire future motion of a non-quantum object. And yes, in that case you could know whether the object is speeding up or slowing down (or neither), but as far as determining the future motion, it doesn't matter. The equations don't care what kind of change the acceleration represents. – David Zaslavsky♦ Sep 9 '11 at 19:45
@Jonathon: as is hinted in dmckee's answer below, another great thing about acceleration is that it doesn't care whether it is being measured from the ground, from a car moving at constant velocity relative to the ground, or from a plane moving at constant velocity in the other direction relative to the ground--all three people will agree on the acceleration of the object. – Jerry Schirmer Sep 10 '11 at 0:41
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The answer depends entirely on what you define as stopped{*}.

Moreover, physics must be the same in all inertial reference frames{+}, which means that it can not depend on the distinction you are trying to make.

Accordingly there is no physical difference, just a linguistic one that relies on a common understanding of "stopped".

{*} Growing up on the surface of a planet with a non-trivial atmosphere you are probably used to thinking of the frame of the ground as special for these purposes, but physics doesn't believe in that.

{+} The principle of relativity.

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so there is no difference between acceleration and direction? I'm doing A-level at school, I know it may not fit your convention but could take it down a level (NPI). (You had to start somewhere too, you weren't just born knowing all about particle physics) – Jonathan. Sep 9 '11 at 16:22
@Jonathan: just a minor nitpick -- this really isn't particle physics at all. This is a pretty basic question about vectors and linguistic conventions. – Colin K Sep 9 '11 at 18:33
@Colin, I know but dmckee is particle physicist. – Jonathan. Sep 9 '11 at 18:57
@Jonathan: There is no (can not be a) difference between accelerations that are parallel, anti-parallel, or orthogonal to the velocity because the there is no unique frame in which to measure velocity (this is not, BTW something new, Galileo relied on this principle). Thus any distinction you make between "acceleration" and "deceleration" is purely a linguistic one based on the frame you have chosen to measure the velocity in. – dmckee♦ Sep 10 '11 at 0:05
Mind you, you can make the distinction in whatever frame you chose and it sometimes makes communication easier, just be aware that the distinction has no physical significance divorced from your choice of reference frame. – dmckee♦ Sep 10 '11 at 0:10

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