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What is the quantity in physics that is advance from speed?

i know one is velocity, one is speed, one is acceleration.

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closed as not a real question by Qmechanic, Ron Maimon, Colin K, dmckee Aug 14 '12 at 17:00

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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It seems logical that speed is related to speed... –  Fabian Aug 12 '12 at 18:15
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I am at a loss to determine what the question here is. In particular I don't understand "is advanced from" as it is used here. Can someone clarify the question? –  dmckee Aug 12 '12 at 21:38
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1 Answer

up vote 2 down vote accepted

To give an overview of the different measurements encountered in kinematics (as there doesn't seem to be a real, precise question asked here).

The most familiar quantity we encounter in kinematics is speed. Speed is abstractly how fast something is going. It does not take into account direction, and is therefore known as a scalar quantity. It is what we measure on the speedometer of a car. Speed is often denoted by $v$ in physics.

Velocity takes into account both speed and direction. It is a vector quantity, and is often denoted by $\vec{v}$ or $\mathbf v$ in physics. Velocity is the rate of change of distance with respect to time, which in calculus notation is written:

$$\vec{v}=\mathbf{v}=\frac{d\vec{s}}{dt}$$

It is also useful to note that if we know the velocity, it is easy to find the speed, as we have the simple relationship:

$$v=\|\vec{v}\|=\|\mathbf{v}\|$$

Where $\|\cdot\|$ denotes the euclidean norm.

Acceleration formally is defined as the rate of change with velocity with respect to time, that is, it's a measurement of how fast the velocity is changing, so it is a vector quantity. That is, in calculus notation:

$$\vec{a}=\mathbf{a}=\frac{d\vec{v}}{dt}=\frac{d^{2}\vec{s}}{dt^{2}}$$

However, it is also sometimes used to refer to the norm of that quantity. So you may also see:

$$a=\left\|\frac{d\vec{v}}{dt}\right\|=\|\vec{a}\|=\|\mathbf{a}\|$$

Bear in mind, there are also various higher derivatives (rates of change) which are occasionally used in kinematics, such as the derivative of acceleration with respect to time, called jerk and the derivative of that with respect to time, called jounce.

I hope this helps clear things up a bit, if you still have questions, feel free to ask them in the comments.

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As far as I know, the norm of $\mathbf{a}$ isn't the same as the derivative of the norm of $\mathbf{v}$. The latter is what is usually called tangential acceleration. –  Javier Badia Aug 12 '12 at 19:18
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@JavierBadia: as written, they have a = the norm of the derivative, not the other way around. –  Jerry Schirmer Aug 13 '12 at 3:58
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