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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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    $\begingroup$ It seems logical that speed is related to speed... $\endgroup$
    – Fabian
    Commented Aug 12, 2012 at 18:15
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    $\begingroup$ 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? $\endgroup$ Commented Aug 12, 2012 at 21:38

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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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    $\begingroup$ 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. $\endgroup$
    – Javier
    Commented Aug 12, 2012 at 19:18
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    $\begingroup$ @JavierBadia: as written, they have a = the norm of the derivative, not the other way around. $\endgroup$ Commented Aug 13, 2012 at 3:58

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