Difference between revisions of "Acceleration"

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Acceleration, the second [[derivative]] of [[displacement]], is defined to be the change of [[velocity]].
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==Definition==
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'''Acceleration''', the second [[derivative]] of [[displacement]], is defined to be the change of [[velocity]] per unit time at a certain instance.
  
 
A common misconception is that acceleration implies a POSITIVE change of velocity, while it could also mean a NEGATIVE one.
 
A common misconception is that acceleration implies a POSITIVE change of velocity, while it could also mean a NEGATIVE one.
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==Formula for Acceleration==
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Let <math>\textbf{v}_1</math> be the velocity of an object at a time <math>t_1</math> and <math>\textbf{v}_2</math> be the velocity of the same object at a time <math>t_2</math>.  If acceleration, <math>\textbf{a}</math>, is known to be constant, then <cmath>\textbf{a} = \frac{\textbf{v}_2 -\textbf{v}_1 }{t_2 - t_1}</cmath> Note that velocity is a vector, so the magnitudes cannot be just subtracted in general.
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If acceleration is not constant, then we can treat velocity as a function of time, <math>v(t)</math>.  Then, at a particular instance, <cmath>\textbf{a} = \lim_{h\to 0} \frac{v(t+h)-v(t)}{(t+h)-t} = v'(t)</cmath>
  
 
[[Category:Physics]]
 
[[Category:Physics]]

Revision as of 13:12, 24 April 2009

Definition

Acceleration, the second derivative of displacement, is defined to be the change of velocity per unit time at a certain instance.

A common misconception is that acceleration implies a POSITIVE change of velocity, while it could also mean a NEGATIVE one.

Formula for Acceleration

Let $\textbf{v}_1$ be the velocity of an object at a time $t_1$ and $\textbf{v}_2$ be the velocity of the same object at a time $t_2$. If acceleration, $\textbf{a}$, is known to be constant, then \[\textbf{a} = \frac{\textbf{v}_2 -\textbf{v}_1 }{t_2 - t_1}\] Note that velocity is a vector, so the magnitudes cannot be just subtracted in general.

If acceleration is not constant, then we can treat velocity as a function of time, $v(t)$. Then, at a particular instance, \[\textbf{a} = \lim_{h\to 0} \frac{v(t+h)-v(t)}{(t+h)-t} = v'(t)\]

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