Difference between revisions of "Acceleration"
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| − | Acceleration, the second [[derivative]] of [[displacement]], is defined to be the change of [[velocity]]. | + | ==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 
be the velocity of an object at a time 
and 
be the velocity of the same object at a time 
. If acceleration, 
, is known to be constant, then ![\[\textbf{a} = \frac{\textbf{v}_2 -\textbf{v}_1 }{t_2 - t_1}\]](http://latex.artofproblemsolving.com/9/d/0/9d0c5c81651dc2902b5a031eed74255573ad95b1.png)
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, 
. Then, at a particular instance, ![\[\textbf{a} = \lim_{h\to 0} \frac{v(t+h)-v(t)}{(t+h)-t} = v'(t)\]](http://latex.artofproblemsolving.com/5/e/2/5e2779497134457ee3c1d38bb730e024007acee8.png)
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