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{excerpt}
The time rate of change of [velocity].{excerpt}

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h2. Mathematical Definition

{latex}\begin{large}\[ \vec{a} = \frac{d\vec{v}}{dt} \]\end{large}{latex}

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h2. One-Dimensional Acceleration

h4. Utility of the One-Dimensional Case

As with all [vector] equations, the equations of kinematics are usually approached by separation into components.  In this fashion, the equations become three simultaneous one-dimensional equations.  Thus, the consideration of motion in one dimension with acceleration can be generalized to the three-dimensional case.

h4. Graphical Representations of Acceleration

The one-dimensional form of the definition of acceleration:

{latex}\begin{large} \[ a = \frac{dv}{dt}\] \end{large}{latex}

indicates that acceleration is the slope of a velocity versus time graph.  

Since velocity is the derivative of position:

{latex}\begin{large}\[ v = \frac{dx}{dt}\]\end{large}{latex}

we can also write:

{latex}\begin{large}\[ a = \frac{d^{2}x}{dt^{2}}\]\end{large}{latex}

Thus, we can also see the effects of acceleration in a position versus time graph like that shown here:

!accel.gif!

Consider the position vs. time graph shown above.  If you were to lay a ruler along the curve of the graph at the origin, the ruler would have to be horizontal to follow the curve, indicating zero slope.  Thus, the velocity is zero at the origin.  As you follow the curve, however, the ruler would have to be held at a steeper and steeper angle (see the lines added in the graph below).  The slope grows with time, indicating that the velocity is becoming more and more positive (the speed is increasing).  This positive change in velocity indicates a positive acceleration.  In calculus terminology, we would say that a graph which is "concave up" or has positive curvature indicates a positive acceleration.

!acceltangents.gif!

Examples that use the graphical representations of acceleration are:

{contentbylabel:graphical_representation,constant_acceleration|maxResults=50|operator=AND|showSpace=false|excerpt=true}

h4.  The Danger of Deceleration

Acceleration poses a problem for the specialized vocabulary of physics.  The other two major kinematical quantities, [velocity] and [position], have a related scalar quantity.  For instance, [distance] is a scalar that is related to [displacement].  [Speed|speed] is a scalar that is related to velocity.  If we are discussing instantaneous velocity, then speed is the magnitude of velocity.   The acceleration can also be discussed in terms of a vector acceleration or simply the magnitude, but for acceleration we have no special term for the magnitude.  The vector is called "the acceleration" and the magnitude is "the magnitude of the acceleration".  This can result in confusion.  

This problem is exacerbated by the fact that in everyday language, we often discuss _distance_, _speed_ and _acceleration_.  The everyday definitions of distance and speed are basically equivalent to their physics definitions, since we rarely discuss direction of travel in everyday speech and these quantities are scalars in physics (no direction).  Unfortunately, in physics, we usually use the term "acceleration" to refer to a vector, while in everyday speech it denotes a magnitude. 

The difficulties do not end there.  Everyday usage _does_ make one concession to the vector nature of motion.  When we discuss acceleration in everyday speech, we usually specify whether the object is "accelerating" (speeding up) or "decelerating" (slowing down).  Both terms imply a change in velocity, and so in physics we can call either case "accelerating".  The physics way of explaining the difference is:

||everyday term || physics equivalent ||
| acceleration | acceleration and velocity point in the same direction |
| deceleration | acceleration points in the direction opposite the velocity |

To understand the physics definition, imagine a child on a playground swing.  If you want to help the child swing faster, you must push them in the same direction as they are currently moving (so the acceleration of your push is in the same direction as the child's velocity).  If you want to help them slow down, you must push them in the direction opposite their current motion (so that the acceleration of the push points opposite to the velocity).  

The difference between acceleration and deceleration (in the everyday sense) can also be illustrated graphically:

{table}{tr}{th:align=center|bgcolor=#F2F2F2}positive acceleration\\positive velocity\\"accelerating"{th}{th:align=center|bgcolor=#F2F2F2}negative acceleration\\negative velocity\\"accelerating"{th}{th:align=center|bgcolor=#F2F2F2}negative acceleration\\positive velocity\\"decelerating"{th}{th:align=center|bgcolor=#F2F2F2}positive acceleration\\negative velocity\\"decelerating"{th}{tr}{tr}{td}!pos1.gif!{td}{td}!neg1.gif!{td}{td}!neg2.gif!{td}{td}!pos2.gif!{td}{tr}{table}

Both the graphs that show "acceleration" have slopes that are steepening with time.  The only difference is that one of the graphs has a steepening _positive_ slope and the other has a steepening _negative slope.  Both graphs showing "deceleration" have slopes that are approaching zero as time evolves.  (Again, one has a negative slope and one has a positive slope.) 

{warning}It is a very common misconception that a negative acceleration *always* slows down the object it acts upon.  This is *not true*.  It is important to note that a graph which has a negative slope approaching zero (slowing down) implies a positive acceleration, and a graph which has a negative slope that is steepening (speeding up) implies a negative acceleration.  It may help you to remember that the concavity of the graph specifies the direction of the acceleration.{warning}

h4. Constant Acceleration

h5. Integration with Respect to Time

The definition of acceleration can be integrated:

{latex}\begin{large}\[ \int_{v_{\rm i}}^{v} dv = \int_{t_{\rm i}}^{t} a\: dt \] \end{large}{latex}

For the special case of constant acceleration, the integral yields:

{latex}\begin{large} \[ v - v_{\rm i} = a(t-t_{\rm i}) \] \end{large}{latex}

which is equivalent to:

{latex}\begin{large} \[ v = v_{\rm i} + a (t-t_{\rm i}) \] \end{large}{latex}

We can now substitute into this equation the definition of velocity,

{latex}\begin{large}\[ v = \frac{dx}{dt}\]\end{large}{latex}

which gives:

{latex} \begin{large} \[ \frac{dx}{dt} = v_{\rm i} + a t - a t_{\rm i} \] \end{large}{latex}

We can now integrate again:

{latex} \begin{large} \[ \int_{x_{\rm i}}^{x} dx = \int_{t_{\rm i}}^{t} \left( v_{\rm i} - at_{\rm i} + a t\right)\:dt \]   \end{large}{latex}

to find:

{latex} \begin{large} \[ x - x_{\rm i} = v_{\rm i} (t-t_{\rm i}) - a t_{\rm i} (t-t_{\rm i}) + \frac{1}{2}a( t^{2} - t_{\rm i}^{2}) \] \end{large}{latex}

We finish up with some algebra:

{latex}\begin{large}\[ x = x_{\rm i} + v_{\rm i} (t-t_{\rm i}) + \frac{1}{2} a (t^{2} - 2 t t_{\rm i} + t_{\rm i}^{2}) \] \end{large}{latex}

which is equivalent to:

{latex}\begin{large} \[ x = x_{\rm i} + v_{\rm i} (t-t_{\rm i}) + \frac{1}{2} a (t - t_{\rm i})^{2} \] \end{large}{latex}

h5. Integration with Respect to Position

The definition of acceleration can also be integrated with respect to position, if we use a calculus trick that relies on the chain rule.  Returning to the definition of acceleration:

{latex}\begin{large}\[ \frac{dv}{dt} = a \] \end{large}{latex}

we would like to find an expression for _v_ as a function of _x_ instead of _t_.  One way to achieve this is to use the chain rule to write:

{latex}\begin{large} \[ \frac{dv}{dx}\frac{dx}{dt} = a \] \end{large}{latex}

We can now elminate _t_ from this expression by using the defnition of velocity to recognize that _dx_/_dt_ = _v_.  Thus:

{latex}\begin{large} \[ \frac{dv}{dx}v = a \] \end{large}{latex}

which is easily integrated for the case of constant acceleration:

{latex}\begin{large} \[ \int_{v_{\rm i}}^{v} v \:dv = \int_{x_{\rm i}}^{x} a \:dx \] \end{large}{latex}

to give:

{latex}\begin{large}\[ v^{2} = v_{\rm i}^{2} + 2 a (x-x_{\rm i}) \] \end{large}{latex}

 





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h2. Relevant Examples

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