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{excerpt} 
The time rate of change of [position|position]. {excerpt}


h3. Mathematical Definition

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

h3. 1-D Motion with Constant Velocity

h4. The Advantages and the Utility of Constant Velocity

Velocity is a [vector|vector].  Thus, if an object is moving with constant velocity, it is moving at a constant rate and in a constant direction.  For the special case of constant velocity, we can simplify the mathematical definition of velocity to:

{latex}\begin{large} \[ v = \frac{\Delta x}{\Delta t} \] \end{large}{latex}

where it is important to note that, although we have removed the arrow over the _v_, the velocity is _still_ a vector.  An object moving with constant velocity is _necessarily_ moving in only one dimension.  For the case of one dimensional motion, vectors are not usually written with an arrow.  

This special case might seem a bit useless, since it is rare for objects to move purely in a straight line with no change in speed.  However, you can often describe a complicated motion with acceptable accuracy by building it out of smaller segments in which the velocity is basically constant.  

h4. Utility of the One-Dimensional Case

As with all [vector|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. Representing Constant Velocity Graphically

Looking at the equation above, it should be clear that the mathematical definition of constant velocity is equivalent to the formula for the slope of a position versus time graph.  To see the utility of this correspondence, consider the following plots:

!posvel.gif! !slowvel.gif! !negvel.gif!

Consider the left plot.  The abscissa is time, and the ordinate is position.  This graph is giving the position of some object (called object A) as a function of time.  Looking at the graph, we can see that for each second of time that elapses, the object changes its position by 2 meters.  This is the same as saying that the slope of the left plot is 2 m / (1 s) or, more simply, 2 m/s.  Object A, then, is moving with a speed of 2 m/s.  

Contrast that with the middle plot.  Object B is only changing its position by 1 meter every second.  Thus, it is moving with a speed of 1 m/s.  

Finally, look at the plot on the right.  Object C is changing its position by 2 meters every second, and so it has a speed of 2 m/s.  Note that objects A and C have the same speed.  The graphs are different, however, because object C is moving in the negative direction.  
{note}That doesn't necessarily mean objects A and C are moving in different _actual_ directions.  Recall from [lesson 1|Lesson 1 (Average Velocity)] that when we are looking at position data, we should always consider what *coordinate system* applies (does the positive x direction point east? west? north?...), and we haven't determined that for any of the graphs yet.{note}

h4. {toggle-cloak:id=ch1} Check Your Understanding 

{cloak:id=ch1}
Suppose that all three of the plots describe objects moving on the same coordinate system.  The positive direction of the system points east, and the negative points west.  Try to describe the motions of the three objects in words, based upon what you have learned from the graphs.

{toggle-cloak:id=ans1} *Answer:*  {cloak:id=ans1}There are many ways to describe the motion, but here are some important features.  At the beginning of the motion, A and C are both at the same location, while B is 5 meters east of A and C.  A begins moving east at 2 m/s, B begins moving west at 2 m/s and C begins moving east at 1 m/s.  By the end of the motion (as plotted) A has caught B (at a position 10 m east of where A and C began).  A and B are now 20 meters east of C.{cloak:ans1}
{cloak:ch1}


h2. Relevant Examples

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\begin{large}\[ f(t)\]\end{large}

\begin{large}\[ \langle f(t)\rangle_{t} \equiv \frac{\int_{t_{i}}^{t_{f}} f(t)\:dt}{t_{f} - t_{i}} \]\end{large}

\begin{large}\[ \langle \vec{v} \rangle_{t} = \frac{\int_{t_{i}}^{t_{f}} \vec{v} \:dt}{t_{f}-t_{i}}
=\frac{\int_{t_{i}}^{t_{f}} \left(\frac{dx}{dt}\hat{x} + \frac{dy}{dt}\hat{y} + \frac{dz}{dt}\hat{z}\right) \:dt}{t_{f}-t_{i}}
\]\end{large}

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

\begin{large}\[ \langle \vec{v} \rangle_{t} = \frac{\int_{x_{i}}^{x_{f}} \hat{x}\:dx + \int_{y_{i}}^{y_{f}} \hat{y}\:dy
+ \int_{z_{i}}^{z_{f}} \hat{z}\:dz}{t_{f} - t_{i}} \]\end{large}

\begin{large}\[ \langle \vec{v} \rangle_{t} = \frac{\vec{r}_{f} - \vec{r}_{i}}{t_{f}-t_{i}} \equiv \frac{\Delta\vec{r}}{\Delta t} \]\end{large}

\begin{large} \[ \langle v\rangle_{t} = \frac{x_{\rm f} - x_{\rm i}}{t_{\rm f} - t_{\rm i}} \] \end{large}

\begin{large} \[ \langle v_{\rm pd}\rangle_{t} = \frac{x_{\rm p} - x_{\rm d}}{t_{\rm p} - t_{\rm d}} = \frac{ +2\:{\rm blocks} - 0\:{\rm blocks}}{2\:{\rm minutes}} = + 1\:{\rm blocks/min}\]\end{large}

One item that is worth noting is that physics problems often do not give actual times. Instead, they give elapsed times. In this situation, for instance, we were not told exactly when the student left the dorm (10:00 AM? 12:00 PM?) or when they arrived at the physics building. We were only told that the difference between the times was 2 minutes. (If the student left the dorm at 11:00 AM, they arrived at 11:02 AM). This information is sufficent to find the average velocity. Because it is so rare to be given initial and final times, the velocity equation is often written:

\begin{large}\[ \langle v \rangle_{t} = \frac{x_{\rm f} - x_{\rm i}}{t}\]\end{large}

where t denotes elapsed time.

It is a common source of confusion that the equations of mechanics often use "final" and "initial" as their subscripts. For the trip described at the beginning of this lesson, it is clear that the (overall) initial position is the dorm (x = 0 m) and the (overall) final position is the library (x = + 3 m), yet we have just used the equation for average velocity with the final position taken to be the physics building. The equations are not required to use the overall final and overall initial positions and times. You are free to break up the motion into as many segments as desired, and apply the equation to the beginning and end of each segment. The only requirement is that the position taken for the "initial" one occurs earlier in the motion than the "final" one.

\begin{large}\[ \langle v_{\rm cp}\rangle_{t} = \frac{x_{\rm c} - x_{\rm p}}{t_{\rm cp}} = \frac{(-1\:{\rm blocks})- (+ 2\:{\rm blocks})}{2\:{\rm minutes}} = -1.5\:{\rm blocks/min}\] \end{large}

Using both methods of reporting direction together in one statement results in confusion. What would it mean if we reported the student had an average velocity of - 1.5 blocks/min west?