[b]Kinematics[/b] is the study of motion and comes from the Greek word κι′νηση (ki'ne-se), which simply means motion. As you do more physics and we use the Greek alphabet, you will likely learn to read Greek as a side effect. You won't know what you're saying, but you will learn the alphabet and basic pronunciation with just a little effort. I personally found out by accident that I could read Greek some time during graduate school in physics when I met a Greek guy at school and read his T-shirt. [br][br]To study motion we need to first establish some definitions. We need to be able to describe where things are, how they move, etc. [b]The position of an object is a vector that describes the object's location with respect to some agreed-upon coordinate system. Both the location of the origin and the orientation of the axes must be known to make any sense of a position vector.[/b][br][br]A position vector in 2 or 3 dimensions is denoted [math]\vec{r}[/math]. An example of a position vector is: [math]\vec{r}=(3\hat{i}+4\hat{j})m[/math]. Since this vector only has an x and y component, it means that the object is in the xy plane (z=0). If a [math]\hat{k}[/math] component isn't present, as in this case, it means that you can assume it's zero.  To leave the xy plane, we'd have to have a non-zero [math]\hat{k}[/math] component. [br][br]In this particular case, an object is located 3 meters along the positive x-axis from the origin and 4 meters along the y-axis from it. As mentioned above, in order for this to make any sense at all, the location of the origin and the orientation of the axes must be known.[br][br]The total distance from the origin is the[b] magnitude [/b]of the position vector, which is obtained just like we find the hypotenuse of a triangle - by taking the root of the sum of the squares of the side lengths: [br][br][center][math]|\vec{r}|=\sqrt{r_x^2+r_y^2+r_z^2}.[/math][/center] [br]Realize that in a two-dimensional case, this is the familiar Pythagorean theorem, or[br][br][center][math]|\vec{r}|=\sqrt{r_x^2+r_y^2}.[/math][/center] [br]Here the x and y components can be seen as the lengths of the sides of the right triangle and the vector's magnitude as the hypotenuse.[br][br]When an object moves, it is said to undergo [b]displacement[/b]. In other words, the displacement is just the change in position. It is denoted [math]\Delta\vec{r}[/math]. The upper case Greek delta (pronounced like our 'd') is always understood to imply the later (or final) condition minus the earlier (or initial) condition.

If something is in motion with respect to the origin of the coordinate system, its position will naturally be a time-dependent vector. A few examples are [math]\vec{r}=2t\hat{i}+\hat{j}[/math], or [math]\vec{r}=(3+cos(t))\hat{i}+(2+sin(t))\hat{j}[/math]. Since computer languages generally don't like units, we are always assuming that the units are SI if they are not specified. Thus, we see that if the position is in meters and time is in seconds, that the first vector should rightfully be written [math]\vec{r}=(2m/s)t\hat{i}+1m\hat{j}[/math] such that each component will have units of meters, but I will not always be so explicit. Take note that GeoGebra, and every other computer language that I can think of, expects radians as the input arguments to trigonometric functions. That means the trig functions above should be seen, for instance, as [math]\cos\left(\frac{1rad}{s}t\right)[/math] , and not just [math]\cos{t}[/math]. With this in mind, can you see in your mind's eye what these time-dependent vectors would look like as time (represented by the t) ticks on? Let's try to make these time-dependent vectors in GeoGebra.