If a particle is moving it will have a certain velocity. The velocity of the particle could be constant within a certain period of time, or it could change within a certain period of time. For right now I am only going to talk about a constant velocity.
Just like when you were looking at the position of a particle, a particles velocity will subjective to an observer. For example, as a car is moving down the freeway an observer on earth would say that car is going at a certain constant speed, such as 60 mph. However, what if I told you that car is moving faster than 60 mph. It is actually moving at the speed the earth is spinning, plus the speed that the earth is revolving around the sun, plus the speed that the solar system is moving through the galaxy, and so on. But since the observer on earth can’t see this they say the car is moving at 60 mph. Remember, humans cannot feel an object that is moving at a constant velocity, we can only visually observe constant velocity. If you say that you can feel velocity you are actually feeling a change in velocity.
Depending on what the particles velocity is the particle position will change within a given time. You can find the particles velocity by using the equation below.
(Eq 1) $v=\frac{ds}{dt}$
ds = Change in displacement
dt = Change in time
Example
For example a particle travels a distance of 100m in 20s what was its velocity?The particle is not accelerating.
Solution
$v = \frac{100m}{20} = 5 \frac{m}{s}$
Remember that we live in a 3-dimensional world which mean you will have to be able to find the velocity in all three dimensions. Expressing velocity like this doesn’t give us an easy value to work with. However, we can find the magnitude of these 3 values by using the equation below.
(Eq 2) $v=\sqrt{v_xi^2+v_yj^2+v_zk^2}$
Example
The velocity of a particle has the following x, y, and z values, 2i$\frac{m}{s}$, 4j$\frac{m}{s}$, 7k$\frac{m}{s}. What is the total velocity of the particle?
Solution
$\sqrt{(2i)^2+(4j)^2+(7k)^2}=8.3\frac{m}{s}$
Above I have only talked about velocity. You can however, use velocity to find the position of a particle within a certain time period. In order to do this you would have to use the following equation.
(Eq 3) $s=vt+s_o$
$s$ = final position
$v$ = velocity
$t$ = time
$s_o$ = initial position
Example
A train is 45 miles away from the station. It is moving at a constant speed of 50mph. The current time is 3 pm. What time will the train arrive at the station. Assume the train does not accelerate or decelerate.
Solution
$ v = 50mph$, $s_0 = o$, $s = 45 miles$
Modify equation 3 to the following.
$t=\frac{s-s_o}{v}=\frac{45~miles}{50~mph}=0.9~hours$
0.9 hours can be converted to 54 minutes, which means the train will arrive at 3:54 pm