Unit 2-Kinematics

Let's say that the long hand is at the top of the hour, just to make things a little easier for this purpose. So we will be tracking this long hand as time goes by. Let's pretend that each tick mark (from 1-12) is a distance of 5 m. The long hand starts at the top of the hour (at 11 o' clock) and eventually travels to the top of the hour once again (at 12 o' clock), traveling a total distance of 60m in 60 minutes (3600 sec).
To find out what the average speed of the long hand is, we would use the equation, speed=distance/time. So 60 m divided by 3600 seconds equals to about an average speed of .02 m/sec.
The long hand is traveling at a constant speed of about .02 m/sec, but what is the velocity of the long hand in this particular situation?
In order to figure this out, we will use the equation, velocity=displacement(or the distance traveled from the starting point)/time. So 0 m/3600sec is equal to a velocity of 0. In one 60 minute period, the long hand traveled a total of 60 m, but had no velocity simply because it ended up where it started within this period. But for example, if we were to track the long hand for 10 minutes, the average speed would be the same at .02 m/sec (because the long hand is traveling at a constant speed), and it would have traveled a total of 10m. Because it traveled from point A to point B in this case, the velocity would be 10m/600 sec, or .02 m/sec to the right.