VELOCITY

 When we describe speed plus the direction of motion, we are specifying velocity. When something moves at a constant velocity or constant speed, then equal distances are covered in equal intervals of time. Constant velocity means constant speed with no change in direction. A car that rounds a curve at a constant speed do not have a constant velocity- its velocity is no constant because the direction of the travel is continuously changing.  Velocity can be constant only when motion is along a straight-line path.

Velocity is a vector. Both direction and quantity must be stated. It one train has a velocity of 100km/h north, and a second train has a velocity of 100km/h south, the two trains have different velocities, even though their speed is the same. Other examples of vectors are force, and field intensity.
Example 1: if a person walked 400 m in a straight line in 5 min, that person’s velocity would be (400 m [forward])÷(5 min) = 80 m/min [forward] .
If the same person walked 100 m [North] then 300 m [South] in 5 minutes, we first find their displacement.
displacement = 200 m [S]
velocity = 200÷5 = 40 m/min [S]
If that person walked 100 m [E] in .75 min, 100 m [N] in 1.50 min, 100 m [W] in 1.00 min and finally 100 m [S] in 1.75 min, that person would end up back where they started. Since their displacement is zero, Their velocity is zero.
Remember,

(average velocity) = displacement ÷ time.
Example 2:
 A hiker traveled 80.0 m [S] at 1.00 m/s, then 80.0 m [S] at 5.00 m/s. What is the hiker’s average velocity?

Answer:
displacement = 160.0 m [S]
time for the first part is 80.0÷1.00= 80.0 s, time for the second part is 80.0 m ÷ 5.00 m/s = 16.0 s.
Total time = 80.0+16.0 = 96.0 s
Therefore, the velocity is (160.0 m [S])÷96.0 s = 1.67 m/s [S]
Problems

Example 3: A train on a straight track traveled 60.0 km/h [E] for 2.00 h, stopped for 15 min, then traveled 100.0 km [W] at 133 km/h.
a. What was the train’s average speed for the whole trip?
b. What was the train’s average velocity for the whole trip?
Answer:
a. To find average speed, we need total distance and total time.
During the first part of the trip, the train covered 60.0×2.00 = 120 km in 2.00 h.
During the second part of the trip the train traveled 0.00 km in 0.25 h.
During the third part of the trip, the train traveled 100.0 km in 0.75 h.
In total the train traveled 220 km in 3 .00 h.
Average speed = (220 km)÷3.00 = 73.3 km/h

b. To find average velocity, we need displacement and total time.
During the first part of the trip, the train covered 60.0×2.00 = 120 km [E] in 2.00 h.
During the second part of the trip the train traveled 0.00 km in 0.25 h.
During the third part of the trip, the train traveled 100.0 km [W] in (100.0 km ÷ 133 km/h) = 0.75 h.
The train’s displacement was (120-100) = 20 km [E] in 3 .00 h.
Average velocity = (20 km [E])÷3.00h = 6.7 km/h [E]
Example 6: A runner covers one lap of a circular track 40.0 m in diameter in 62.5 s. For that lap, what were her average speed and average velocity?

Solution:

average speed = (total distance)/(total time) = (π*40.0)/(62.5) = 2.01m/s
average velocity = displacement/time = 0/62.5 = 0 m/s

 
Problems Reference:
http://www.physics247.com/physics-homework-help/speed-velocity-acceleration.php