Examples with explanations on the concepts of average speed and average velocity of moving object. More problems and their
solutions can be found in this website.
The average speed is a scalar quantity (magnitude) that describes the rate of change (with the time) of the distance of a moving object.
The average velocity is a vector quantity (magnitude and direction) that describes the rate of change (with the time) of the position of a moving object.
Example 1: An object moves from A to D along the red path as shown below in 41 minutes and 40 seconds.
a) Find the average speed of the object in m/s
b) Find the average velocity of the object in m/s
Solution:
a) Using the given scale (1km per division); the total distance d is given by
d = AB + BC + CD = 2 + 5 + 2 = 9 km
average speed =


=


=


=


= 3.6 m/s

b) The final and initial and positions of the moving object are used to find the displacement. The distance from A (initial position) to D (final position) is equal to AD = 5 km. The displacement is the vector AD whose magnitude if 5 km and its direction is to the east.
average velocity =


=


=


= 2.5 m/s

The average velocity is a vector whose magnitude is 2.5 m/s and its direction is to the east.
Example 2: An object moves, along a line, from point A to B to C and then back to B again as shown in the figure below in half an hour.
a) Find the average speed of the moving object in km/h.
b) Find the magnitude of the average velocity of the object in km/h.
Solution:
a) The total distance d covered by the object is
d = AB + BC + CB = 5 km + 4 km + 4 km = 13 km
average speed =


=


= 26 km/h

b) The magnitude of the displacement is equal to the distance from A (initial position) to B (final position) which is equal to 5 km.
average velocity =


=


= 10 km/h

Example 3: An fast object moves from point A to B to C to D and then back to A along the rectangle shown in the figure below in 5 seconds.
a) Find the average speed of the moving object in m/s.
b) Find the velocity of the object in m/s.
Solution:
a) The total distance d is equal to the perimeter of the rectangle. Using the given scale,
d = 2 AB + 2 BC = 10 + 6 = 16 km
average speed =


=


=


= 3200 m/s

b) Since the moving object starts at point A and finish at A, there is no change in the position of the object and therefore the displacement is equal to zero.
Example 4: A person walks, for two hours, from point A to B to C along a circular field as shown in the figure below.
a) Find the average speed of the person in km/h.
b) Find the velocity of the person.
Solution:
a) The total distance d is equal to half the circumference of the circle and given by
d = (1/2)(2 * Pi * 3) = 3 Pi
average speed =


=


= 1.5 Pi km/h = 4.7 km/h

b) The magnitude of the displacement D is equal to the diameter AC of the circle and is given by
D = 2 * 3 = 6 Km with direction to the East
average velocity =


=


= 3 km/h

Example 5: A person walks for one hour and 12 minutes, from point A to point B, along a circular field as shown in the figure.
a) Find the average speed of the person in km/h.
b) Find the magnitude of the displacement of the person in km/h.
Solution:
a) The total distance d is equal to the quarter the circumference of the circle and given by
d = (1/4)(2 * Pi * 3) = 1.5 Pi
average speed =


=

1.5 Pi km
1 hour + 12 minutes

=

1.5 Pi km
1 hour + 12/60 hour

=


= 1.25 Pi km/h = 3.9 km/h
b) The magnitude of the displacement D is equal to the hypotenuse AB of the right angle ABO as shown below
Use Pythagora's theorem to find AB as follows
AB^{2} = 3^{2} + 3^{2} = 18
D = AB = 3√2 km
average velocity =


=

3√2 km
1 hour + 12 minutes

=


= 2.5√2 km/h = 3.5 km/h