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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.

average speed =  
distance
time

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.


average velocity =  
change in position
time
=  
displacement
time


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

displacement and distance - Example 1
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 =  
distance
time
=
9 km
41 mn + 40 s
=  
9000 m
(41*60 + 40) s
=  
9000 m
2500 s
= 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 =  
displacement
time
=  
5 km
41 mn + 40 s
=  
5000 m
2500 s
= 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.

displacement and distance - Example 2

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 =  
distance
time
=
13 km
0.5 hour
= 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 =  
displacement
time
=  
5 km
0.5 hour
= 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.

displacement and distance - Example 3
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 =  
distance
time
=
16 km
5 seconds
=
16000 m
5 seconds
= 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.


average velocity =  
displacement
time
=  
0
5 second
= 0


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.

displacement and distance - Example 4
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 =  
distance
time
=
3 Pi km
2 hours
= 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 =  
displacement
time
=  
6 km
2 hours
= 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.

displacement and distance - Example 5
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 =  
distance
time
=
1.5 Pi km
1 hour + 12 minutes
=
1.5 Pi km
1 hour + 12/60 hour
=
1.5 Pi km
1.2 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

displacement and distance - Solution to Example 5
Use Pythagora's theorem to find AB as follows

AB2 = 32 + 32 = 18

D = AB = 3√2 km


average velocity =  
displacement
time
=  
3√2 km
1 hour + 12 minutes
=  
3√2 km
1.2 hours

= 2.5√2 km/h = 3.5 km/h



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Updated: 26 September 2015

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