We all know that when we move, we cover a distance. Distances are part of our daily lives in getting around school, playing sport and travel. But what if we walk in a complete circle? We've moved through a distance but overall have returned to where we started.
This page will show how to add vectors, or to resolve a vector into two perpendicular components. You might need to revisit Pythagoras and trigonometry first.
Key Concepts
Distance is a scalar quantity and displacement is a vector quantity.
Displacement is a vector quantity, which means it has magnitude and direction, unlike its scalar equivalent distance (see Vectors). This makes it possible for a body to move through a distance without having an overall displacement.
Displacement should be represented by an arrow, not just a line. Defining a positive direction is also important. Let's say that up and right are positive directions; a displacement down and to the left would have a negative value.
A 2-dimensional displacement can be represented by vectors. Added nose to tail, the original vector is the resultant.
It is possible to resolving (split) a displacement vector into horizontal and vertical components. This is a useful technique when horiztonal and vertical motion can be treated independently, such as Projectile motion.
This Geogebra animation demonstrates how we can represent the motion of a particle using a displacement-time graph to show how the gradient and sign of the line relates to the direction and velocity of the particle.
Use quizzes to practise application of theory.
START QUIZ!
A body travels along the path shown
Fill in the following quantities
Distance travelled m (1sf)
Magnitude of displacement m
Displacement East m
Displacement North m
The distance travelled is an estimate.
The triangle made by the components is a 345 triangle
The animation shows the motion of a blue dot.
Fill in values for the displacement from starting point at the following times
0s
2s with direction (NSEW)
4s
10s
Taking π to be 3, estimate the distance travelled at the following times
2s
4s
8s
10s
The circumference = 2πr = 12 (approximating π to 3)
The following blue dot only moves in 1 dimension
Enter the values for the displacement from the original position at the following times (no units):
3s
4s
6s
Enter the values for distance travelled at these times:
3s
4s
6s
Enter the values for the displacement from the origin at the following times:
0s
3s
4s
6s
Left is negative right is positive
Observe the motion of the blue dot and deduce which of the graphs represents the motion.
+ displacement followed by -
The end part is the fastest so steepest.
How much of Distance and displacement have you understood?
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