Applications of vectors - displacement and velocity
Put vectors to work on motion in Year 11 Specialist Mathematics in Queensland (QCAA). This subtopic uses component vectors to model displacement and velocity in the plane: where something is, how far it has moved, and how fast and in which direction it is travelling.
You will learn to find displacement between points, distance and speed as vector magnitudes, position under constant velocity, resultant displacement of successive legs and the bearing of a velocity — the modelling skills that lead into relative velocity and force problems.
Theory
Displacement and velocity are the everyday jobs of vectors in Year 11 Specialist Mathematics (QCAA, Queensland). A displacement vector records a change in position, a velocity vector records how fast and in which direction something moves, and their magnitudes give the scalar distance and speed. This page shows how to model constant-velocity motion in the plane using component (i, j) vectors.
Every position, movement and motion in the plane can be described with a two-dimensional vector written in component form \(x\mathbf{i}+y\mathbf{j}\), where \(\mathbf{i}\) points east and \(\mathbf{j}\) points north.
The position vector \(\mathbf{r}\) locates a point relative to the origin. The displacement from \(A\) to \(B\) is the change in position \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\); its magnitude \(|\overrightarrow{AB}|\) is the straight-line distance between the points.
A velocity vector \(\mathbf{v}\) gives both the direction of motion and the speed \(|\mathbf{v}|\). Under constant velocity the position after time \(t\) is \(\mathbf{r}=\mathbf{r}_0+t\mathbf{v}\), so the displacement in that time is \(t\mathbf{v}\).
Keep the vector and its magnitude apart: displacement and velocity are vectors (direction matters), while distance and speed are scalars (size only). Successive movements add head to tail to give a resultant displacement or velocity.
Displacement from \(A\) to \(B\) is the difference of position vectors:
Under constant velocity, the position after time \(t\) is:
Distance and speed are the magnitudes of the displacement and velocity vectors:
Average velocity is displacement divided by time; distance is speed times time:
How to model displacement and velocity
- Write each quantity as a vector in component form \(x\mathbf{i}+y\mathbf{j}\) (east–north), and note the times involved.
- Choose the relationship: displacement \(=\mathbf{b}-\mathbf{a}\); position \(=\mathbf{r}_0+t\mathbf{v}\); average velocity \(=\) displacement \(\div\) time; resultant \(=\) sum of the parts.
- Substitute and simplify component by component, one operation per line.
- Take the magnitude \(\sqrt{v_1^{2}+v_2^{2}}\) whenever a scalar distance or speed is asked, and a bearing if a direction is asked.
Displacement is the change in position \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\):
| \(\overrightarrow{AB}\) | \(=\) | \(\mathbf{b}-\mathbf{a}\) |
| \(=\) | \((7\mathbf{i}+6\mathbf{j})-(\mathbf{i}-2\mathbf{j})\) | |
| \(=\) | \((7-1)\mathbf{i}+(6-(-2))\mathbf{j}\) | |
| \(=\) | \(6\mathbf{i}+8\mathbf{j}\) |
Distance is the magnitude of the displacement:
| \(|\overrightarrow{AB}|\) | \(=\) | \(\sqrt{6^2+8^2}\) |
| \(=\) | \(\sqrt{36+64}\) | |
| \(=\) | \(\sqrt{100}\) | |
| \(=\) | \(10\) |
\(\overrightarrow{AB}=6\mathbf{i}+8\mathbf{j}\) km, and the distance flown is \(10\) km.
Constant velocity gives \(\mathbf{r}=\mathbf{r}_0+t\mathbf{v}\) with \(t=3\):
| \(\mathbf{r}\) | \(=\) | \(\mathbf{r}_0+3\mathbf{v}\) |
| \(=\) | \((\mathbf{i}+2\mathbf{j})+3(2\mathbf{i}+3\mathbf{j})\) | |
| \(=\) | \((\mathbf{i}+2\mathbf{j})+(6\mathbf{i}+9\mathbf{j})\) | |
| \(=\) | \((1+6)\mathbf{i}+(2+9)\mathbf{j}\) | |
| \(=\) | \(7\mathbf{i}+11\mathbf{j}\) |
The displacement in that time is \(3\mathbf{v}\):
| \(3\mathbf{v}\) | \(=\) | \(3(2\mathbf{i}+3\mathbf{j})\) |
| \(=\) | \(6\mathbf{i}+9\mathbf{j}\) |
The position is \(7\mathbf{i}+11\mathbf{j}\) and the displacement is \(6\mathbf{i}+9\mathbf{j}\) m.
Average velocity \(=\) displacement \(\div\) time; first find the displacement:
| \(\overrightarrow{PQ}\) | \(=\) | \(\mathbf{q}-\mathbf{p}\) |
| \(=\) | \((14-2)\mathbf{i}+(-13-3)\mathbf{j}\) | |
| \(=\) | \(12\mathbf{i}-16\mathbf{j}\) |
Divide by the \(4\) hours:
| \(\mathbf{v}\) | \(=\) | \(\dfrac{12\mathbf{i}-16\mathbf{j}}{4}\) |
| \(=\) | \(3\mathbf{i}-4\mathbf{j}\) |
Speed is the magnitude of the velocity:
| \(|\mathbf{v}|\) | \(=\) | \(\sqrt{3^2+(-4)^2}\) |
| \(=\) | \(\sqrt{9+16}\) | |
| \(=\) | \(\sqrt{25}\) | |
| \(=\) | \(5\) |
The average velocity is \(3\mathbf{i}-4\mathbf{j}\) km/h and the speed is \(5\) km/h.
Speed is the magnitude of the velocity:
| \(|\mathbf{v}|\) | \(=\) | \(\sqrt{5^2+12^2}\) |
| \(=\) | \(\sqrt{25+144}\) | |
| \(=\) | \(\sqrt{169}\) | |
| \(=\) | \(13\) |
Bearing is measured clockwise from north, so \(\tan\theta=\dfrac{\text{east}}{\text{north}}\):
| \(\tan\theta\) | \(=\) | \(\dfrac{5}{12}\) |
| \(\theta\) | \(=\) | \(\tan^{-1}\!\left(\dfrac{5}{12}\right)\) |
| \(=\) | \(22.62^\circ\) | |
| \(\approx\) | \(23^\circ\) |
The speed is \(13\) km/h and the bearing is \(023^\circ\).
Common pitfalls
Frequently asked questions
What is the difference between distance and displacement?
Displacement is a vector \(\mathbf{b}-\mathbf{a}\) that records both how far and in which direction the position changed. Distance is the scalar magnitude \(|\mathbf{b}-\mathbf{a}|\) — just the straight-line length.
How do you find the displacement between two points?
Subtract the position vectors, finish minus start: \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\). Do it component by component, so \((9\mathbf{i}+7\mathbf{j})-(2\mathbf{i}+3\mathbf{j})=7\mathbf{i}+4\mathbf{j}\).
How do you find speed from a velocity vector?
Speed is the magnitude \(|\mathbf{v}|=\sqrt{v_1^{2}+v_2^{2}}\). For \(\mathbf{v}=8\mathbf{i}+6\mathbf{j}\), the speed is \(\sqrt{64+36}=10\).
What is the formula for position under constant velocity?
The position after time \(t\) is \(\mathbf{r}=\mathbf{r}_0+t\mathbf{v}\), where \(\mathbf{r}_0\) is the starting position and \(\mathbf{v}\) is the constant velocity. The displacement in that time is \(t\mathbf{v}\).
How do you find average velocity?
Average velocity is the displacement divided by the time taken, \(\mathbf{v}=\dfrac{\mathbf{b}-\mathbf{a}}{t}\). It is a vector, so keep it in component form.
How do you find the bearing of a velocity?
With \(\mathbf{i}\) east and \(\mathbf{j}\) north, the bearing is measured clockwise from north: \(\tan\theta=\dfrac{\text{east}}{\text{north}}\). Then write the answer as a three-figure bearing such as \(037^\circ\).