Displacement, velocity, acceleration and force as vectorsEdexcel International A Level Maths: Revision notes

Section 1

Vector quantities in a plane

Displacement, velocity, acceleration and force are all vectors, so each has a magnitude and a direction and is written in component form $x\mathbf{i}+y\mathbf{j}$ . Speed is the magnitude of velocity, $\sqrt{x^2+y^2}$ , and is a scalar. Position is a displacement from a fixed origin. Standard units: m, m s⁻¹, m s⁻², N.

Key termsdisplacementvelocityspeedposition vector

Common mistake

Giving velocity as the answer for speed. Speed is a number found with Pythagoras.

Section 2

Constant velocity and displacement

For constant velocity, $\mathbf{v}=\dfrac{\text{change of displacement}}{\text{time}}$ , so $\text{displacement}=\mathbf{v}t$ . The position at time $t$ is $\mathbf{r}=\mathbf{r}_0+\mathbf{v}t$ . Example: velocity $(3\mathbf{i}-4\mathbf{j})$ m s⁻¹ from $(2\mathbf{i}+5\mathbf{j})$ gives a displacement of $(18\mathbf{i}-24\mathbf{j})$ in 6 s and a position of $(20\mathbf{i}-19\mathbf{j})$ . Speed $=\sqrt{3^2+4^2}=5$ m s⁻¹.

Key termsconstant velocity

Exam tip

Displacement and position are different. Position includes the starting point.

Section 3

Two moving objects and positions

With $\mathbf{r}=\mathbf{r}_0+\mathbf{v}t$ , to find when an object is due north (or east) of a point, equate the $\mathbf{i}$ (or $\mathbf{j}$ ) coordinates and solve for $t$ . Distance between points $A$ and $B$ is $|\overrightarrow{AB}|=|\mathbf{r}_B-\mathbf{r}_A|$ . Example: $\mathbf{r}=(4t-8)\mathbf{i}+(3t-5)\mathbf{j}$ is due north of $(12\mathbf{i}+4\mathbf{j})$ when $4t-8=12$ , so $t=5$ .

Key termsdue north

Common mistake

Equating the wrong coordinates. Due north means the same $\mathbf{i}$ (east) coordinate.

Section 4

Constant acceleration

For constant acceleration, $\mathbf{a}=\dfrac{\mathbf{v}-\mathbf{u}}{t}$ (change of velocity divided by time), so $\mathbf{v}=\mathbf{u}+\mathbf{a}t$ . Example: velocity changes from $(2\mathbf{i}+\mathbf{j})$ to $(14\mathbf{i}-8\mathbf{j})$ in 4 s: $\mathbf{a}=\dfrac{12\mathbf{i}-9\mathbf{j}}{4}=(3\mathbf{i}-2.25\mathbf{j})$ m s⁻², of magnitude $3.75$ m s⁻². If a particle moves due north, the $\mathbf{i}$ component of its velocity is zero, which gives its time.

Key termsacceleration

Common mistake

Subtracting in the wrong order. Change in velocity is final minus initial.

Section 5

Forces as vectors and direction

Forces are added as vectors, component by component, and the same methods apply to give magnitude ( $\sqrt{x^2+y^2}$ ) and direction ( $\tan\theta=y/x$ ). State directions from a named line, for example ' $81.9^\circ$ north of east' or ' $80.1^\circ$ above the direction of $-\mathbf{i}$ '. Check the signs of the components to place the vector in the correct quadrant.

Key termsdirection

Exam tip

Sketch the vector so you know which quadrant it lies in.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Displacement, velocity, acceleration and force as vectors

A particle moves in a horizontal plane with constant velocity $(3\mathbf{i}-4\mathbf{j})$ m s $^{-1}$ , where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors in the directions east and north. At time $t=0$ its position vector relative to a fixed origin $O$ is $(2\mathbf{i}+5\mathbf{j})$ m.
Find the position vector of the particle at $t=6$ s.2 marks
A particle moves in a plane with constant acceleration. At time $t=0$ its velocity is $(2\mathbf{i}+\mathbf{j})$ m s $^{-1}$ and 4 s later its velocity is $(14\mathbf{i}-8\mathbf{j})$ m s $^{-1}$ , where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors.
Find the magnitude of the acceleration.2 marks
Relative to a fixed origin $O$ , a ship $S$ moves with constant velocity $(4\mathbf{i}+3\mathbf{j})$ km h $^{-1}$ , where $\mathbf{i}$ and $\mathbf{j}$ are unit vectors due east and due north. At time $t=0$ (hours) the ship is at the point with position vector $(-8\mathbf{i}-5\mathbf{j})$ km. A lighthouse $L$ is at the point with position vector $(12\mathbf{i}+4\mathbf{j})$ km.
Find the position vector of $S$ at time $t$ hours. Hence find the distance of $S$ from $O$ when $t=2$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).