Vectors in Mechanics

Edexcel International A-Level Mathematics· M1 §2.1, M1 §2.2· 25 min read

1. 2D Vector Basics: Magnitude, Direction and i,j Notation★★☆☆☆⏱ 6 min

✓ Calculator OK

All vectors in M1 are 2-dimensional, represented using perpendicular unit vectors $i$ (horizontal, right) and $j$ (vertical, up). A vector written as $x i + y j$ has x-component $x$ and y-component $y$ .

📘 Definition

Vector Magnitude and Direction

The magnitude of a vector $v = x i + y j$ is calculated using Pythagoras' theorem. Direction is given as an angle relative to a reference axis (usually the positive horizontal axis).

|\boldsymbol{v}| = \sqrt{x^2 + y^2}

📐 Worked Example

Calculate the magnitude and direction of the vector $v = 3 i + 4 j$ . Give direction to 1 decimal place, relative to the positive horizontal axis.

1
Step 1: Calculate magnitude using Pythagoras:
2
$|\boldsymbol{v}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$
3
Step 2: Calculate direction using arctangent of (y-component / x-component):
4
$\theta = \arctan\left(\frac{4}{3}\right) = 53.1^\circ$
5
Final answer: Magnitude = 5, direction = 53.1° above positive horizontal axis

2. Vector Resolution and Resultant Calculation★★★☆☆⏱ 7 min

✓ Calculator OK

Resolving a vector means splitting it into perpendicular i and j components. To find the resultant of multiple vectors, add their i components together and their j components together separately.

📘 Definition

Vector Resolution

A vector of magnitude $∣ v ∣$ at angle $θ$ to the positive horizontal axis has components $∣ v ∣ cos θ$ (i) and $∣ v ∣ sin θ$ (j).

📐 Worked Example

Two forces act on a particle: $F_{1} = 2 i + 5 j$ N and $F_{2} = 4 i - 1 j$ N. Find the resultant force $R$ and its magnitude to 3 significant figures.

1
Step 1: Add i components of the two forces:
2
$x = 2 + 4 = 6$
3
Step 2: Add j components of the two forces:
4
$y = 5 + (-1) = 4$
5
Step 3: Write resultant vector:
6
$\boldsymbol{R} = 6\boldsymbol{i} + 4\boldsymbol{j} \text{ N}$
7
Step 4: Calculate magnitude:
8
$|\boldsymbol{R}| = \sqrt{6^2 + 4^2} = \sqrt{52} = 7.21 \text{ N (3 sf)}$

3. Vectors for Displacement, Velocity and Acceleration★★★☆☆⏱ 7 min

✓ Calculator OK

All motion quantities in M1 can be represented as vectors. For constant velocity, velocity is change in displacement over time. For constant acceleration, acceleration is change in velocity over time. No calculus is used for these calculations in M1.

📘 Definition

Constant Vector Motion Formulae

For constant velocity $v$ : $s = s_{0} + v t$ , where $s$ is final displacement, $s_{0}$ is initial displacement, $t$ is time. For constant acceleration $a$ : $v = u + a t$ , where $u$ is initial velocity, $v$ is final velocity.

📐 Worked Example

A particle has initial velocity $u = 2 i + 3 j$ m/s, and accelerates at $a = 1 i - 0.5 j$ m/s² for 4 seconds. Find the final velocity of the particle.

1
Step 1: Use the constant acceleration formula for vectors:
2
$\boldsymbol{v} = \boldsymbol{u} + \boldsymbol{a}t$
3
Step 2: Substitute values:
4
$\boldsymbol{v} = 2\boldsymbol{i} + 3\boldsymbol{j} + 4(1\boldsymbol{i} - 0.5\boldsymbol{j})$
5
Step 3: Expand and simplify components:
6
$\boldsymbol{v} = 2\boldsymbol{i} + 3\boldsymbol{j} + 4\boldsymbol{i} - 2\boldsymbol{j} = 6\boldsymbol{i} + 1\boldsymbol{j} \text{ m/s}$

4. Exam-Style Vector Problem Solving★★★★☆⏱ 8 min

✓ Calculator OK

Most exam vector questions combine resolution, resultant calculation and motion formulae. Start every problem by sketching a quick vector diagram to confirm direction of components before calculating values.

📐 Worked Example

A hiker walks 10 km at 30° above the positive horizontal axis, then walks 8 km at 60° below the positive horizontal axis. Find the total displacement vector of the hiker, to 3 significant figures.

1
Step 1: Resolve first displacement into components:
2
$\boldsymbol{s_1} = 10\cos30^\circ \boldsymbol{i} + 10\sin30^\circ \boldsymbol{j} = 8.66\boldsymbol{i} + 5\boldsymbol{j} \text{ km}$
3
Step 2: Resolve second displacement into components (note negative sine value for downward angle):
4
$\boldsymbol{s_2} = 8\cos(-60^\circ) \boldsymbol{i} + 8\sin(-60^\circ) \boldsymbol{j} = 4\boldsymbol{i} - 6.93\boldsymbol{j} \text{ km}$
5
Step 3: Add components to find total displacement:
6
$\boldsymbol{s_{total}} = (8.66 + 4)\boldsymbol{i} + (5 - 6.93)\boldsymbol{j} = 12.7\boldsymbol{i} - 1.93\boldsymbol{j} \text{ km (3 sf)}$

Exam tip:

Edexcel examiners award 1 mark for correctly resolving each vector into i and j components, even if your final answer is wrong. Always write down each resolution step clearly to pick up method marks.

5. Common Pitfalls

Wrong move:

Forgetting to state direction relative to a reference axis when asked for vector direction

Why:

Mark schemes require explicit reference to a reference axis (e.g. positive horizontal, North) to award full marks

Correct move:

Always write direction as e.g. "53° above the positive horizontal axis" or "27° clockwise from North"

Wrong move:

Using calculus to differentiate/integrate vectors for velocity/displacement

Why:

Calculus of vectors is out of scope for M1, all problems use constant velocity/acceleration so linear ratio formulae only are required

Correct move:

Use $v = Δ s /Δ t$ for constant velocity, $v = u + a Δ t$ for constant acceleration

Wrong move:

Mixing up sine and cosine when resolving vector components

Why:

The component adjacent to the angle uses cosine, opposite uses sine, swapping gives incorrect values

Correct move:

Label the angle between the vector and the x-axis, confirm adjacent = $∣ v ∣ cos θ$ , opposite = $∣ v ∣ sin θ$ before calculating

Wrong move:

Forgetting negative signs for components pointing left or down

Why:

$i$ is positive right, $j$ is positive up, so vectors pointing opposite directions have negative components

Correct move:

Assign sign convention at the start of every problem, check component signs align with direction

Wrong move:

Calculating magnitude by adding components directly instead of using Pythagoras

Why:

Vector components are perpendicular, arithmetic sum does not give the correct resultant length

Correct move:

Always calculate magnitude as $(i component)^{2} + (j component)^{2}$

6. Quick Reference Cheatsheet

Concept	Formula / Rule	Units (where applicable)
Magnitude of vector $v = x i + y j$	$∣ v ∣ = x^{2} + y^{2}$	N, m/s, m, m/s²
Resolve vector of magnitude $∣ v ∣$ at angle $θ$ to x-axis	$v = ∣ v ∣ cos θ i + ∣ v ∣ sin θ j$
Resultant of vectors $v_{1}, v_{2}$	$R = (x_{1} + x_{2}) i + (y_{1} + y_{2}) j$	Same as input vectors
Constant velocity motion	$s = s_{0} + v t$	m, m/s
Constant acceleration motion	$v = u + a t$	m/s, m/s²

7. Frequently Asked

Do I need to use i,j notation for all vector answers in M1?

Unless the question specifies otherwise, you can answer using either magnitude-direction form or i,j notation, but i,j is almost always faster for calculation and reduces error risk.

Can I use bearing notation for vector directions?

Yes, as long as you clearly state you are using 3-figure bearings relative to North, otherwise state direction relative to the positive horizontal (x) axis.

What's Next

Now that you have mastered vector basics for M1, you can apply these skills to a wide range of mechanics problems in your Edexcel IAL exam. Vectors are foundational for force equilibrium, projectile motion, and connected particle problems that make up the rest of the M1 syllabus. Be sure to practice mixing vector calculations with Newton's laws of motion, as this is a common high-mark question format on past papers. Remember that all M1 vector problems use constant acceleration or velocity, so you never need to use calculus for these questions. When solving exam problems, always start by sketching a quick vector diagram to confirm component directions before you perform calculations, as this will help you avoid sign errors and lost marks.