What are vectors in the context of Mechanics 1 for Edexcel International A Levels?

In the context of Mechanics 1 for Edexcel International A Levels, vectors are mathematical entities used to represent quantities that have both magnitude and direction, such as force, velocity, and displacement. Unlike scalars, which only have magnitude, vectors are crucial for solving problems involving direction and movement in mechanics.

How do you add and subtract vectors in Mechanics 1?

To add vectors in Mechanics 1, you use the head-to-tail method or the parallelogram method. In the head-to-tail method, you place the tail of the second vector at the head of the first vector and draw the resultant vector from the tail of the first vector to the head of the second vector. For subtraction, you add the negative of the vector to be subtracted. Algebraically, vectors are added or subtracted by adding or subtracting their corresponding components.

What is the significance of the dot product in vector mechanics?

The dot product, or scalar product, is significant in vector mechanics as it provides a measure of how much one vector extends in the direction of another. It is calculated as the product of the magnitudes of the two vectors and the cosine of the angle between them. The dot product is used to determine angles between vectors and to find projections, which are important in analyzing forces and motion in mechanics.

How do you find the magnitude and direction of a vector in Mechanics 1?

To find the magnitude of a vector in Mechanics 1, you use the Pythagorean theorem on its components. For a vector with components (x, y), the magnitude is calculated as the square root of (x² + y²). To find the direction, you calculate the angle the vector makes with the positive x-axis using the inverse tangent function, tan⁻¹(y/x). This process is essential for understanding the vector's orientation in space.

Why are vectors important in solving mechanics problems in Edexcel International A Levels?

Vectors are important in solving mechanics problems in Edexcel International A Levels because they provide a comprehensive way to represent and analyze physical quantities that have both magnitude and direction. This includes forces, velocities, and accelerations, which are fundamental to understanding motion and equilibrium in mechanics. Vectors allow for precise calculations and predictions of physical behavior in two and three dimensions.

Why is "Quoting $\arctan(v_y / v_x)$ directly as the bearing" a common mistake in Vectors in Mechanics?

Why it happens: Calculators return values in (-90^, 90^) — only correct for the NE quadrant. How to avoid it: Sketch the vector. Identify the quadrant. For SE, bearing = 180^ - (|v_x|/|v_y|). For SW, 180^ + (). For NW, 360^ - (). ALWAYS use a three-figure bearing. Source: WME01/01 examiner reports

Why is "Concluding a collision after checking only one component" a common mistake in Vectors in Mechanics?

Why it happens: Equating i components yields one value of t — students assume that's the collision time. How to avoid it: A collision requires r_A(t) = r_B(t), i.e. BOTH components agree at the SAME t. Always check the second component matches before declaring a collision. Source: WME01/01 examiner reports — flagged

Why is "Differentiating $|\mathbf{r}_{AB}|$ (with the square root) by hand" a common mistake in Vectors in Mechanics?

Why it happens: Trying to optimise distance directly. How to avoid it: Minimise |r_AB|^2 instead — the same t gives the minimum, but the calculus is cleaner (no chain rule on the square root). Or complete the square.

Why is "Writing 'bearing $-30^{\circ}$' or 'bearing $400^{\circ}$'" a common mistake in Vectors in Mechanics?

Why it happens: Treating bearing as just an angle. How to avoid it: Bearings are always in [0^, 360^) and quoted as three-figure numbers (e.g. 045^, 127^, 315^).

Why is "Forgetting the $\tfrac{1}{2}$ in $\tfrac{1}{2}\mathbf{a}t^{2}$" a common mistake in Vectors in Mechanics?

Why it happens: In a vector context, students sometimes carry the 12 only on one component. How to avoid it: Apply the formula in full vector form first: r = r_0 + ut + 12at^2. The 12 multiplies the ENTIRE acceleration vector.

Mechanics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Vectors in Mechanics

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Vectors in Mechanics Study Notes — Edexcel IAL Mathematics M1 (WME01/01, 2018 spec — 2026 onwards)

Two-dimensional kinematics in vector form: position $\mathbf{r}(t)$ , velocity $\mathbf{v}(t)$ and displacement $\mathbf{s}$ in $\mathbf{i}$ , $\mathbf{j}$ form. Closest-approach and collision problems analysed by minimising the relative position vector.

At a glance

Constant velocity: $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}t$ .
Constant acceleration: $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}$ ; $\mathbf{v}(t) = \mathbf{u} + \mathbf{a}t$ .
Magnitude / speed: $|\mathbf{v}| = \sqrt{v_x^{2} + v_y^{2}}$ .
Direction / bearing: SKETCH first; use $\arctan$ only after fixing the quadrant.
Relative position: $\mathbf{r}_{AB} = \mathbf{r}_B - \mathbf{r}_A$ .
Closest approach: minimise $|\mathbf{r}_{AB}|^{2}$ (avoid the square root).
Collision: both $\mathbf{i}$ AND $\mathbf{j}$ components must agree at the same $t$ .
Bearings always three-figure, clockwise from North.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XMA01-YMA01 syllabus (2026 onwards).

M1 3.1 — Use vector notation $\mathbf{r} = x\mathbf{i} + y\mathbf{j}$ for position, velocity and acceleration.
M1 3.2 — Apply $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}$ for constant-acceleration kinematics in 2-D.
M1 3.3 — Solve problems involving closest approach and collisions of two particles moving in a plane.

Vector form of suvat

Apply the suvat equations component-wise in vector form.

In two dimensions, every kinematic quantity is a vector with $\mathbf{i}$ and $\mathbf{j}$ components. The scalar suvat equations carry over directly:

$\mathbf{v} = \mathbf{u} + \mathbf{a}\,t, \qquad \mathbf{s} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}, \qquad \mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}.$

Apply COMPONENT-WISE. If $\mathbf{u} = u_x\mathbf{i} + u_y\mathbf{j}$ and $\mathbf{a} = a_x\mathbf{i} + a_y\mathbf{j}$ :

$v_x = u_x + a_x t$
$v_y = u_y + a_y t$

Worked example. $\mathbf{r}_0 = \mathbf{i} - 3\mathbf{j}$ , $\mathbf{u} = 2\mathbf{i} + 4\mathbf{j}$ , $\mathbf{a} = \mathbf{i} - 2\mathbf{j}$ at $t = 3$ :

$\mathbf{r}(3) = (1 + 6 + 4.5)\mathbf{i} + (-3 + 12 - 9)\mathbf{j} = 11.5\mathbf{i}\,\text{m}.$

$\mathbf{v}(3) = (2 + 3)\mathbf{i} + (4 - 6)\mathbf{j} = 5\mathbf{i} - 2\mathbf{j}\,\text{m s}^{-1}.$

Speed:

$|\mathbf{v}(3)| = \sqrt{25 + 4} = \sqrt{29} \approx 5.39\,\text{m s}^{-1}.$

A 2-D position vector r = x i + y j has magnitude √(x² + y²) and direction set by its bearing from North.

Apply suvat component-wise.
$\mathbf{v} = \mathbf{u} + \mathbf{a}t$ ; $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}$ .
Speed = magnitude of velocity vector.
Don't forget the $\tfrac{1}{2}$ on the ENTIRE acceleration vector.

Direction and bearings

Sketch first; arctan second. Always three-figure bearing.

Bearings are angles measured clockwise from North. They live in $[0^{\circ}, 360^{\circ})$ and are quoted in three-figure form ( $045^{\circ}$ , $127^{\circ}$ , $315^{\circ}$ ).

Bearing $\to$ velocity components: if $\mathbf{i}$ = East and $\mathbf{j}$ = North, then speed $v$ at bearing $\theta$ has components

$v_E = v\sin\theta, \qquad v_N = v\cos\theta.$

(Note the swap: $\sin\theta$ for East, $\cos\theta$ for North, because bearings are from North.)

Velocity components $\to$ bearing: SKETCH the vector. Identify the quadrant. Then compute the angle from North using basic trigonometry — NEVER quote $\arctan(v_y/v_x)$ directly.

Quadrant	Velocity	Bearing
NE	$v_E > 0$ , $v_N > 0$	$\theta = \arctan(v_E/v_N)$
SE	$v_E > 0$ , $v_N < 0$	$\theta = 180^{\circ} - \arctan(v_E/
SW	$v_E < 0$ , $v_N < 0$	$\theta = 180^{\circ} + \arctan(
NW	$v_E < 0$ , $v_N > 0$	$\theta = 360^{\circ} - \arctan(

Example. $\mathbf{v} = 4\mathbf{i} - 3\mathbf{j}$ km/h (with $\mathbf{i}$ East, $\mathbf{j}$ North).

$v_E = 4 > 0$ , $v_N = -3 < 0$ $\Rightarrow$ SE quadrant.
Angle from North: $180^{\circ} - \arctan(4/3) \approx 180^{\circ} - 53.13^{\circ} \approx 126.87^{\circ}$ .
Bearing $\approx 127^{\circ}$ .

Speed (magnitude): $|\mathbf{v}| = \sqrt{4^{2} + 3^{2}} = 5\,\text{km h}^{-1}$ .

Bearings: clockwise from North, three figures.
Bearing $\to$ components: $v_E = v\sin\theta$ , $v_N = v\cos\theta$ .
Components $\to$ bearing: sketch, identify quadrant, then compute.
Never quote $\arctan(v_y/v_x)$ as a bearing directly.

Relative position and closest approach

Minimise $|\mathbf{r}_{AB}|^{2}$ to find when two particles are closest.

Relative position vector: $\mathbf{r}_{AB}(t) = \mathbf{r}_B(t) - \mathbf{r}_A(t)$ . This points from $A$ to $B$ at time $t$ . Its magnitude is the distance between the particles.

Closest approach is the minimum value of $|\mathbf{r}_{AB}|$ over all $t \ge 0$ .

Method 1 — minimise $|\mathbf{r}_{AB}|^{2}$ (recommended):

Write $\mathbf{r}_{AB}(t)$ in component form.
Compute $d^{2}(t) = |\mathbf{r}_{AB}|^{2} = (\mathbf{r}_{AB} \cdot \mathbf{r}_{AB}) =$ quadratic in $t$ .
Minimise: differentiate $d(d^{2})/dt = 0$ , OR complete the square.
Substitute back to find the minimum distance.

Why $d^{2}$ and not $d$ ? $d = \sqrt{d^{2}}$ — minimising one minimises the other (square root is monotonic for $d \ge 0$ ). But $d^{2}$ is a clean quadratic; $d$ involves a square root and chain rule.

Worked example. Ship $A$ at $(0,0)$ with velocity $(3, 4)$ ; ship $B$ at $(10, 0)$ with velocity $(-1, 2)$ , both km/h.

$\mathbf{r}_A = (3t, 4t)$ .
$\mathbf{r}_B = (10 - t, 2t)$ .
$\mathbf{r}_{AB} = (10 - 4t, -2t)$ .
$d^{2} = (10 - 4t)^{2} + 4 t^{2} = 100 - 80 t + 16 t^{2} + 4 t^{2} = 20 t^{2} - 80 t + 100$ .
Complete the square: $20(t - 2)^{2} + 20$ .
Minimum at $t = 2$ h. $d^{2}_{\min} = 20$ , so $d_{\min} = 2\sqrt{5} \approx 4.47$ km.

Relative position: $\mathbf{r}_{AB} = \mathbf{r}_B - \mathbf{r}_A$ .
Closest approach: minimise $|\mathbf{r}_{AB}|^{2}$ .
Differentiate or complete the square.
Square root only once, at the end.

Do the particles collide?

Check BOTH components agree at the SAME $t$ .

Two particles collide if their position vectors are equal at the same instant:

$\mathbf{r}_A(t) = \mathbf{r}_B(t).$

Since these are two-dimensional vectors, this requires TWO scalar equations to hold simultaneously:

$x_A(t) = x_B(t)$
$y_A(t) = y_B(t)$

Procedure:

Equate the $\mathbf{i}$ components; solve for $t$ .
Check the $\mathbf{j}$ components agree at that $t$ .
If both match, they collide at the found $t$ . If not, they don't collide.

Example — $\mathbf{r}_P = (1 + 3t)\mathbf{i} + (2 + t)\mathbf{j}$ , $\mathbf{r}_Q = (7 - t)\mathbf{i} + (-1 + 2t)\mathbf{j}$ .

$\mathbf{i}$ components: $1 + 3t = 7 - t \Rightarrow 4t = 6 \Rightarrow t = 1.5$ .
Check $\mathbf{j}$ at $t = 1.5$ : $2 + 1.5 = 3.5$ vs $-1 + 3 = 2$ . Disagree.
So $P$ and $Q$ do NOT collide.

Note: if a question SAYS the particles collide and asks 'find the time / show that...', then both components SHOULD agree — if your numbers disagree, recheck arithmetic.

Collide $\Leftrightarrow$ position vectors equal at same $t$ .
TWO scalar conditions in 2-D — both must hold.
Equate $\mathbf{i}$ components $\to$ find $t$ ; check $\mathbf{j}$ at that $t$ .
If $\mathbf{j}$ disagrees, they don't collide.

Finding velocity from two position observations

For constant velocity, $\mathbf{v} = (\mathbf{r}_1 - \mathbf{r}_0)/t$ .

If a particle moves with constant velocity and you know its positions at two different times, you can find the velocity directly:

$\mathbf{v} = \dfrac{\mathbf{r}_1 - \mathbf{r}_0}{t_1 - t_0}.$

Then the speed is $|\mathbf{v}|$ and the direction follows by sketch + trigonometry.

Worked example. $\mathbf{r}_0 = 3\mathbf{i} - \mathbf{j}$ at $t = 0$ , $\mathbf{r}_1 = 11\mathbf{i} + 7\mathbf{j}$ at $t = 4$ .

$\mathbf{v} = \dfrac{(11 - 3)\mathbf{i} + (7 - (-1))\mathbf{j}}{4} = \dfrac{8\mathbf{i} + 8\mathbf{j}}{4} = 2\mathbf{i} + 2\mathbf{j}.$

Speed: $|\mathbf{v}| = \sqrt{4 + 4} = 2\sqrt{2} \approx 2.83\,\text{m s}^{-1}$ .
Direction: NE quadrant, angle from North $= \arctan(2/2) = 45^{\circ}$ . Bearing $045^{\circ}$ .

This is just the vector form of 'velocity = displacement / time'.

Constant velocity: $\mathbf{v} = \Delta\mathbf{r}/\Delta t$ .
Speed = $|\mathbf{v}|$ .
Bearing: sketch + arctan + quadrant correction.

How it’s examined

Vectors in mechanics typically appears as a $10$ - $14$ mark question in the second half of the WME01/01 paper. Common setups: two ships / boats / particles with given initial positions and constant velocities, asking for closest approach (worth $5$ - $7$ marks) or collision check ( $4$ - $5$ marks) or finding when they're at a specified separation. Bearings are nearly always required for the final mark — three-figure form, with explicit sketch to justify the quadrant. The minimisation of $|\mathbf{r}_{AB}|^{2}$ via differentiation or completing the square is the standard mark-scheme method.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WME01/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Vectors in Mechanics

Step-by-step solutions to past-paper-style questions on vectors in mechanics, written exactly the way a tutor would explain them at the board.

1Position vector under constant acceleration (7 marks, M1)
Core• Adapted from WME01/01 January 2024 Q3• vectors, position, velocity, acceleration
Question
A particle $P$ has initial position $\mathbf{r}_0 = (\mathbf{i} - 3\mathbf{j})\,\text{m}$ and initial velocity $\mathbf{u} = (2\mathbf{i} + 4\mathbf{j})\,\text{m s}^{-1}$ . It moves with constant acceleration $\mathbf{a} = (\mathbf{i} - 2\mathbf{j})\,\text{m s}^{-2}$ . Find (a) the velocity of $P$ when $t = 3\,\text{s}$ , (b) the position of $P$ when $t = 3\,\text{s}$ , (c) the speed at $t = 3\,\text{s}$ .
Step-by-step solution
1. Step 1
  Draw a coordinate diagram and resolve. Use the vector suvat $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ component-wise.
  $\mathbf{v}(3) = (2\mathbf{i} + 4\mathbf{j}) + 3(\mathbf{i} - 2\mathbf{j}) = 5\mathbf{i} - 2\mathbf{j}\,\text{m s}^{-1}$
2. Step 2
  Use $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}$ .
  $\mathbf{r}(3) = (\mathbf{i} - 3\mathbf{j}) + 3(2\mathbf{i} + 4\mathbf{j}) + \tfrac{1}{2}(9)(\mathbf{i} - 2\mathbf{j})$
3. Step 3
  Simplify component by component.
  $\mathbf{r}(3) = (1 + 6 + 4.5)\mathbf{i} + (-3 + 12 - 9)\mathbf{j} = 11.5\mathbf{i} + 0\mathbf{j}\,\text{m}$
4. Step 4
  Speed = magnitude of velocity at $t = 3$ .
  $|\mathbf{v}(3)| = \sqrt{5^{2} + (-2)^{2}} = \sqrt{29} \approx 5.39\,\text{m s}^{-1}$
Answer
(a) $\mathbf{v}(3) = (5\mathbf{i} - 2\mathbf{j})\,\text{m s}^{-1}$ . (b) $\mathbf{r}(3) = 11.5\mathbf{i}\,\text{m}$ . (c) Speed $= \sqrt{29} \approx 5.39\,\text{m s}^{-1}$ .
Examiner tip
M1 for $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ . A1 for velocity. M1 for $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}$ . A1A1 for both components. M1 for $|\mathbf{v}|$ . A1 for speed. Examiner reports note that students who skip writing the formula in vector form and try to do each component in their head make sign errors on the $\tfrac{1}{2}\mathbf{a}t^{2}$ term.
2Velocity given as speed and bearing (6 marks, M1)
Core• Adapted from WME01/01 June 2023 Q2• vectors, bearing, magnitude and direction
Question
A ship sails with constant velocity of $15\,\text{km h}^{-1}$ on a bearing of $120^{\circ}$ . Taking $\mathbf{i}$ East and $\mathbf{j}$ North, find (a) the velocity vector, (b) the position vector at time $t$ hours given that the ship was at $(\mathbf{i} + 2\mathbf{j})\,\text{km}$ when $t = 0$ .
Step-by-step solution
1. Step 1
  Bearing $120^{\circ}$ is measured clockwise from North. That puts the velocity in the SE quadrant, $30^{\circ}$ South of East.
2. Step 2
  Resolve the velocity. The angle to the East-axis is $90^{\circ} - 120^{\circ}$ in standard bearing convention; geometrically, $v_E = 15\sin 120^{\circ}$ (East component) and $v_N = 15\cos 120^{\circ}$ (North component).
3. Step 3
  Compute components.
  $v_E = 15\sin 120^{\circ} = 15 \cdot \tfrac{\sqrt{3}}{2} \approx 12.99$
4. Step 4
  And the North component (note negative — moving South).
  $v_N = 15\cos 120^{\circ} = 15 \cdot (-\tfrac{1}{2}) = -7.5$
5. Step 5
  Velocity vector:
  $\mathbf{v} = 12.99\mathbf{i} - 7.5\mathbf{j}\,\text{km h}^{-1}$
6. Step 6
  Position at time $t$ :
  $\mathbf{r}(t) = (\mathbf{i} + 2\mathbf{j}) + t(12.99\mathbf{i} - 7.5\mathbf{j}) = (1 + 12.99 t)\mathbf{i} + (2 - 7.5 t)\mathbf{j}$
Answer
(a) $\mathbf{v} \approx (13.0\mathbf{i} - 7.5\mathbf{j})\,\text{km h}^{-1}$ . (b) $\mathbf{r}(t) = (1 + 12.99 t)\mathbf{i} + (2 - 7.5 t)\mathbf{j}\,\text{km}$ .
Examiner tip
B1 for sketch showing bearing $120^{\circ}$ correctly. M1 for resolving using $\sin$ / $\cos$ of the appropriate angle. A1 for $v_E$ . A1 for $v_N$ (with correct sign). M1A1 for position formula. The bearing-to-vector conversion is examined every series; a quick sketch resolves which trig ratio goes with which component.
3Closest approach of two ships (10 marks, M1)
Challenge• Adapted from WME01/01 June 2024 Q5• vectors, closest approach, minimise
Question
At $t = 0$ , ship $A$ is at $(0)\mathbf{i} + 0\mathbf{j}$ moving with velocity $(3\mathbf{i} + 4\mathbf{j})\,\text{km h}^{-1}$ . Ship $B$ is at $(10\mathbf{i} + 0\mathbf{j})$ moving with velocity $(-\mathbf{i} + 2\mathbf{j})\,\text{km h}^{-1}$ . Find the time at which the ships are closest, and the minimum distance between them.
Step-by-step solution
1. Step 1
  Position vectors at time $t$ :
  $\mathbf{r}_A = (3t)\mathbf{i} + (4t)\mathbf{j}, \quad \mathbf{r}_B = (10 - t)\mathbf{i} + (2t)\mathbf{j}$
2. Step 2
  Relative position $\mathbf{AB} = \mathbf{r}_B - \mathbf{r}_A$ .
  $\mathbf{AB} = (10 - 4t)\mathbf{i} + (-2t)\mathbf{j}$
3. Step 3
  Distance squared (avoid the square root for cleaner differentiation):
  $d^{2} = (10 - 4t)^{2} + (-2t)^{2} = 100 - 80 t + 16 t^{2} + 4 t^{2} = 20 t^{2} - 80 t + 100$
4. Step 4
  Method 1 — complete the square. $20(t^{2} - 4 t) + 100 = 20[(t-2)^{2} - 4] + 100 = 20(t-2)^{2} + 20$ .
5. Step 5
  Minimum at $t = 2$ . Minimum $d^{2} = 20$ , so $d_{\min} = \sqrt{20} = 2\sqrt{5} \approx 4.47\,\text{km}$ .
6. Step 6
  Method 2 — differentiate. $\dfrac{d(d^{2})}{dt} = 40 t - 80 = 0 \Rightarrow t = 2$ . Same answer.
Answer
Closest approach at $t = 2$ h with minimum distance $2\sqrt{5} \approx 4.47\,\text{km}$ .
Examiner tip
M1 for position vectors. M1 for relative vector $\mathbf{AB}$ . M1 for $|\mathbf{AB}|^{2}$ . A1 for the quadratic in $t$ . M1 for minimisation (differentiate OR complete-the-square). A1 for $t = 2$ . M1A1 for minimum distance. ECF generous. Many candidates differentiate $|\mathbf{AB}|$ instead of $|\mathbf{AB}|^{2}$ — both work but $d^{2}$ is cleaner because there is no square root.
4Do two particles collide? (8 marks, M1)
Extended• Adapted from WME01/01 January 2023 Q4• vectors, collision, position vectors
Question
Two particles $P$ and $Q$ move with constant velocities. At time $t = 0$ , $P$ is at $(\mathbf{i} + 2\mathbf{j})\,\text{m}$ with velocity $(3\mathbf{i} + \mathbf{j})\,\text{m s}^{-1}$ ; $Q$ is at $(7\mathbf{i} - \mathbf{j})\,\text{m}$ with velocity $(-\mathbf{i} + 2\mathbf{j})\,\text{m s}^{-1}$ . (a) Show that the particles collide and find the time and position of collision.
Step-by-step solution
1. Step 1
  Position vectors at time $t$ :
  $\mathbf{r}_P = (1 + 3t)\mathbf{i} + (2 + t)\mathbf{j}, \quad \mathbf{r}_Q = (7 - t)\mathbf{i} + (-1 + 2t)\mathbf{j}$
2. Step 2
  Collision requires $\mathbf{r}_P = \mathbf{r}_Q$ at the same time $t$ . Equate $\mathbf{i}$ components:
  $1 + 3t = 7 - t \Rightarrow 4t = 6 \Rightarrow t = 1.5\,\text{s}$
3. Step 3
  Check $\mathbf{j}$ components at $t = 1.5$ :
  $2 + 1.5 = 3.5, \quad -1 + 2(1.5) = 2$
4. Step 4
  $3.5 \ne 2$ — the $\mathbf{j}$ components do not match. So $P$ and $Q$ are not at the same position at $t = 1.5$ , and there is no time at which all components match. The particles do not collide.
5. Step 5
  Note: the question as stated says 'show that the particles collide', but our check shows they do not. In a real exam this would mean rechecking — but the technique is the standard one: equate $\mathbf{i}$ components, find $t$ , then check $\mathbf{j}$ components agree at that $t$ .
Answer
Equating $\mathbf{i}$ gives $t = 1.5\,\text{s}$ , but the $\mathbf{j}$ components disagree ( $3.5 \ne 2$ ), so the particles do not collide. If a question claims they do, recheck the data or assumptions.
Examiner tip
M1 for position vectors. M1 for equating $\mathbf{i}$ components. A1 for $t = 1.5$ . M1 for checking $\mathbf{j}$ components. A1 for the conclusion. B1B1 for the position vector and clear reasoning. Most exam questions DO have a collision (so both equations give the same $t$ ); if they don't, you must say so explicitly — examiners credit the reasoning over the numerical 'answer'.
5Finding velocity from two position observations (7 marks, M1)
Extended• Adapted from WME01/01 January 2024 Q6• vectors, velocity, from observations
Question
At $t = 0$ , a particle $P$ is at $\mathbf{r}_0 = (3\mathbf{i} - \mathbf{j})\,\text{m}$ . At $t = 4\,\text{s}$ it is at $\mathbf{r}_1 = (11\mathbf{i} + 7\mathbf{j})\,\text{m}$ . Given that the velocity is constant, find (a) the velocity, (b) the speed, (c) the bearing on which $P$ is moving.
Step-by-step solution
1. Step 1
  Constant velocity: $\mathbf{v} = (\mathbf{r}_1 - \mathbf{r}_0)/t$ .
  $\mathbf{v} = \dfrac{(11 - 3)\mathbf{i} + (7 - (-1))\mathbf{j}}{4} = \dfrac{8\mathbf{i} + 8\mathbf{j}}{4} = 2\mathbf{i} + 2\mathbf{j}\,\text{m s}^{-1}$
2. Step 2
  Speed = magnitude.
  $|\mathbf{v}| = \sqrt{4 + 4} = 2\sqrt{2} \approx 2.83\,\text{m s}^{-1}$
3. Step 3
  Bearing: $\mathbf{v} = 2\mathbf{i} + 2\mathbf{j}$ — $2$ East and $2$ North (NE quadrant). Angle from North: $\arctan(2/2) = 45^{\circ}$ .
4. Step 4
  Bearing = $045^{\circ}$ (three-figure form, measured clockwise from North).
Answer
(a) $\mathbf{v} = (2\mathbf{i} + 2\mathbf{j})\,\text{m s}^{-1}$ . (b) Speed $= 2\sqrt{2} \approx 2.83\,\text{m s}^{-1}$ . (c) Bearing $= 045^{\circ}$ .
Examiner tip
M1 for velocity = displacement/time. A1 for velocity vector. M1 for magnitude. A1. M1 for bearing argument with sketch. A1 for $045^{\circ}$ . Use three-figure bearings (e.g. $045^{\circ}$ not $45^{\circ}$ ) — examiners deduct a B-mark for missing leading zero.

Key Formulae — Vectors in Mechanics

The formulae you need to memorise for vectors in mechanics on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Vector form: $\mathbf{v} = \mathbf{u} + \mathbf{a}t$
$\mathbf{v} = \mathbf{u} + \mathbf{a}\,t$
$\mathbf{u}, \mathbf{v}$
initial and final velocity vectors
$\mathbf{a}$
constant acceleration vector
$t$
time (s)
When to use
Component-wise application of suvat. Useful when motion is in two dimensions and acceleration has both $\mathbf{i}$ and $\mathbf{j}$ components. NOT in the IAL formula booklet but follows from the scalar version.
Example
$\mathbf{u} = 2\mathbf{i} + 4\mathbf{j}$ , $\mathbf{a} = \mathbf{i} - 2\mathbf{j}$ at $t = 3$ : $\mathbf{v} = 5\mathbf{i} - 2\mathbf{j}$ .
Vector form: $\mathbf{s} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}$
$\mathbf{s} = \mathbf{u}\,t + \tfrac{1}{2}\mathbf{a}\,t^{2}$
$\mathbf{s}$
displacement vector from start
$\mathbf{u}$
initial velocity
$\mathbf{a}$
acceleration
When to use
Find displacement (and hence position) under constant acceleration in vector form. NOT in the booklet but routinely used.
Example
$\mathbf{u} = (2, 4)$ , $\mathbf{a} = (1, -2)$ , $t = 3$ : $\mathbf{s} = (6, 12) + \tfrac{1}{2}(9, -18) = (10.5, 3)$ .
Position vector under constant acceleration
$\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}$
$\mathbf{r}(t)$
position at time $t$
$\mathbf{r}_0$
initial position
$\mathbf{u}, \mathbf{a}$
initial velocity, constant acceleration
When to use
Most common starting point for vector kinematics. Reduces to $\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t$ when $\mathbf{a} = \mathbf{0}$ (constant velocity).
Example
$\mathbf{r}_0 = \mathbf{i} - 3\mathbf{j}$ , $\mathbf{u} = 2\mathbf{i} + 4\mathbf{j}$ , $\mathbf{a} = \mathbf{i} - 2\mathbf{j}$ at $t = 3$ : $\mathbf{r} = 11.5\mathbf{i}$ .
Relative position vector
$\mathbf{r}_{AB}(t) = \mathbf{r}_B(t) - \mathbf{r}_A(t)$
$\mathbf{r}_{AB}$
position of $B$ relative to $A$
$\mathbf{r}_A, \mathbf{r}_B$
positions of $A$ and $B$ at time $t$
When to use
First step in closest-approach and collision problems. The distance between the particles is $|\mathbf{r}_{AB}|$ — minimise this to find the closest approach.
Example
$\mathbf{r}_A = (3t, 4t)$ , $\mathbf{r}_B = (10 - t, 2t)$ : $\mathbf{r}_{AB} = (10 - 4t, -2t)$ .
Magnitude and direction of a 2-D vector
$|\mathbf{a}| = \sqrt{a_x^{2} + a_y^{2}}, \quad \tan\theta = a_y / a_x$
$\mathbf{a}$
vector in 2-D
$\theta$
angle from positive $\mathbf{i}$ -direction
When to use
Convert vector form to speed + bearing. Always sketch to pick the correct quadrant — calculators give the principal value of $\arctan$ , which may need $\pm 180^{\circ}$ adjustment.
Example
$\mathbf{v} = -\mathbf{i} + \mathbf{j}$ : $|\mathbf{v}| = \sqrt{2}$ , $\theta = 135^{\circ}$ (NW quadrant) $\Rightarrow$ bearing $315^{\circ}$ .

Key Definitions and Keywords — Vectors in Mechanics

Definitions to memorise and the exact keywords mark schemes credit for vectors in mechanics answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Position vector
Examiner keyword
The vector $\mathbf{r}$ from a fixed origin $O$ to the particle. Changes with time as the particle moves. Velocity is the time derivative: $\mathbf{v} = \dot{\mathbf{r}}$ .
Velocity vector
Examiner keyword
The rate of change of position with respect to time. A vector with magnitude = speed and direction = direction of motion. In $\mathbf{i}$ , $\mathbf{j}$ form: $\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j}$ .
Relative velocity
The velocity of one particle as seen from another: $\mathbf{v}_{B|A} = \mathbf{v}_B - \mathbf{v}_A$ . Used in closest-approach problems where the 'observer-particle frame' simplifies the geometry.
Closest approach
Examiner keyword
The minimum value of $|\mathbf{r}_B - \mathbf{r}_A|$ over time. Found by differentiating $|\mathbf{r}_{AB}|^{2}$ with respect to $t$ and setting equal to zero, or by completing the square.
Collide
Two particles collide if their position vectors are equal at the same instant. Requires BOTH the $\mathbf{i}$ and $\mathbf{j}$ components to agree at the same $t$ .
Unit vectors $\mathbf{i}$ and $\mathbf{j}$
Standard 2-D basis: $\mathbf{i}$ points along the positive $x$ -axis, $\mathbf{j}$ along the positive $y$ -axis. In navigation: $\mathbf{i}$ East, $\mathbf{j}$ North.

Common Mistakes and Misconceptions — Vectors in Mechanics

The traps other students keep falling into on vectors in mechanics questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Quoting $\arctan(v_y / v_x)$ directly as the bearing
WME01/01 examiner reports
Why it happens
Calculators return values in $(-90^{\circ}, 90^{\circ})$ — only correct for the NE quadrant.
How to avoid it
Sketch the vector. Identify the quadrant. For SE, bearing $= 180^{\circ} - \arctan(|v_x|/|v_y|)$ . For SW, $180^{\circ} + \arctan(\ldots)$ . For NW, $360^{\circ} - \arctan(\ldots)$ . ALWAYS use a three-figure bearing.
✕Concluding a collision after checking only one component
WME01/01 examiner reports — flagged
Why it happens
Equating $\mathbf{i}$ components yields one value of $t$ — students assume that's the collision time.
How to avoid it
A collision requires $\mathbf{r}_A(t) = \mathbf{r}_B(t)$ , i.e. BOTH components agree at the SAME $t$ . Always check the second component matches before declaring a collision.
✕Differentiating $|\mathbf{r}_{AB}|$ (with the square root) by hand
Why it happens
Trying to optimise distance directly.
How to avoid it
Minimise $|\mathbf{r}_{AB}|^{2}$ instead — the same $t$ gives the minimum, but the calculus is cleaner (no chain rule on the square root). Or complete the square.
✕Writing 'bearing $-30^{\circ}$ ' or 'bearing $400^{\circ}$ '
Why it happens
Treating bearing as just an angle.
How to avoid it
Bearings are always in $[0^{\circ}, 360^{\circ})$ and quoted as three-figure numbers (e.g. $045^{\circ}$ , $127^{\circ}$ , $315^{\circ}$ ).
✕Forgetting the $\tfrac{1}{2}$ in $\tfrac{1}{2}\mathbf{a}t^{2}$
Why it happens
In a vector context, students sometimes carry the $\tfrac{1}{2}$ only on one component.
How to avoid it
Apply the formula in full vector form first: $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}$ . The $\tfrac{1}{2}$ multiplies the ENTIRE acceleration vector.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Vectors in Mechanics — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Vectors in Mechanics

Short Study Notes in the form of Slides

Short Study Notes — Vectors in Mechanics

Detailed Notes

What you’ll learn

How it’s examined

1Position vector under constant acceleration (7 marks, M1)

2Velocity given as speed and bearing (6 marks, M1)

3Closest approach of two ships (10 marks, M1)

4Do two particles collide? (8 marks, M1)

5Finding velocity from two position observations (7 marks, M1)

Vector form: v=u+at\mathbf{v} = \mathbf{u} + \mathbf{a}tv=u+at

Vector form: s=ut+12at2\mathbf{s} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}s=ut+21​at2

Position vector under constant acceleration

Relative position vector

Magnitude and direction of a 2-D vector

Position vector

Velocity vector

Relative velocity

Closest approach

Collide

Unit vectors i\mathbf{i}i and j\mathbf{j}j

✕Quoting arctan⁡(vy/vx)\arctan(v_y / v_x)arctan(vy​/vx​) directly as the bearing

✕Concluding a collision after checking only one component

✕Differentiating ∣rAB∣|\mathbf{r}_{AB}|∣rAB​∣ (with the square root) by hand

✕Writing 'bearing −30∘-30^{\circ}−30∘' or 'bearing 400∘400^{\circ}400∘'

✕Forgetting the 12\tfrac{1}{2}21​ in 12at2\tfrac{1}{2}\mathbf{a}t^{2}21​at2

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Vectors in Mechanics — frequently asked questions

Vector form: $\mathbf{v} = \mathbf{u} + \mathbf{a}t$

Vector form: $\mathbf{s} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^{2}$

Unit vectors $\mathbf{i}$ and $\mathbf{j}$

✕Quoting $\arctan(v_y / v_x)$ directly as the bearing

✕Differentiating $|\mathbf{r}_{AB}|$ (with the square root) by hand

✕Writing 'bearing $-30^{\circ}$ ' or 'bearing $400^{\circ}$ '

✕Forgetting the $\tfrac{1}{2}$ in $\tfrac{1}{2}\mathbf{a}t^{2}$