Vectors in Mechanics

This lesson covers the use of vectors in mechanics as required by the Edexcel 9MA0 A-Level Mathematics specification. You need to work with position vectors, velocity vectors and force vectors in two dimensions, using both column-vector and i-j notation.

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Vector Notation

In two dimensions, vectors can be written as:

• Column notation: v = (3, -2) or as a column
• i-j notation: v = 3i - 2j

where i is the unit vector in the x-direction and j is the unit vector in the y-direction.

The magnitude of v = ai + bj is:

|v| = √(a² + b²)

The direction (angle θ with the positive x-axis) is:

θ = arctan(b/a)

Exam Tip: When finding the direction, always draw a sketch to check which quadrant the vector is in. The arctan function only gives angles in the range -90° to 90°, so you may need to add or subtract 180°.

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Position, Velocity and Acceleration Vectors

If a particle has position vector r, velocity vector v and acceleration vector a:

v = dr/dt (differentiate each component with respect to t) a = dv/dt (differentiate again)

And conversely: v = ∫a dt r = ∫v dt

Example: A particle has position vector r = (3t² + 1)i + (4t - t²)j.

v = dr/dt = 6t i + (4 - 2t)j a = dv/dt = 6i - 2j (constant acceleration)

At t = 2: r = 13i + 4j, v = 12i + 0j (moving purely in the x-direction).

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Displacement and Distance

The displacement from point A to point B is a vector: AB = rB - rA

The distance from A to B is the magnitude of this vector: |AB| = √((xB - xA)² + (yB - yA)²)

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Constant Acceleration in Vector Form

The SUVAT equations can be written in vector form:

v = u + at r = r₀ + ut + ½at²

where r₀ is the initial position, u is the initial velocity, and a is the constant acceleration.

Example: A particle starts at position (2i + 3j) m with velocity (4i - j) m s⁻¹ and has acceleration (-2i + 3j) m s⁻².

At time t: r = (2i + 3j) + (4i - j)t + ½(-2i + 3j)t² r = (2 + 4t - t²)i + (3 - t + 1.5t²)j

At t = 3: r = (2 + 12 - 9)i + (3 - 3 + 13.5)j = 5i + 13.5j

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Force Vectors

Forces are vectors and can be added using vector addition.

Resultant force: The sum of all force vectors.

F_resultant = F₁ + F₂ + F₃ + ...

Newton's Second Law in vector form:

F = ma

This gives both the magnitude and direction of the acceleration.

Example: Forces (3i + 4j) N and (-1i + 2j) N act on a 2 kg particle. Find the acceleration.

Resultant = (3 - 1)i + (4 + 2)j = 2i + 6j N a = F/m = (2i + 6j)/2 = i + 3j m s⁻²

|a| = √(1 + 9) = √10 ≈ 3.16 m s⁻²

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Equilibrium in Vector Form

A particle is in equilibrium when the resultant force is the zero vector:

F₁ + F₂ + ... + Fₙ = 0i + 0j

This gives two scalar equations (one for the i-component, one for the j-component).

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Finding Speed and Direction

If v = ai + bj, then:

• Speed = |v| = √(a² + b²)
• Direction: angle θ = arctan(b/a) measured from the positive x-axis (or as a bearing if specified).

Example: v = -3i + 4j m s⁻¹. Speed = √(9 + 16) = 5 m s⁻¹. Direction: arctan(4/(-3)) — the vector is in the second quadrant, so θ = 180° - arctan(4/3) ≈ 180° - 53.1° = 126.9° from the positive x-axis.

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Unit Vectors

A unit vector in the direction of v is:

v-hat = v / |v|

Example: v = 6i + 8j. |v| = 10. Unit vector = (6i + 8j)/10 = 0.6i + 0.8j.

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Summary

• Vectors in mechanics can be expressed as i-j notation or column vectors.
• v = dr/dt and a = dv/dt; integrate to reverse the process.
• SUVAT extends to vector form: v = u + at and r = r₀ + ut + ½at².
• Newton's Second Law: F = ma gives both magnitude and direction.
• Speed = |v|, and direction from arctan (always check the quadrant).

A-Level Deep Dive: Vectors in Mechanics

Spec mapping

Edexcel 9MA0-03 specification, Paper 3 — Statistics and Mechanics, Sections 9-10 (Kinematics and Forces). Section 9 covers "Understand and use the language of kinematics: position, displacement, distance travelled, velocity, speed, acceleration" with explicit vector treatment in two dimensions, including $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$r=r0​+ut+21​at2 and the related $\mathbf{v} = \mathbf{u} + \mathbf{a}t$v=u+at. Section 10 covers Newton's laws in vector form, with $\mathbf{F} = m\mathbf{a}$F=ma applied to particles moving under constant force in a plane. Vector mechanics is examined on Paper 3 only, but it draws synoptically on Paper 1 Section 10 (Vectors as pure objects) for the algebra of $\mathbf{i}$i, $\mathbf{j}$j components, dot product, magnitude, and unit vectors. The Edexcel formula booklet provides scalar SUVAT but does not re-state the vector forms — students must transcribe them from $s = ut + \tfrac{1}{2}at^2$s=ut+21​at2 themselves.

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Worked example with full mark scheme

Question (8 marks):

A ship $S$S moves in a straight line with constant velocity. At time $t = 0$t=0 hours, $S$S is at the position with vector $(3\mathbf{i} + 2\mathbf{j})$(3i+2j) km relative to a fixed origin $O$O, where $\mathbf{i}$i and $\mathbf{j}$j point east and north respectively. The velocity of $S$S is $(4\mathbf{i} - \mathbf{j})$(4i−j) km/h.

A lighthouse $L$L is at the fixed position $(15\mathbf{i} - 4\mathbf{j})$(15i−4j) km.

(a) Write down the position vector $\mathbf{r}(t)$r(t) of $S$S at time $t$t. (1)

(b) Find the value of $t$t at which $S$S is closest to $L$L, and the distance $|SL|$∣SL∣ at that moment. (7)

Solution with mark scheme:

(a) Using $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u}t$r(t)=r0​+ut with zero acceleration:

$\mathbf{r}(t) = (3 + 4t)\mathbf{i} + (2 - t)\mathbf{j}$r(t)=(3+4t)i+(2−t)j

B1 — correct vector position function with both components.

(b) Step 1 — write the displacement $\overrightarrow{LS}(t)$LS(t).

$\overrightarrow{LS}(t) = \mathbf{r}(t) - \mathbf{r}_L = (3 + 4t - 15)\mathbf{i} + (2 - t - (-4))\mathbf{j} = (4t - 12)\mathbf{i} + (6 - t)\mathbf{j}$LS(t)=r(t)−rL​=(3+4t−15)i+(2−t−(−4))j=(4t−12)i+(6−t)j

M1 — subtracting position vectors in the correct order. Common error: subtracting $\mathbf{r}_S - \mathbf{r}_L$rS​−rL​ vs $\mathbf{r}_L - \mathbf{r}_S$rL​−rS​ — the magnitude is unchanged so this does not lose marks here, but on questions involving direction (bearing) the order matters absolutely.

A1 — correct components for the displacement.

Step 2 — square the magnitude (avoid square roots until the end).

$|\overrightarrow{LS}|^2 = (4t - 12)^2 + (6 - t)^2$∣LS∣2=(4t−12)2+(6−t)2

Expand:

$= 16t^2 - 96t + 144 + 36 - 12t + t^2 = 17t^2 - 108t + 180$=16t2−96t+144+36−12t+t2=17t2−108t+180

M1 — squaring magnitude correctly (using $|a\mathbf{i} + b\mathbf{j}|^2 = a^2 + b^2$∣ai+bj∣2=a2+b2). This is the AO2 reasoning move: examiners reward minimising $|LS|^2$∣LS∣2 rather than $|LS|$∣LS∣ because the square root has the same minimum at the same $t$t.

A1 — correct quadratic in $t$t.

Step 3 — minimise.

Differentiate or complete the square. Using calculus:

$\frac{d}{dt}|LS|^2 = 34t - 108 = 0 \implies t = \frac{108}{34} = \frac{54}{17}$dtd​∣LS∣2=34t−108=0⟹t=34108​=1754​

M1 — valid minimisation method (calculus or completing the square).

A1 — correct $t = \dfrac{54}{17}$t=1754​ hours.

Step 4 — substitute back to find minimum distance.

$|LS|^2_{\min} = 17 \cdot \frac{54^2}{17^2} - 108 \cdot \frac{54}{17} + 180 = \frac{2916}{17} - \frac{5832}{17} + \frac{3060}{17} = \frac{144}{17}$∣LS∣min2​=17⋅172542​−108⋅1754​+180=172916​−175832​+173060​=17144​

So $|LS|_{\min} = \dfrac{12}{\sqrt{17}} = \dfrac{12\sqrt{17}}{17}$∣LS∣min​=17​12​=171217​​ km.

A1 — correct minimum distance, rationalised.

Total: 8 marks (B1 M3 A4).

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Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): A particle $P$P of mass $2$2 kg moves under the action of a single constant force $\mathbf{F}$F N. At time $t = 0$t=0 s the particle is at the origin moving with velocity $(3\mathbf{i} + \mathbf{j})$(3i+j) m/s. At time $t = 4$t=4 s the particle has velocity $(-\mathbf{i} + 5\mathbf{j})$(−i+5j) m/s.

(a) Find the acceleration of $P$P. (2)

(b) Hence find $\mathbf{F}$F and the magnitude of $\mathbf{F}$F. (2)

(c) Find the position vector of $P$P at $t = 4$t=4 s. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — applying $\mathbf{a} = (\mathbf{v} - \mathbf{u})/t = ((-1 - 3)\mathbf{i} + (5 - 1)\mathbf{j})/4$a=(v−u)/t=((−1−3)i+(5−1)j)/4.
• A1 (AO1.1b) — correct $\mathbf{a} = (-\mathbf{i} + \mathbf{j})$a=(−i+j) m/s$^2$2.

(b)

• M1 (AO1.1b) — applying $\mathbf{F} = m\mathbf{a} = 2 \cdot (-\mathbf{i} + \mathbf{j})$F=ma=2⋅(−i+j).
• A1 (AO1.1b) — $\mathbf{F} = -2\mathbf{i} + 2\mathbf{j}$F=−2i+2j N, magnitude $|\mathbf{F}| = 2\sqrt{2}$∣F∣=22​ N.

(c)

• M1 (AO3.1a) — applying $\mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$r=ut+21​at2 with $\mathbf{r}_0 = \mathbf{0}$r0​=0.
• A1 (AO1.1b) — $\mathbf{r}(4) = 4(3\mathbf{i} + \mathbf{j}) + 8(-\mathbf{i} + \mathbf{j}) = (4\mathbf{i} + 12\mathbf{j})$r(4)=4(3i+j)+8(−i+j)=(4i+12j) m.

Total: 6 marks split AO1 = 5, AO3 = 1. The AO3 mark is for selecting the correct vector SUVAT formula given the data available (initial velocity and acceleration, not initial and final velocity).

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Synoptic links

Connects to:

1. Paper 1 Section 10 — Vectors (pure): the algebra of $\mathbf{i}\mathbf{j}$ij components, magnitude $|a\mathbf{i} + b\mathbf{j}| = \sqrt{a^2 + b^2}$∣ai+bj∣=a2+b2​, and unit-vector direction $\hat{\mathbf{r}} = \mathbf{r}/|\mathbf{r}|$r^=r/∣r∣ are imported wholesale into mechanics. Without confidence in pure vector arithmetic, mechanics calculations stall at the first subtraction.

2. Paper 3 Section 9 — Scalar kinematics (SUVAT): every vector SUVAT equation reduces to two scalar SUVAT equations, one per component. $\mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$r=ut+21​at2 is just $x = u_x t + \tfrac{1}{2}a_x t^2$x=ux​t+21​ax​t2 and $y = u_y t + \tfrac{1}{2}a_y t^2$y=uy​t+21​ay​t2 run in parallel. Recognising this collapses "scary" 2D problems into routine 1D ones.

3. Paper 3 Section 10 — Forces and Newton's laws: $\mathbf{F} = m\mathbf{a}$F=ma in vector form means each component of force produces the corresponding component of acceleration independently. This is why projectile motion under gravity has horizontal velocity unchanged: $F_x = 0$Fx​=0 implies $a_x = 0$ax​=0.

4. Relative velocity: $\mathbf{v}_{A/B} = \mathbf{v}_A - \mathbf{v}_B$vA/B​=vA​−vB​ is the velocity of $A$A as seen from $B$B. Closest-approach problems are cleanest in the frame of one of the moving objects, where the other moves in a straight line and the minimum separation is the perpendicular distance from the origin to that line.

5. Bearings and navigation: mechanics questions often dress vectors in compass bearings — $\mathbf{i}$i east, $\mathbf{j}$j north — and require converting between bearing form (e.g. $030°$030°) and component form. A bearing $\theta$θ measured clockwise from north corresponds to velocity $v(\sin\theta\,\mathbf{i} + \cos\theta\,\mathbf{j})$v(sinθi+cosθj).

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Mark-scheme literacy

Vectors-in-mechanics questions on 9MA0-03 typically split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Applying $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$r=r0​+ut+21​at2, computing magnitudes, applying $\mathbf{F} = m\mathbf{a}$F=ma component-wise
AO2 (reasoning / interpretation)	20–30%	Justifying use of $
AO3 (problem-solving / modelling)	15–25%	Setting up the displacement vector for closest-approach problems; selecting between SUVAT variants given partial data; interpreting "collide" or "intercept" as a system of vector equations

Examiner-rewarded phrasing: "the displacement of $S$S from $L$L at time $t$t is …"; "since $|LS|^2$∣LS∣2 is minimised at the same $t$t as $|LS|$∣LS∣, we minimise the simpler quadratic"; "equating $\mathbf{i}$i components and $\mathbf{j}$j components separately gives the simultaneous equations …". Phrases that lose marks: "the distance is" without specifying which two objects; missing units (km, m, m/s, m/s$^2$2); leaving answers as $\sqrt{144/17}$144/17​ rather than $12/\sqrt{17}$12/17​ rationalised.