A boat heads with velocity (6i) km h^-1 in a current (2i + 3j) km h^-1. Find its actual velocity and speed. [2 marks]

Two particles have relative position r_B/A = (4 - 2t)i + (3 - 2t)j m. Do they collide? [3 marks]

SQA AH style: relative velocity-style practice: Ship A moves with velocity (3i + 4j) km h^-1 and ship B with velocity (-i + 2j) km h^-1. Find the velocity of B relative to A, and hence the speed of B relative to A.

The velocity of B relative to A is v_B/A = v_B - v_A (1 mark). v_B/A = (-1 - 3)i + (2 - 4)j = -4i - 2j km h^-1 (2 marks). The relative speed is the magnitude: |v_B/A| = sqrt((-4)^2 + (-2)^2) = sqrt(20) = 2sqrt(5) km h^-1 (2 marks). Markers reward the subtraction in the correct order (B minus A) and the magnitude.

SQA AH style: closest approach-style practice: At t = 0 particle P is at (0, 0) moving with velocity (2i + j) m s^-1 and particle Q is at (7, 1) moving with velocity (-i + j) m s^-1. Find the time of closest approach and the least distance between them.

Relative position of Q from P at time t: positions are r_P = (2t, t) and r_Q = (7 - t, 1 + t), so r_Q/P = (7 - 3t)i + (1)j (2 marks). Distance squared D^2 = (7 - 3t)^2 + 1. Closest approach minimises D^2: ddt(D^2) = 2(7 - 3t)(-3) = 0, so 7 - 3t = 0, giving t = 73 s (2 marks). Least distance D = sqrt(0 + 1) = 1 m (2 marks). Markers reward forming the relative position vector, differentiating the squared distance, and evaluating the minimum.

ScotlandMathematics of MechanicsSyllabus dot point

How do you describe motion in two or three dimensions using vector functions, and how do you find relative velocity, closest approach and whether two bodies collide?

Use position, velocity and acceleration vectors as functions of time; calculate relative velocity; and find the closest approach of two moving bodies and the condition for collision.

A focused answer to the SQA Advanced Higher Mathematics of Mechanics content on vector kinematics: differentiating and integrating position vectors, computing relative velocity and relative position, and finding the time and distance of closest approach or the condition for a collision.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Write each body's position as a vector function $\mathbf{r}(t)$ ; then velocity is $\mathbf{v} = \dot{\mathbf{r}}$ and acceleration is $\mathbf{a} = \ddot{\mathbf{r}}$ , differentiating component by component. The velocity of B relative to A is $\mathbf{v}_{B/A} = \mathbf{v}_B - \mathbf{v}_A$ and the position of B relative to A is $\mathbf{r}_{B/A} = \mathbf{r}_B - \mathbf{r}_A$ . For closest approach, write the relative position $\mathbf{r}_{B/A}(t)$ , form the squared distance $D^2 = \mathbf{r}_{B/A}\cdot\mathbf{r}_{B/A}$ , and minimise it by setting $\dfrac{d}{dt}(D^2) = 0$ . A collision occurs if $\mathbf{r}_{B/A} = \mathbf{0}$ at some time, which forces both components to vanish simultaneously.

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What this dot point is asking
Vector functions of time
Relative velocity
Closest approach and collision
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What this dot point is asking

Real motion is rarely confined to a line. The SQA extends kinematics to two and three dimensions using vectors: position, velocity and acceleration each become vector functions of time. The headline applications are relative velocity (how one body moves as seen from another) and closest approach (the smallest gap between two moving bodies, and whether that gap is zero, meaning a collision).

Vector functions of time

In two or three dimensions, position is a vector $\mathbf{r}(t)$ written in $\mathbf{i}, \mathbf{j}, \mathbf{k}$ components. Differentiation and integration act component by component, exactly as in one dimension.

Velocity and speed from a vector position function

A particle has position $\mathbf{r} = (t^2)\mathbf{i} + (3t - t^2)\mathbf{j}$ m. Find its velocity and speed at $t = 2$ s.

Step 1: Differentiate each component to get velocity

Velocity is the rate of change of position. Differentiating component by component treats $\mathbf{i}$ and $\mathbf{j}$ as fixed unit vectors:

\mathbf{v} = \dot{\mathbf{r}} = (2t)\mathbf{i} + (3 - 2t)\mathbf{j}.

Step 2: Evaluate at $t = 2$

Substitute $t = 2$ into the velocity expression:

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

Step 3: Find the speed

Speed is the magnitude of the velocity vector, found by Pythagoras on the components:

|\mathbf{v}| = \sqrt{4^2 + (-1)^2} = \sqrt{17}\ \text{m s}^{-1}.

Final answer: at $t = 2\ \text{s}$ the velocity is $4\mathbf{i} - \mathbf{j}$ m s $^{-1}$ and the speed is $\sqrt{17}\ \text{m s}^{-1}$ .

Relative velocity

The velocity of B relative to A is what an observer moving with A would measure for B. It is the difference of the two velocities, and the order matters.

This single idea handles "wind relative to a moving cyclist", "current relative to a ship", and the apparent course of one vessel from another. Once you have the relative velocity, its magnitude is the relative speed and its direction (often as a bearing) is the apparent direction of motion.

Closest approach and collision

To find how close two bodies get, track their relative position, not each absolute position. If their separation is least at some instant, that is the closest approach.

Closest approach from the relative position vector

Find the closest approach for the relative position $\mathbf{r}_{B/A} = (5 - t)\mathbf{i} + (2t - 1)\mathbf{j}$ .

Step 1: Form the squared distance

Working with $D^2$ rather than $D$ avoids differentiating a square root, which is messier and more error-prone. Since $D \ge 0$ , the minimum of $D$ and the minimum of $D^2$ occur at the same time:

D^2 = (5 - t)^2 + (2t - 1)^2.

Step 2: Differentiate and set to zero

Differentiate $D^2$ with respect to $t$ , apply the chain rule to each squared term, and set the result to zero to find the time of closest approach:

\frac{d}{dt}(D^2) = 2(5 - t)(-1) + 2(2t - 1)(2) = -10 + 2t + 8t - 4 = 10t - 14 = 0,

giving $t = 1.4\ \text{s}$ .

Step 3: Evaluate the least distance

Substitute $t = 1.4$ back into the relative position and take its magnitude:

\mathbf{r}_{B/A} = (3.6)\mathbf{i} + (1.8)\mathbf{j}, \quad D = \sqrt{3.6^2 + 1.8^2} = \sqrt{16.2} \approx 4.02\ \text{m.}

Final answer: the closest approach is approximately $4.02\ \text{m}$ , occurring at $t = 1.4\ \text{s}$ .

A collision is the special case $D = 0$ . It happens only if the relative position vector becomes exactly $\mathbf{0}$ , which requires every component to be zero at the same time $t$ . Solve one component for $t$ and check whether the other components also vanish there: if they do, the bodies collide; if not, they pass at the closest-approach distance.

Try this

Q1. A boat heads with velocity $(6\mathbf{i})$ km h $^{-1}$ in a current $(2\mathbf{i} + 3\mathbf{j})$ km h $^{-1}$ . Find its actual velocity and speed. [2 marks]

Cue. Add the vectors: $(8\mathbf{i} + 3\mathbf{j})$ km h $^{-1}$ ; speed $\sqrt{64 + 9} = \sqrt{73}$ km h $^{-1}$ .

Q2. Two particles have relative position $\mathbf{r}_{B/A} = (4 - 2t)\mathbf{i} + (3 - 2t)\mathbf{j}$ m. Do they collide? [3 marks]

Cue. $\mathbf{i}$ -component zero at $t = 2$ ; at $t = 2$ the $\mathbf{j}$ -component is $3 - 4 = -1 \ne 0$ , so no collision.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: relative velocity5 marksShip A moves with velocity

(3\mathbf{i} + 4\mathbf{j})

km h

^{-1}

and ship B with velocity

(-\mathbf{i} + 2\mathbf{j})

km h

^{-1}

. Find the velocity of B relative to A, and hence the speed of B relative to A.

Show worked answer →

The velocity of B relative to A is $\mathbf{v}_{B/A} = \mathbf{v}_B - \mathbf{v}_A$ (1 mark).

$\mathbf{v}_{B/A} = (-1 - 3)\mathbf{i} + (2 - 4)\mathbf{j} = -4\mathbf{i} - 2\mathbf{j}$ km h $^{-1}$ (2 marks).

The relative speed is the magnitude: $|\mathbf{v}_{B/A}| = \sqrt{(-4)^2 + (-2)^2} = \sqrt{20} = 2\sqrt{5}$ km h $^{-1}$ (2 marks). Markers reward the subtraction in the correct order (B minus A) and the magnitude.

AH style: closest approach6 marksAt

t = 0

particle P is at

(0, 0)

moving with velocity

(2\mathbf{i} + \mathbf{j})

m s

^{-1}

and particle Q is at

(7, 1)

moving with velocity

(-\mathbf{i} + \mathbf{j})

m s

^{-1}

. Find the time of closest approach and the least distance between them.

Show worked answer →

Relative position of Q from P at time $t$ : positions are $\mathbf{r}_P = (2t, t)$ and $\mathbf{r}_Q = (7 - t, 1 + t)$ , so $\mathbf{r}_{Q/P} = (7 - 3t)\mathbf{i} + (1)\mathbf{j}$ (2 marks).

Distance squared $D^2 = (7 - 3t)^2 + 1$ . Closest approach minimises $D^2$ : $\dfrac{d}{dt}(D^2) = 2(7 - 3t)(-3) = 0$ , so $7 - 3t = 0$ , giving $t = \tfrac{7}{3}$ s (2 marks).

Least distance $D = \sqrt{0 + 1} = 1$ m (2 marks). Markers reward forming the relative position vector, differentiating the squared distance, and evaluating the minimum.

Related dot points

Sources & how we know this

Advanced Higher Mathematics of Mechanics Course Specification (C802 77) — SQA (2019)