Vector kinematics AI HL

IB Mathematics: Applications and Interpretation HL · Unit 3

Lesson 3.12 · Vector kinematics

Model motion in two and three dimensions, compare moving objects, and describe paths with changing velocity.

Praise along the journey

“The LORD shall preserve thy going out and thy coming in from this time forth, and even for evermore.” Psalm 121:8 (KJV)

Kinematics helps us describe where an object is going and how its motion changes. This psalm invites us to praise the Lord for his care in our own going out and coming in. Whether your path today is familiar or uncertain, you can bring it to God in prayer.

Learning objectives

By the end of this lesson, you should be able to:

model constant-velocity motion in two and three dimensions;
distinguish position, displacement, velocity, speed, and acceleration;
determine whether moving objects meet and find their closest separation;
differentiate a position vector to obtain velocity and acceleration;
find position from a variable velocity and an initial position; and
interpret projectile and circular motion using vector models.

1. Position, velocity, and acceleration

A position vector $\mathbf{r}(t)$ gives an object’s location relative to an origin at time $t$ . Velocity is the rate at which position changes; acceleration is the rate at which velocity changes:

$\boxed{\mathbf{v}(t)=\frac{d\mathbf{r}}{dt}, \qquad \mathbf{a}(t)=\frac{d\mathbf{v}}{dt} =\frac{d^2\mathbf{r}}{dt^2}}$

Differentiate each component separately. Speed is the magnitude of velocity, $|\mathbf{v}(t)|$ . Speed is a scalar; velocity is a vector.

Worked example · Reading a motion model

A particle’s position, in metres, is $\mathbf{r}(t)=\begin{pmatrix}2+3t\\5-t^2\end{pmatrix}$ for $t\geq0$ seconds. Find its velocity and acceleration.

$\mathbf{v}(t)=\begin{pmatrix}3\\-2t\end{pmatrix}\text{ m s}^{-1}, \qquad \mathbf{a}(t)=\begin{pmatrix}0\\-2\end{pmatrix}\text{ m s}^{-2}$

The horizontal velocity is constant, while the vertical velocity changes by $-2$ m s $^{-1}$ each second.

2. Constant velocity in two or three dimensions

If an object starts at $\mathbf{r}_0$ and maintains a constant velocity $\mathbf{v}$ , its position after time $t$ is:

$\boxed{\mathbf{r}(t)=\mathbf{r}_0+t\mathbf{v}}$

This traces a straight line. When $t$ represents elapsed time, usually $t\geq0$ ; the complete mathematical line would allow every real value of $t$ .

Worked example · Three-dimensional flight

A helicopter starts at $(20,-25,0)$ km relative to a station and travels at velocity $\begin{pmatrix}4.2\\5.8\\-0.5\end{pmatrix}$ km h $^{-1}$ . Find its position after $2$ hours and its speed.

$\mathbf{r}(2)= \begin{pmatrix}20\\-25\\0\end{pmatrix} +2\begin{pmatrix}4.2\\5.8\\-0.5\end{pmatrix} = \begin{pmatrix}28.4\\-13.4\\-1.0\end{pmatrix}\text{ km}$

$|\mathbf{v}|= \sqrt{4.2^2+5.8^2+(-0.5)^2} \approx\boxed{7.18\text{ km h}^{-1}}$

A negative third coordinate means the helicopter is below the chosen reference height; it does not necessarily mean below sea level.

Units: If positions are in kilometres and $t$ is in hours, each velocity component is in km h $^{-1}$ . Keep units consistent when comparing distances and times.

3. Relative position and meeting

For moving objects $A$ and $B$ , the position of $B$ relative to $A$ is:

$\boxed{\overrightarrow{AB}(t)=\mathbf{r}_B(t)-\mathbf{r}_A(t)}$

Their separation is $|\overrightarrow{AB}(t)|$ . They meet only when their positions are equal at the same time. An intersection of their paths alone does not prove a meeting.

Worked example · Do the objects meet?

Two robots start at $(1,2)$ and $(10,-1)$ metres. Their velocities are $\begin{pmatrix}3\\1\end{pmatrix}$ and $\begin{pmatrix}-2\\2\end{pmatrix}$ m s $^{-1}$ , respectively.

$\mathbf{r}_A(t)=\begin{pmatrix}1+3t\\2+t\end{pmatrix}, \qquad \mathbf{r}_B(t)=\begin{pmatrix}10-2t\\-1+2t\end{pmatrix}$

Equal $x$ -coordinates require $1+3t=10-2t$ , giving $t=1.8$ . Equal $y$ -coordinates require $2+t=-1+2t$ , giving $t=3$ . Since these times differ, the robots do not meet.

4. Closest approach for constant velocities

Subtract the position vectors to form a separation vector, then minimize its length. It is often easier to minimize the squared distance, because a nonnegative number and its square reach their minimum at the same time:

$D(t)=|\mathbf{r}_B(t)-\mathbf{r}_A(t)|^2$

Expand $D(t)$ into a quadratic and find its minimum. If the question restricts time to $t\geq0$ or to a finite interval, check that the calculated time lies in the allowed domain.

Worked example · Closest separation

At $t=0$ , object $A$ is at $(0,0)$ and travels at $\begin{pmatrix}2\\1\end{pmatrix}$ m s $^{-1}$ . Object $B$ is at $(12,4)$ and travels at $\begin{pmatrix}-1\\1\end{pmatrix}$ m s $^{-1}$ . Find their closest separation for $t\geq0$ .

$\mathbf{r}_B(t)-\mathbf{r}_A(t) =\begin{pmatrix}12-3t\\4\end{pmatrix}$

$D(t)=(12-3t)^2+4^2$

This is smallest when $12-3t=0$ , or $t=4$ s. The minimum separation is $\boxed{4\text{ m}}$ . The objects never meet because their vertical separation remains $4$ m.

5. Variable velocity

When velocity changes with time, differentiate the position vector to find velocity, or integrate velocity to recover position. An initial position determines the integration constants.

Worked example · Recover position from velocity

A particle has velocity $\mathbf{v}(t)=\begin{pmatrix}7\\6-4t\end{pmatrix}$ m s $^{-1}$ and starts at $\mathbf{r}(0)=\begin{pmatrix}2\\-3\end{pmatrix}$ m. Find $\mathbf{r}(t)$ and $\mathbf{a}(t)$ .

$\mathbf{r}(t) =\begin{pmatrix}2\\-3\end{pmatrix} +\int_0^t\begin{pmatrix}7\\6-4u\end{pmatrix}\,du = \boxed{\begin{pmatrix}2+7t\\-3+6t-2t^2\end{pmatrix}\text{ m}}$

$\boxed{\mathbf{a}(t) =\begin{pmatrix}0\\-4\end{pmatrix}\text{ m s}^{-2}}$

The particle’s vertical velocity is zero at $t=1.5$ s. This is when its vertical coordinate reaches a maximum.

6. Projectile motion as a vector model

If horizontal acceleration is negligible and vertical acceleration is the constant $-g$ , a projectile launched from $(x_0,y_0)$ with initial velocity $(u_x,u_y)$ has:

$\mathbf{r}(t)= \begin{pmatrix} x_0+u_xt\\ y_0+u_yt-\frac12gt^2 \end{pmatrix}, \qquad \mathbf{v}(t)= \begin{pmatrix} u_x\\u_y-gt \end{pmatrix}$

Here $g$ is a positive gravitational acceleration, usually expressed in m s $^{-2}$ . To find a landing time, solve the equation for the height of the landing surface and select a physically meaningful value of $t$ .

Worked example · Maximum height

A ball leaves ground level with initial velocity $\begin{pmatrix}12\\18\end{pmatrix}$ m s $^{-1}$ . Take $g=9.8$ m s $^{-2}$ . Find when it reaches its maximum height.

$v_y=18-9.8t=0 \quad\Longrightarrow\quad t=\frac{18}{9.8}\approx\boxed{1.84\text{ s}}$

At that moment the vertical velocity is zero, but the ball is still moving horizontally at $12$ m s $^{-1}$ .

7. Circular and elliptical paths

Some position vectors contain sine and cosine instead of linear or quadratic terms. A model $\mathbf{r}(t)=\begin{pmatrix}R\cos(\omega t)\\R\sin(\omega t)\end{pmatrix}$ traces a circle of radius $R$ about the origin. The parameter $\omega$ determines how quickly the angle changes, in radians per unit time.

$\mathbf{v}(t)= \begin{pmatrix} -R\omega\sin(\omega t)\\ R\omega\cos(\omega t) \end{pmatrix}, \qquad \mathbf{a}(t)=-\omega^2\mathbf{r}(t)$

The acceleration points toward the origin. A phase shift such as $\cos\!\bigl(\omega(t-a)\bigr)$ shifts the motion in time by $a$ .

Worked example · Verify a circular path

A marker’s position is $\mathbf{r}(t)=\begin{pmatrix}1.5\cos(0.1t^2)\\1.5\sin(0.1t^2)\end{pmatrix}$ metres, for $t\geq0$ . Show that it stays on a circle and find its velocity.

$|\mathbf{r}(t)|^2 =1.5^2\bigl(\cos^2(0.1t^2)+\sin^2(0.1t^2)\bigr) =1.5^2$

The radius is $1.5$ m. Using the chain rule:

$\boxed{\mathbf{v}(t)= \begin{pmatrix} -0.3t\sin(0.1t^2)\\ 0.3t\cos(0.1t^2) \end{pmatrix}\text{ m s}^{-1}}$

Its speed is $0.3t$ m s $^{-1}$ , so the marker moves around the circle at a changing speed.

IB-style Paper 1 · Skill practice

A particle has $\mathbf{r}(t)=\begin{pmatrix}4-2t\\1+5t\end{pmatrix}$ metres. Find its velocity and position at $t=3$ s. [3 marks]

Show solution

$\mathbf{v}=\begin{pmatrix}-2\\5\end{pmatrix}$ m s $^{-1}$ and $\mathbf{r}(3)=\boxed{\begin{pmatrix}-2\\16\end{pmatrix}\text{ m}}$ .
A particle has $\mathbf{r}(t)=\begin{pmatrix}2t^2\\3t-t^2\end{pmatrix}$ metres. Find its velocity, acceleration, and speed at $t=2$ s. [5 marks]

Show solution

$\mathbf{v}(t)=\begin{pmatrix}4t\\3-2t\end{pmatrix}, \qquad \mathbf{a}(t)=\begin{pmatrix}4\\-2\end{pmatrix}$

At $t=2$ , velocity is $(8,-1)$ m s $^{-1}$ and speed is $\boxed{\sqrt{65}\approx8.06\text{ m s}^{-1}}$ .
Two objects have positions $\mathbf{r}_A(t)=\begin{pmatrix}1+2t\\4-t\end{pmatrix}$ and $\mathbf{r}_B(t)=\begin{pmatrix}7-t\\1+t\end{pmatrix}$ metres. Determine whether they meet for $t\geq0$ . [4 marks]

Show solution

The $x$ -coordinates agree at $1+2t=7-t$ , giving $t=2$ . The $y$ -coordinates agree at $4-t=1+t$ , giving $t=1.5$ . Since the times differ, they do not meet.
Given $\mathbf{v}(t)=\begin{pmatrix}4t\\-3\end{pmatrix}$ m s $^{-1}$ and $\mathbf{r}(0)=\begin{pmatrix}5\\2\end{pmatrix}$ m, find $\mathbf{r}(t)$ . [4 marks]

Show solution

Integrate each velocity component and apply the initial position:

$\boxed{\mathbf{r}(t)=\begin{pmatrix}5+2t^2\\2-3t\end{pmatrix}\text{ m}}$
For $\mathbf{r}(t)=\begin{pmatrix}3\cos(2t)\\3\sin(2t)\end{pmatrix}$ m, find the path radius and speed. [4 marks]

Show solution

$|\mathbf{r}(t)|=3$ , so the radius is $\boxed{3\text{ m}}$ . Velocity is $\begin{pmatrix}-6\sin(2t)\\6\cos(2t)\end{pmatrix}$ m s $^{-1}$ , whose magnitude is $\boxed{6\text{ m s}^{-1}}$ .
A projectile has $\mathbf{r}(t)=\begin{pmatrix}8t\\2+12t-5t^2\end{pmatrix}$ m. Find its initial position, initial velocity, and vertical acceleration. [4 marks]

Show solution

Initial position is $\boxed{(0,2)\text{ m}}$ . Since $\mathbf{v}(t)=\begin{pmatrix}8\\12-10t\end{pmatrix}$ , initial velocity is $\boxed{(8,12)\text{ m s}^{-1}}$ . Vertical acceleration is $\boxed{-10\text{ m s}^{-2}}$ .

IB-style applications · Linked questions

These connected questions progress from forming a model to interpreting its result.

Application 1 · Helicopter navigation

A helicopter’s position relative to a communications tower is modelled by

$\mathbf{r}(t)= \begin{pmatrix}20\\-25\\0\end{pmatrix} +t\begin{pmatrix}4.2\\5.8\\-0.5\end{pmatrix},\qquad 0\leq t\leq4.$

Time $t$ is measured in hours. The components represent eastward, northward, and vertical displacements in kilometres.

Find the helicopter’s speed, to three significant figures. [3 marks]
Find its distance from the tower at $t=0$ . [3 marks]
Find its horizontal bearing of travel, measured clockwise from north, to the nearest degree. Explain why its downward component does not affect the bearing. [4 marks]

Show worked solution · Application 1

(a)

$|\mathbf{v}|= \sqrt{4.2^2+5.8^2+(-0.5)^2} =\sqrt{51.53} \approx\boxed{7.18\text{ km h}^{-1}}$

(b)

$|\mathbf{r}(0)|=\sqrt{20^2+(-25)^2+0^2} =\sqrt{1025}\approx\boxed{32.0\text{ km}}$

(c) Both horizontal components of velocity are positive, so the direction is northeast. The angle clockwise from north is $\tan^{-1}(4.2/5.8)\approx\boxed{036^\circ}$ . A bearing describes direction in the horizontal plane; the vertical component describes descent separately.

Application 2 · Aircraft separation

Two aircraft travel at constant velocities in a horizontal coordinate system. At $t=0$ , aircraft $A$ is at $(0,0)$ km and aircraft $B$ is at $(30,40)$ km. Their velocities are $\begin{pmatrix}10\\5\end{pmatrix}$ km h $^{-1}$ and $\begin{pmatrix}4\\-3\end{pmatrix}$ km h $^{-1}$ , respectively. Time $t$ is in hours, with $t\geq0$ .

Write the position vector of each aircraft at time $t$ . [3 marks]
Find their closest separation and when it occurs. Give answers to three significant figures. [6 marks]
Safety rules require a separation of at least $10$ km. Determine whether this rule is broken. [2 marks]

Show worked solution · Application 2

(a)

$\mathbf{r}_A(t)=\begin{pmatrix}10t\\5t\end{pmatrix}, \qquad \mathbf{r}_B(t)=\begin{pmatrix}30+4t\\40-3t\end{pmatrix}$

(b) The relative position is $\begin{pmatrix}30-6t\\40-8t\end{pmatrix}$ . Hence:

$D(t)=(30-6t)^2+(40-8t)^2 =100(t-5)^2$

The minimum occurs at $\boxed{t=5.00\text{ h}}$ , with separation $\boxed{0\text{ km}}$ . The aircraft meet at $(50,25)$ km.

(c) Yes. Their separation falls below $10$ km before they meet, so the rule is broken.

Application 3 · Projectile from a platform

A ball is launched from a platform $20$ m above level ground with initial velocity $(15,10)$ m s $^{-1}$ . Take $g=10$ m s $^{-2}$ , neglect air resistance, and measure $t$ in seconds.

Write the position vector $\mathbf{r}(t)$ and velocity vector $\mathbf{v}(t)$ , taking the ground directly below the launch point as the origin. [5 marks]
Find the greatest height of the ball above the ground. [4 marks]
Find the time and horizontal distance when the ball lands. Give numerical answers to three significant figures. [5 marks]

Show worked solution · Application 3

(a)

$\boxed{\mathbf{r}(t)= \begin{pmatrix}15t\\20+10t-5t^2\end{pmatrix}\text{ m}, \qquad \mathbf{v}(t)= \begin{pmatrix}15\\10-10t\end{pmatrix}\text{ m s}^{-1}}$

(b) At greatest height, $10-10t=0$ , so $t=1$ s. Height is $20+10-5=\boxed{25\text{ m}}$ .

(c) At landing, $20+10t-5t^2=0$ , giving $t^2-2t-4=0$ . The positive solution is $t=1+\sqrt5\approx\boxed{3.24\text{ s}}$ . Horizontal distance is $15(1+\sqrt5)\approx\boxed{48.5\text{ m}}$ .

Application 4 · Variable-velocity vehicle

A small vehicle has velocity $\mathbf{v}(t)=\begin{pmatrix}7\\6-4t\end{pmatrix}$ m s $^{-1}$ for $0\leq t\leq4$ , and starts at $(2,-3)$ m.

Find the position vector and acceleration vector. [5 marks]
Find the time when the vehicle’s northward coordinate is greatest, and calculate that coordinate. [4 marks]
Find the vehicle’s speed at this time and its displacement from its starting point. [4 marks]

Show worked solution · Application 4

(a) Integrating and applying the initial position gives:

$\boxed{\mathbf{r}(t)= \begin{pmatrix}2+7t\\-3+6t-2t^2\end{pmatrix}\text{ m}, \qquad \mathbf{a}(t)=\begin{pmatrix}0\\-4\end{pmatrix}\text{ m s}^{-2}}$

(b) The northward velocity $6-4t$ is zero when $t=\boxed{1.5\text{ s}}$ . The corresponding coordinate is $-3+6(1.5)-2(1.5)^2=\boxed{1.5\text{ m}}$ .

(c) At $t=1.5$ , velocity is $\begin{pmatrix}7\\0\end{pmatrix}$ , so speed is $\boxed{7\text{ m s}^{-1}}$ . Displacement from the start is $\mathbf{r}(1.5)-\mathbf{r}(0)=\boxed{\begin{pmatrix}10.5\\4.5\end{pmatrix}\text{ m}}$ .

Application 5 · Spinning sensor

A sensor attached to a rotating arm has position vector

$\mathbf{r}(t)= \begin{pmatrix} 2\cos(3t)\\ 2\sin(3t) \end{pmatrix}\text{ m},\qquad t\geq0,$

where $t$ is measured in seconds and the angle $3t$ is in radians.

Show that the sensor travels on a circle of radius $2$ m. [3 marks]
Find its velocity and speed. [4 marks]
Find its acceleration and explain how its direction relates to its position. [4 marks]

Show worked solution · Application 5

(a) $|\mathbf{r}(t)|^2=4\cos^2(3t)+4\sin^2(3t)=4$ . Thus its distance from the origin is always $\boxed{2\text{ m}}$ .

(b)

$\mathbf{v}(t)= \begin{pmatrix}-6\sin(3t)\\6\cos(3t)\end{pmatrix}\text{ m s}^{-1}, \qquad |\mathbf{v}(t)|=\boxed{6\text{ m s}^{-1}}$

(c)

$\mathbf{a}(t)= \begin{pmatrix}-18\cos(3t)\\-18\sin(3t)\end{pmatrix} =-9\mathbf{r}(t)\text{ m s}^{-2}$

The negative multiple means acceleration always points toward the origin, opposite to the position vector.

Reflection · Through changing circumstances

An object’s direction and speed may change, but its position can still be described at each moment. Psalm 121:8 reminds us to praise the Lord for his care through our going out and coming in. What part of today’s journey would you like to place in his hands?

Lesson 3.12 · Vector kinematics

Praise along the journey

Learning objectives

1. Position, velocity, and acceleration

2. Constant velocity in two or three dimensions

3. Relative position and meeting

4. Closest approach for constant velocities

5. Variable velocity

6. Projectile motion as a vector model

7. Circular and elliptical paths

IB-style Paper 1 · Skill practice

IB-style applications · Linked questions

Application 1 · Helicopter navigation

Application 2 · Aircraft separation

Application 3 · Projectile from a platform

Application 4 · Variable-velocity vehicle

Application 5 · Spinning sensor

Reflection · Through changing circumstances

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Vector kinematics AI HL

Lesson 3.12 · Vector kinematics

Praise along the journey

Learning objectives

1. Position, velocity, and acceleration

2. Constant velocity in two or three dimensions

3. Relative position and meeting

4. Closest approach for constant velocities

5. Variable velocity

6. Projectile motion as a vector model

7. Circular and elliptical paths

IB-style Paper 1 · Skill practice

IB-style applications · Linked questions

Application 1 · Helicopter navigation

Application 2 · Aircraft separation

Application 3 · Projectile from a platform

Application 4 · Variable-velocity vehicle

Application 5 · Spinning sensor

Reflection · Through changing circumstances

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