Vector Applications to Kinematics

IB Mathematics AI HL· 6 min read

1. Modelling Motion with Position Vectors★★☆☆☆⏱ 15 min

📘 Definition

Position Vector of a Moving Object

$r (t) = r_{0} + t v$

An object that starts at position $r_{0}$ and moves at constant velocity $v$ has position vector $r (t)$ at time $t$ . This is the vector equation of a line where the time $t$ plays the role of the parameter and the velocity $v$ plays the role of the direction vector. The object's speed is $∣ v ∣$ , the magnitude of the velocity vector.

Example:

A boat starting at $(2, 1)$ km with velocity $(34)$ km/h has position $r (t) = (21) + t (34)$ and speed $∣ v ∣ = 3^{2} + 4^{2} = 5$ km/h.

The velocity vector carries two pieces of information at once: its direction is the direction of travel, and its magnitude is the speed. To find where an object is at a particular time, substitute that value of $t$ . To find when an object reaches a given point, set $r (t)$ equal to that point and solve for $t$ .

📐 Worked Example

A drone starts at the point $(1, 2, 5)$ (in metres) and flies with constant velocity $4 - 3 12$ m/s. Find its position after 3 seconds and its speed.

1
Write the position vector as a function of time:
2
$\mathbf{r}(t) = \begin{pmatrix}1 \\ 2 \\ 5\end{pmatrix} + t\begin{pmatrix}4 \\ -3 \\ 12\end{pmatrix}$
3
Substitute $t = 3$ to find the position after 3 seconds:
4
$\mathbf{r}(3) = \begin{pmatrix}1 + 12 \\ 2 - 9 \\ 5 + 36\end{pmatrix} = \begin{pmatrix}13 \\ -7 \\ 41\end{pmatrix}$
5
The drone is at $(13, - 7, 41)$ metres. The speed is the magnitude of the velocity vector:
6
$|\mathbf{v}| = \sqrt{4^2 + (-3)^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13 \text{ m/s}$

2. Relative Position and Closest Approach★★★☆☆⏱ 20 min

📘 Definition

Relative Position and Closest Approach

For two objects with position vectors $r_{A} (t)$ and $r_{B} (t)$ , the relative position of $B$ from $A$ is $d (t) = r_{B} (t) - r_{A} (t)$ . The distance between them at time $t$ is $∣ d (t) ∣$ . The closest approach is the smallest value of this distance, found by minimising $∣ d (t) ∣$ (or, more easily, $∣ d (t) ∣^{2}$ ) over time.

The distance between two moving objects is itself a function of time. To find the closest approach, form the relative position vector, write the squared distance $∣ d (t) ∣^{2}$ as a function of $t$ , then minimise it. On a GDC you can graph the distance and read off the minimum directly, which is the standard AI HL approach.

📐 Worked Example

Two ships leave port. Ship A starts at $(0, 0)$ km with velocity $(32)$ km/h. Ship B starts at $(10, 0)$ km with velocity $(13)$ km/h. Find the closest distance between the ships and the time at which it occurs.

1
Write the position vector of each ship as a function of time $t$ (hours):
2
$\mathbf{r}_A(t) = \begin{pmatrix}0 \\ 0\end{pmatrix} + t\begin{pmatrix}3 \\ 2\end{pmatrix}, \quad \mathbf{r}_B(t) = \begin{pmatrix}10 \\ 0\end{pmatrix} + t\begin{pmatrix}1 \\ 3\end{pmatrix}$
3
Form the relative position vector of B from A:
4
$\mathbf{d}(t) = \mathbf{r}_B(t) - \mathbf{r}_A(t) = \begin{pmatrix}10 \\ 0\end{pmatrix} + t\begin{pmatrix}1 - 3 \\ 3 - 2\end{pmatrix} = \begin{pmatrix}10 - 2t \\ t\end{pmatrix}$
5
Write the squared distance as a function of $t$ (avoids the square root):
6
$|\mathbf{d}(t)|^2 = (10 - 2t)^2 + t^2 = 100 - 40t + 4t^2 + t^2 = 5t^2 - 40t + 100$
7
Minimise using the vertex of the quadratic (or graph it on the GDC). The minimum is at:
8
$t = \frac{-(-40)}{2 \times 5} = \frac{40}{10} = 4 \text{ hours}$
9
Substitute $t = 4$ to find the minimum squared distance, then take the square root:
10
$|\mathbf{d}(4)|^2 = 5(16) - 40(4) + 100 = 20 \implies |\mathbf{d}(4)| = \sqrt{20} = 2\sqrt{5} \approx 4.47 \text{ km}$
11
The ships are closest after 4 hours, at a distance of about 4.47 km.

3. Projectile and Circular Motion as Vectors★★★☆☆⏱ 15 min

Not every object moves at constant velocity. When the velocity changes, the position is still described by a position vector, but each component is now a function of time. Two important cases appear in applied problems: projectile motion and circular motion.

📘 Definition

Projectile Motion (Vector Form)

$r (t) = (x_{0} + u t y_{0} + w t - \frac{1}{2} g t^{2})$

A projectile launched from $(x_{0}, y_{0})$ with initial horizontal speed $u$ and initial vertical speed $w$ has a position vector whose horizontal component increases at a constant rate while the vertical component is pulled down by gravity $g$ . The horizontal and vertical motions are handled separately inside the one vector.

Example:

A ball kicked from ground level $(0, 0)$ with horizontal speed $8$ m/s and vertical speed $10$ m/s, taking $g = 9.8$ , has $r (t) = (8 t 10 t - 4.9 t^{2})$ .

📘 Definition

Circular Motion (Vector Form)

$r (t) = (a + R cos (ω t) b + R sin (ω t))$

An object moving in a circle of radius $R$ about the centre $(a, b)$ at constant angular speed $ω$ has this position vector. As $t$ increases the point traces the circle; the value $ω$ controls how fast it goes round.

Example:

A rider on a Ferris wheel of radius $15$ m centred at $(0, 18)$ turning at $ω = 0.2$ rad/s has $r (t) = (15 cos (0.2 t) 18 + 15 sin (0.2 t))$ .

📐 Worked Example

A drone follows the path $r (t) = (20 cos (0.5 t) 20 sin (0.5 t))$ (metres, seconds), circling a tower at the origin. Find its position at $t = π$ seconds and its distance from the tower.

1
Substitute $t = π$ into each component (use radians on the GDC):
2
$\mathbf{r}(\pi) = \begin{pmatrix}20\cos(0.5\pi) \\ 20\sin(0.5\pi)\end{pmatrix} = \begin{pmatrix}20\cos(90^\circ) \\ 20\sin(90^\circ)\end{pmatrix} = \begin{pmatrix}0 \\ 20\end{pmatrix}$
3
The drone is at $(0, 20)$ metres. Its distance from the tower at the origin is the magnitude of the position vector:
4
$|\mathbf{r}(\pi)| = \sqrt{0^2 + 20^2} = 20 \text{ m}$
5
This confirms the radius: on a circular path centred at the origin, the drone stays a constant 20 m from the tower.

4. Common Pitfalls

Wrong move:

Confusing the velocity vector with speed

Why:

Velocity is a vector (direction and magnitude); speed is only the magnitude. Reporting a vector where a single number is asked, or vice versa, loses marks

Correct move:

Give speed as the scalar $∣ v ∣ = v_{1}^{2} + v_{2}^{2} + v_{3}^{2}$ , and keep the velocity as a column vector

Wrong move:

Minimising the distance $∣ d (t) ∣$ with the square root still in place

Why:

The square root makes the algebra and calculator work harder and error-prone

Correct move:

Minimise the squared distance $∣ d (t) ∣^{2}$ , which has its minimum at the same time, then square root only the final value

Wrong move:

Adding the two velocity vectors when comparing two moving objects

Why:

Relative motion is a difference, not a sum; adding them describes something physically meaningless

Correct move:

Subtract the position vectors, $d (t) = r_{B} (t) - r_{A} (t)$ , to get the relative position

Wrong move:

Leaving the GDC in degree mode for circular motion problems

Why:

Angular speed $ω t$ is in radians, so evaluating $cos (ω t)$ in degrees gives wrong positions

Correct move:

Switch the GDC to radian mode whenever an angle is generated by $ω t$

5. Quick Reference Cheatsheet

Concept	Rule/Formula
Position at time t	$r (t) = r_{0} + t v$
Speed	$∣ v ∣ = v_{1}^{2} + v_{2}^{2} + v_{3}^{2}$
Relative position of B from A	$d (t) = r_{B} (t) - r_{A} (t)$
Distance between two objects	$∣ d (t) ∣$
Closest approach	Minimise $∣ d (t) ∣^{2}$ over $t$
Circular motion	$r (t) = (a + R cos (ω t) b + R sin (ω t))$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Model two boats with position vectors and find their closest approach
2023 · 2
Aircraft tracking: find position at a given time and minimum separation

What's Next

Vector kinematics turns the vector equation of a line into a tool for describing real motion, and it is a favourite context for IB AI HL Paper 2 questions about ships, aircraft and drones. The closest-approach method — form the relative position vector, minimise the squared distance on your GDC — is a reliable, repeatable procedure worth mastering. From here you can connect these ideas to bearings and trigonometry for navigation problems, and to modelling with functions when the motion is not a straight line. Because AI HL allows a GDC throughout, focus on setting up the vectors correctly and letting the calculator handle the graphing and minimisation.

→Vector equation of a line, intersections