# Vector Applications to Kinematics

> IB Mathematics AI HL · Geometry and trigonometry
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u3-vector-kinematics/

This sub-topic shows how vectors describe motion. You will model an object moving in a straight line at constant velocity using a position vector that depends on time, calculate its speed, compare two moving objects with relative position vectors, find their closest approach, and describe projectile and circular motion with position vectors. A GDC is used throughout for numerical work.

**Prerequisites:** [Vector equation of a line and its parametric form](https://www.owlsprep.com/study/ib-math-ai-hl-u3-vector-equation-of-a-line/)

## Learning objectives

- Model the motion of an object moving at constant velocity using the position vector $\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}$
- Find the position of a moving object at any time, and calculate its speed as the magnitude of its velocity vector
- Use relative position vectors to compare the motion of two objects
- Find the closest approach (minimum distance) between two moving objects and the time it occurs
- Describe projectile and circular motion using a position vector that varies with time

## Modelling Motion with Position Vectors

**Position Vector of a Moving Object** — An object that starts at position $\mathbf{r}_0$ and moves at constant velocity $\mathbf{v}$ has position vector $\mathbf{r}(t)$ at time $t$. This is the vector equation of a line where the time $t$ plays the role of the parameter and the velocity $\mathbf{v}$ plays the role of the direction vector. The object's **speed** is $|\mathbf{v}|$, the magnitude of the velocity vector.

*Notation:* $\mathbf{r}(t) = \mathbf{r}_0 + t\,\mathbf{v}$

*Example:* A boat starting at $(2, 1)$ km with velocity $\begin{pmatrix}3 \\ 4\end{pmatrix}$ km/h has position $\mathbf{r}(t) = \begin{pmatrix}2 \\ 1\end{pmatrix} + t\begin{pmatrix}3 \\ 4\end{pmatrix}$ and speed $|\mathbf{v}| = \sqrt{3^2 + 4^2} = 5$ 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 $\mathbf{r}(t)$ equal to that point and solve for $t$.

> **tip**
>
> Keep units consistent. If position is in kilometres and time in hours, then velocity is in km/h and speed is in km/h.

**Worked example:** A drone starts at the point $(1, 2, 5)$ (in metres) and flies with constant velocity $\begin{pmatrix}4 \\ -3 \\ 12\end{pmatrix}$ 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. $$|\mathbf{v}| = \sqrt{4^2 + (-3)^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13 \text{ m/s}$$

## Relative Position and Closest Approach

**Relative Position and Closest Approach** — For two objects with position vectors $\mathbf{r}_A(t)$ and $\mathbf{r}_B(t)$, the **relative position** of $B$ from $A$ is $\mathbf{d}(t) = \mathbf{r}_B(t) - \mathbf{r}_A(t)$. The distance between them at time $t$ is $|\mathbf{d}(t)|$. The **closest approach** is the smallest value of this distance, found by minimising $|\mathbf{d}(t)|$ (or, more easily, $|\mathbf{d}(t)|^2$) 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 $|\mathbf{d}(t)|^2$ 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 $\begin{pmatrix}3 \\ 2\end{pmatrix}$ km/h. Ship B starts at $(10, 0)$ km with velocity $\begin{pmatrix}1 \\ 3\end{pmatrix}$ 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.

## Projectile and Circular Motion as Vectors

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.

**Projectile Motion (Vector Form)** — 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.

*Notation:* $\mathbf{r}(t) = \begin{pmatrix} x_0 + u\,t \\ y_0 + w\,t - \tfrac{1}{2} g t^2 \end{pmatrix}$

*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 $\mathbf{r}(t) = \begin{pmatrix}8t \\ 10t - 4.9t^2\end{pmatrix}$.

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

*Notation:* $\mathbf{r}(t) = \begin{pmatrix} a + R\cos(\omega t) \\ b + R\sin(\omega t) \end{pmatrix}$

*Example:* A rider on a Ferris wheel of radius $15$ m centred at $(0, 18)$ turning at $\omega = 0.2$ rad/s has $\mathbf{r}(t) = \begin{pmatrix}15\cos(0.2t) \\ 18 + 15\sin(0.2t)\end{pmatrix}$.

> **tip**
>
> For any position vector written component-by-component, you can still find the object's location at a chosen time by substituting $t$, and the distance between two objects by subtracting their position vectors and taking the magnitude. Use your GDC to evaluate and to graph the distance.

**Worked example:** A drone follows the path $\mathbf{r}(t) = \begin{pmatrix}20\cos(0.5t) \\ 20\sin(0.5t)\end{pmatrix}$ (metres, seconds), circling a tower at the origin. Find its position at $t = \pi$ seconds and its distance from the tower.

1. Substitute $t = \pi$ 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.

## Common pitfalls

- **Wrong:** Confusing the velocity vector with speed
  - Why it fails: 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: Give speed as the scalar $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$, and keep the velocity as a column vector
- **Wrong:** Minimising the distance $|\mathbf{d}(t)|$ with the square root still in place
  - Why it fails: The square root makes the algebra and calculator work harder and error-prone
  - Correct: Minimise the squared distance $|\mathbf{d}(t)|^2$, which has its minimum at the same time, then square root only the final value
- **Wrong:** Adding the two velocity vectors when comparing two moving objects
  - Why it fails: Relative motion is a difference, not a sum; adding them describes something physically meaningless
  - Correct: Subtract the position vectors, $\mathbf{d}(t) = \mathbf{r}_B(t) - \mathbf{r}_A(t)$, to get the relative position
- **Wrong:** Leaving the GDC in degree mode for circular motion problems
  - Why it fails: Angular speed $\omega t$ is in radians, so evaluating $\cos(\omega t)$ in degrees gives wrong positions
  - Correct: Switch the GDC to radian mode whenever an angle is generated by $\omega t$

## Cheatsheet

| Concept | Rule/Formula |
| --- | --- |
| Position at time t | $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$ |
| Speed | $\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ |
| Relative position of B from A | $\mathbf{d}(t) = \mathbf{r}_B(t) - \mathbf{r}_A(t)$ |
| Distance between two objects | $\|\mathbf{d}(t)\|$ |
| Closest approach | Minimise $\|\mathbf{d}(t)\|^2$ over $t$ |
| Circular motion | $\mathbf{r}(t) = \begin{pmatrix} a + R\cos(\omega t) \\ b + R\sin(\omega t) \end{pmatrix}$ |

## 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](https://www.owlsprep.com/study/ib-math-ai-hl-u3-vector-equation-of-a-line/)

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