# Motion in Vector Form

> Describe moving objects with position vectors. Find velocity and acceleration with calculus, and handle boats in currents and planes in crosswinds.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/motion-in-vector-form/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The ferry and the river
A ferry captain points her boat straight across a river and goes full speed. She lands well downstream of the jetty, because the river was moving the whole time. Next trip she aims a little upstream, and the boat glides straight across.

The boat has its own velocity and the river has another. What you see from the bank is the two added together as vectors. Once motion is written as vectors, adding velocities, finding speeds and using calculus all become routine.

**In short:** In vector form, an object's position is $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$. Differentiate each part to get the velocity and acceleration, or integrate to go back, and the speed is the magnitude of the velocity, $|\mathbf{v}|$.

## Starting point: what you already know

- **Vectors in component form:** $\mathbf{a} = x\mathbf{i} + y\mathbf{j}$, with magnitude $|\mathbf{a}| = \sqrt{x^2 + y^2}$ (Introduction to Vectors).
- **Parametric equations:** a path can be given as $x = f(t)$, $y = g(t)$, and you can remove $t$ to get a Cartesian equation (Parametric Equations).
- **Calculus of motion** from Mathematics Advanced: velocity is the derivative of displacement, $v = \dfrac{dx}{dt}$, and acceleration is the derivative of velocity, $a = \dfrac{dv}{dt}$.
- **Speed** is the size of the velocity, so it is never negative.

## Position as a vector

Put the origin somewhere fixed. At time $t$ the object is at the point $\big(x(t), y(t)\big)$, so its **position vector** is

$$\mathbf{r}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j}$$

The two parts $x(t)$ and $y(t)$ are exactly the parametric equations of the path, with time as the parameter.

- **Displacement** from time $t_1$ to $t_2$ is $\mathbf{r}(t_2) - \mathbf{r}(t_1)$.
- **Distance between the two positions** is the magnitude of that displacement.

## Constant velocity

If an object starts at $\mathbf{r}_0$ and moves with a fixed velocity $\mathbf{v}$, then after $t$ seconds

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

### Worked example: A drone flying in a straight line
A drone starts at $3\mathbf{i} + \mathbf{j}$ (metres) and flies with velocity $2\mathbf{i} + 4\mathbf{j}$ m/s.

After 5 s: $\mathbf{r}(5) = (3 + 10)\mathbf{i} + (1 + 20)\mathbf{j} = 13\mathbf{i} + 21\mathbf{j}$.

Its speed is $|2\mathbf{i} + 4\mathbf{j}| = \sqrt{4 + 16} = 2\sqrt{5} \approx 4.47$ m/s.

## Velocity and acceleration by calculus

Differentiate each part on its own:

### Key idea: Calculus with vectors
$$\mathbf{v}(t) = \dot{\mathbf{r}} = \dot{x}\,\mathbf{i} + \dot{y}\,\mathbf{j} \qquad \mathbf{a}(t) = \ddot{\mathbf{r}} = \ddot{x}\,\mathbf{i} + \ddot{y}\,\mathbf{j}$$
Speed $= |\mathbf{v}(t)|$. A dot over a letter means "derivative with respect to time".

### Worked example: A curved path
$\mathbf{r}(t) = 2t\,\mathbf{i} + (t^2 - 1)\,\mathbf{j}$.

- $\mathbf{v}(t) = 2\mathbf{i} + 2t\,\mathbf{j}$ and $\mathbf{a}(t) = 2\mathbf{j}$.
- At $t = 1$: position $2\mathbf{i}$, velocity $2\mathbf{i} + 2\mathbf{j}$, speed $\sqrt{8} = 2\sqrt{2}$.
- Path: $x = 2t$ gives $t = \dfrac{x}{2}$, so $y = \dfrac{x^2}{4} - 1$. A parabola.

Going the other way, **integrate** each part, and use the starting values to find the constants.

### Worked example: From acceleration to position
$\mathbf{a} = -10\mathbf{j}$, $\mathbf{v}(0) = 3\mathbf{i} + 4\mathbf{j}$, $\mathbf{r}(0) = \mathbf{0}$.

Integrate: $\mathbf{v} = 3\mathbf{i} + (4 - 10t)\mathbf{j}$. Integrate again: $\mathbf{r} = 3t\,\mathbf{i} + (4t - 5t^2)\mathbf{j}$.

This is a ball thrown near the ground, which is the next chapter.

### Calculator tip: Speed and direction with Pol
Once you have the velocity $v_x\mathbf{i} + v_y\mathbf{j}$ at a time, `Pol(vx, vy)` gives the speed as $r$ and the direction as $\theta$ (the angle above the positive $x$-axis) in one step.

## Relative velocity: currents and crosswinds

What an observer on the ground sees is the vector sum:

$$\mathbf{v}_{\text{ground}} = \mathbf{v}_{\text{boat in still water}} + \mathbf{v}_{\text{current}}$$

The same rule works for a plane and the wind.

### Worked example: Crossing the river
A river is 120 m wide and flows east at 3 m/s. A boat heads straight north at 4 m/s through the water.

- Ground velocity: $3\mathbf{i} + 4\mathbf{j}$, speed $\sqrt{9 + 16} = 5$ m/s.
- Time to cross: only the north part moves it across, so $t = \dfrac{120}{4} = 30$ s.
- Drift downstream: $3 \times 30 = 90$ m.

### Watch out: Use the right part for the time
The crossing time uses the part of the velocity that points across the river, not the full speed of 5 m/s.

**Mind map**

```text
Motion in vector form
  Position
    r(t) = x(t) i + y(t) j
    Parametric path
    Remove t for Cartesian
  Calculus
    v = dr/dt
    a = dv/dt
    Integrate back with starting values
  Size
    Speed = |v|
    Distance = |displacement|
  Relative velocity
    Ground = own + current
    Plane = own + wind
```

### Year 11: Aiming upstream to go straight across
Same river (3 m/s east), same boat (4 m/s in still water). To land directly opposite, the boat must point upstream at angle $\theta$ from north so the current is cancelled:

$4\sin\theta = 3$, so $\theta = \sin^{-1}\dfrac{3}{4} \approx 48.6^\circ$ west of north.

Speed across $= 4\cos\theta = \sqrt{16 - 9} = \sqrt{7}$ m/s, so the trip takes $\dfrac{120}{\sqrt{7}} \approx 45.4$ s. Going straight costs time.

**Moving straight up or down** means the $\mathbf{i}$ part of the velocity is zero. For $\mathbf{r} = (t^2 - 4t)\mathbf{i} + t\mathbf{j}$, $\dot{x} = 2t - 4 = 0$ at $t = 2$.

**Direction of motion.** The velocity $3\mathbf{i} + 4\mathbf{j}$ points at $\tan^{-1}\frac{4}{3} \approx 53^\circ$ above the positive $x$-axis.

**Least speed.** With $\mathbf{v} = 10\mathbf{i} + (8 - 10t)\mathbf{j}$, the speed is least when the $\mathbf{j}$ part is zero ($t = 0.8$), and that least speed is 10.

**Closest approach.** For two moving objects, write $\mathbf{r}_2 - \mathbf{r}_1$ and minimise the **square** of its length. If $\mathbf{r}_2 - \mathbf{r}_1 = (6 - 2t)\mathbf{i} + (3 - t)\mathbf{j}$, the squared distance is $5t^2 - 30t + 45$, smallest at $t = 3$ where it is 0: they collide. If the smallest value is not 0, they only come close.

### Year 12: Moving in a circle
$\mathbf{r}(t) = 5\cos t\,\mathbf{i} + 5\sin t\,\mathbf{j}$.

- Path: $x^2 + y^2 = 25\cos^2 t + 25\sin^2 t = 25$, a circle of radius 5.
- $\mathbf{v} = -5\sin t\,\mathbf{i} + 5\cos t\,\mathbf{j}$, so the speed is always 5.
- $\mathbf{r} \cdot \mathbf{v} = -25\sin t\cos t + 25\sin t\cos t = 0$: the velocity is always at right angles to the radius, along the tangent.
- $\mathbf{a} = -5\cos t\,\mathbf{i} - 5\sin t\,\mathbf{j} = -\mathbf{r}$: the acceleration points to the centre, even though the speed never changes.

**A circle anywhere.** $\mathbf{r}(t) = (1 + 4\cos t)\mathbf{i} + (-2 + 4\sin t)\mathbf{j}$ gives $(x - 1)^2 + (y + 2)^2 = 16$, centre $(1, -2)$ and radius 4. With $\cos 3t$ and $\sin 3t$ the speed triples but the circle is the same.

## Quick summary

- Position: $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$. Its parts are parametric equations of the path.
- Constant velocity: $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$.
- Differentiate each part to get velocity and acceleration. Integrate each part to go back, using starting values.
- Speed is $|\mathbf{v}|$. Distance between two positions is the size of the displacement.
- Ground velocity = own velocity + current (or wind).

## Quiz

Try these questions from the chapter quiz.

1. An object has $\mathbf{r}(t) = 3t\,\mathbf{i} + 4t\,\mathbf{j}$. What is its velocity?
   - A. $3\mathbf{i} + 4\mathbf{j}$
   - B. $3\mathbf{i}$
   - C. $7t$
   - D. 5
   - Answer: A. $3\mathbf{i} + 4\mathbf{j}$. Differentiate each part: $3$ and $4$.

2. For the same object, what is its speed?
   - A. 7
   - B. 5
   - C. 12
   - D. $\sqrt{7}$
   - Answer: B. 5. $\sqrt{3^2 + 4^2} = 5$.

3. A cyclist starts at $2\mathbf{i}$ and moves with velocity $\mathbf{i} - \mathbf{j}$ m/s. Where is she after 4 s?
   - A. $4\mathbf{i} - 4\mathbf{j}$
   - B. $6\mathbf{i} - 4\mathbf{j}$
   - C. $3\mathbf{i} - \mathbf{j}$
   - D. $2\mathbf{i} - 4\mathbf{j}$
   - Answer: B. $6\mathbf{i} - 4\mathbf{j}$. $\mathbf{r} = 2\mathbf{i} + 4(\mathbf{i} - \mathbf{j})$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/motion-in-vector-form/#quiz