In Specialist Mathematics, motion in a plane is analyzed using vector functions. This allows for the simultaneous tracking of a particle’s horizontal and vertical components of motion relative to a fixed origin.
A particle’s position at any time $t$ is defined by a vector function $\mathbf{r}(t)$. In two dimensions, this is expressed in terms of the unit vectors $\mathbf{i}$ (horizontal) and $\mathbf{j}$ (vertical).
If the position vector is $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$, then:
EXAM TIP: VCAA often requires the speed at a specific time. Ensure you differentiate the components correctly before substituting the value of $t$ into the magnitude formula.
While the vector function $\mathbf{r}(t)$ describes where and when a particle is at a point, the Cartesian equation describes the shape of the path by eliminating the parameter $t$.
| Vector Components | Trigonometric Identity to Use | Cartesian Shape |
|---|---|---|
| $x = a\cos(t), y = b\sin(t)$ | $\cos^2(t) + \sin^2(t) = 1$ | Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ |
| $x = a\sec(t), y = b\tan(t)$ | $\sec^2(t) - \tan^2(t) = 1$ | Hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ |
| $x = f(t), y = g(t)$ | Rearrange for $t$ and substitute | Parabola, Line, etc. |
COMMON MISTAKE: When sketching the path, pay close attention to the domain of $t$. For example, if $t \in [0, \pi]$, a circular path may only be a semi-circle. Always check if there are asymptotes (common in hyperbola paths involving $\sec(t)$ or $\tan(t)$).
Relative motion describes the motion of one object (A) as observed from another moving object (B).
The position of particle $A$ relative to particle $B$ is given by:
$$\mathbf{r}{A/B} = \mathbf{r}_A - \mathbf{r}_B$$
If $B$ is at the origin, then $\mathbf{r}{A/B} = \mathbf{r}_A$.
The velocity of particle $A$ relative to particle $B$ is the rate of change of the relative position:
$$\mathbf{v}_{A/B} = \mathbf{v}_A - \mathbf{v}_B$$
If particles move with constant velocities $\mathbf{u}$ and $\mathbf{v}$, their positions at time $t$ are:
$$\mathbf{r}A(t) = \mathbf{r}{A_0} + \mathbf{u}t$$
$$\mathbf{r}B(t) = \mathbf{r}{B_0} + \mathbf{v}t$$
where $\mathbf{r}{A_0}$ and $\mathbf{r}{B_0}$ are initial positions.
KEY TAKEAWAY: In relative velocity notation, $\mathbf{v}_{A/B}$ is “Velocity of A minus Velocity of B”. Think of it as the velocity vector you would see if you were “sitting” on particle B.
In 2D motion problems, two particles $A$ and $B$ are often analyzed to see if they collide or how close they pass to one another.
For a collision to occur, the particles must be at the same place at the same time:
$$\mathbf{r}_A(t) = \mathbf{r}_B(t)$$
This requires solving the simultaneous equations:
1. $x_A(t) = x_B(t)$
2. $y_A(t) = y_B(t)$
Note: Both equations must yield the same value for $t$.
The distance $D$ between two particles at time $t$ is the magnitude of the relative position vector:
$$D(t) = |\mathbf{r}_{A/B}(t)| = |\mathbf{r}_A(t) - \mathbf{r}_B(t)|$$
To find the minimum distance:
1. Define the squared distance function $f(t) = [D(t)]^2$. (Minimizing $D^2$ is easier than minimizing $D$).
2. Solve $f’(t) = 0$ for $t$.
3. Alternatively, use the geometric property: at the time of closest approach, the relative position vector is perpendicular to the relative velocity vector.
$$\mathbf{r}{A/B}(t) \cdot \mathbf{v}{A/B}(t) = 0$$
VCAA FOCUS: Problems involving a “police boat” intercepting a “motorboat” are common. You may be asked to find a specific velocity vector (or a scalar multiplier $u$) that ensures an interception. This is solved by setting $\mathbf{r}_A(t) = \mathbf{r}_B(t)$ and solving for both $t$ and the unknown velocity component.
REMEMBER: Speed is a scalar ($|\mathbf{v}|$), while velocity is a vector ($\mathbf{v}$). If an exam question asks for velocity, provide the answer in $\mathbf{i}, \mathbf{j}$ form or as a magnitude and direction (bearing). If it asks for speed, provide a magnitude only.
