Game plan 4: Motion in a plane | Specialist external game plans

Part 1

Naming the path, then converting it

Each value of $t$ gives one position vector. The curve the tips of those vectors sweep out is the path the question asks you to name.

The particle traces the path, and the position vector is the arrow from the origin at each instant. Eliminating $t$ turns the trace into a Cartesian equation.

Game plan 4.1

Naming and converting a path

1Name the curve by sight, before any algebra. Sine and cosine with the same coefficient give a circle, and with different coefficients an ellipse. Secant and tangent, or cotangent and cosecant, give a hyperbola. The angle must be the same in both components.
2Write the components as $x=\dots$ and $y=\dots$, both in terms of $t$.
3For an algebraic path, solve one component for $t$ and substitute into the other.
4For a trigonometric path, isolate the trig functions. Subtract any shift and divide out the coefficient, so each component reads like $\cos t=\frac{x-h}{a}$, then substitute into the matching identity.
5Name the curve, and state the centre, radius or axes if the question asks. Note any restriction on the path from the range of $t$.

You see	What it means	Path	Identity to use
$a\cos(t)\,\boldsymbol i+a\sin(t)\,\boldsymbol j$	same coefficients	Circle	$\sin^2A+\cos^2A=1$
$a\cos(t)\,\boldsymbol i+b\sin(t)\,\boldsymbol j$	different coefficients	Ellipse	$\sin^2A+\cos^2A=1$
$a\sec(t)\,\boldsymbol i+b\tan(t)\,\boldsymbol j$	not sine and cosine	Hyperbola	$\tan^2A+1=\sec^2A$
$a\cot(t)\,\boldsymbol i+b\operatorname{cosec}(t)\,\boldsymbol j$	the other trig pair	Hyperbola	$\cot^2A+1=\operatorname{cosec}^2A$

Name the shape from the parametric pair first. The identity follows from it. Neither the order of the components nor which one carries the sine matters, only the coefficients and the trig pair.

Circle	Ellipse	Hyperbola
$(x-h)^2+(y-k)^2=r^2$	$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$	$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$

Watch for

The angles must match. $3\cos(2t)\,\boldsymbol i+3\sin(t)\,\boldsymbol j$ is not a circle. With different angles no Pythagorean identity applies, and the path is none of the three named curves.

Tip

The QCAA asks this as a multiple-choice question most years, and naming the curve is the whole answer.

In the Topic Guides

Part 2

Two particles: collide or cross?

Game plan 4.2

Collide or cross?

1Collide means the same position at the same time. Set $\boldsymbol r_1(t)=\boldsymbol r_2(t)$, solve one component for $t$, and check that the same $t$ works in the other component.
2Paths cross means the same position, possibly at different times. Set $\boldsymbol r_1(t_1)=\boldsymbol r_2(t_2)$ and solve the system.
3Answer in context: the time, the point, or “the paths cross but the particles do not collide”.

Watch for

“Collide” means the same place at the same time. “Paths cross” means the same place, possibly at different times. Set them up differently.

In the Topic Guides

Unit 3 Topic 4.2 · Collisions and crossing paths

Part 3

Projectile motion

Horizontal velocity never changes, and gravity only acts on the vertical velocity. That is why you can treat the two components as separate one-dimensional problems.

Game plan 4.3

Solving a projectile question

1Find the three starting facts: the initial position $(x_0,y_0)$, the initial velocity (launch speed $|\boldsymbol v|$ and launch angle $\theta$), and the constant acceleration due to gravity. No air resistance is assumed.
2Build $\boldsymbol a(t)$, integrate for $\boldsymbol v(t)$ using $\boldsymbol v(0)=|\boldsymbol v|\cos\theta\,\boldsymbol i+|\boldsymbol v|\sin\theta\,\boldsymbol j$, then integrate again for $\boldsymbol r(t)$ using the launch point. Or quote the three equations below.
3Flight time: set $r_j(t)=0$ and solve for $t$. Range: substitute that $t$ into $r_i(t)$.
4Maximum height: set $v_j(t)=0$, solve for $t$, and substitute into $r_j(t)$.
5Landing angle: if the projectile is launched from ground level, it equals the launch angle. Otherwise, substitute the flight time into $\theta=\tan^{-1}\left(\frac{v_j(t)}{v_i(t)}\right)$.
6Path: eliminate $t$ between the components for the Cartesian equation. Speed: $|\boldsymbol v(t)|$.

The three equations of projectile motion · Memorise

$$\boldsymbol a(t)=0\,\boldsymbol i-g\,\boldsymbol j$$

$$\boldsymbol v(t)=|\boldsymbol v|\cos\theta\,\boldsymbol i+\left(-gt+|\boldsymbol v|\sin\theta\right)\boldsymbol j$$

$$\begin{aligned}\boldsymbol r(t)=&\left(|\boldsymbol v|\cos\theta\,t+x_0\right)\boldsymbol i\\&+\left(-\tfrac12gt^2+|\boldsymbol v|\sin\theta\,t+y_0\right)\boldsymbol j\end{aligned}$$

$(x_0,y_0)$ is often $(0,0)$, but not always. Quote these straight into a question, or derive them by integrating twice, fixing the constants from $\boldsymbol v(0)$ and the launch point.

$$y=x\tan\theta-\frac{gx^2\sec^2\theta}{2|\boldsymbol v|^2}\qquad\text{(launched from the origin)}$$

Try it

Launch a projectile

Launch speed Launch angle Launch height Time

Move the time slider. The black horizontal arrow never changes, because gravity only acts on the vertical component. This uses $g=9.8$, as on Paper 2.

Watch for

Not every projectile is thrown upwards. A bullet fired horizontally has $\theta=0^\circ$, an object rolling off a roof has a negative launch angle, and a projectile launched back over its start point has $\theta\gt90^\circ$. The equations stay the same. Only the sign and size of $\theta$ change.

In the Topic Guides

Part 4

Circular motion

Game plan 4.4

Circular motion

1Find $r$ and $\omega$. The radius is $r$, and the angular velocity is $\omega=\frac{\text{speed}}{\text{radius}}$. Widening the circle at the same speed lowers $\omega$.
2Write the position vector $\boldsymbol r(t)=r\cos(\omega t)\,\boldsymbol i+r\sin(\omega t)\,\boldsymbol j$.
3Differentiate for velocity: $\dot{\boldsymbol r}(t)=-r\omega\sin(\omega t)\,\boldsymbol i+r\omega\cos(\omega t)\,\boldsymbol j$.
4Differentiate again for acceleration: $\ddot{\boldsymbol r}(t)=-\omega^2\boldsymbol r(t)$. It points back at the centre.
5Then read off what the question asks for: $|\boldsymbol v|=r\omega$, $|\boldsymbol a|=r\omega^2$ or $\text{period}=\frac{2\pi}{\omega}$. For tangency, show $\dot{\boldsymbol r}(t)\cdot\boldsymbol r(t)=0$.

Differentiating the position twice gives $\boldsymbol a=-\omega^2\boldsymbol r$. The acceleration points the opposite way to the position vector, which is what “points at the centre” means. To prove the velocity is a tangent, show $\boldsymbol v\cdot\boldsymbol r=0$.

In the Topic Guides

Unit 3 Topic 4.3 · Circular motion

Formulas

Formula panel

Each formula is labelled with how to treat it in the exam.

Circle, ellipse and hyperbola equations, under Graph equations for Specialist Mathematics

On the sheet

$\sin^2t+\cos^2t=1$ and $1+\tan^2t=\sec^2t$, under Pythagorean identities

On the sheet

$g=10$ on Paper 1, $g=9.8$ on Paper 2

Memorise

The three equations of projectile motion, from $\boldsymbol a(t)=0\,\boldsymbol i-g\,\boldsymbol j$. Quote them, or rebuild them by integrating twice.

Memorise

$\omega=\frac{\text{speed}}{\text{radius}}$, $\text{period}=\frac{2\pi}{\omega}$, $|\boldsymbol v|=r\omega$, $|\boldsymbol a|=r\omega^2$. Differentiate $\boldsymbol r(t)$ twice.

Derive

QCAA language

What the question says, and what it means

Question type	What you’ll see	What it means
Path	“Determine the Cartesian equation of the path”	Game plan 4.1. Trig components lead to an identity
Meeting	“Collide”, “paths cross”, “meet”	Game plan 4.2. Read which one, then set it up
Projectile	A launch speed and angle, “greatest height”, “range”	Game plan 4.3
Circular	$r\cos(\omega t)$ and $r\sin(\omega t)$ components	Game plan 4.4