Vectors in Kinematics

In Specialist Mathematics, vectors are used to describe the motion of particles in two and three dimensions. Unlike linear kinematics, vector kinematics accounts for both the magnitude and the direction of motion simultaneously.

1. Position and Displacement

The position vector of a particle at any time \(t\) (where \(t \ge 0\)) is defined relative to a fixed origin \(O\). It is expressed as a vector function of time.

Position Vector

In three dimensions, the position vector \(\mathbf{r}(t)\) is given by:
\$\(\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k}\)\$
Where:
* \(x(t), y(t),\) and \(z(t)\) are the component functions (parametric equations) for the particle’s coordinates at time \(t\).
* \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are unit vectors in the directions of the \(x, y,\) and \(z\) axes respectively.

Displacement

Displacement is the change in the position vector over a specific time interval \([t_1, t_2]\).
\$\(\Delta \mathbf{r} = \mathbf{r}(t_2) - \mathbf{r}(t_1)\)\$
* The magnitude of displacement is the straight-line distance between the initial and final positions: \(|\mathbf{r}(t_2) - \mathbf{r}(t_1)|\).

KEY TAKEAWAY: Position is a vector relative to the origin, while displacement is a vector representing the change from one position to another. Displacement does not depend on the path taken, only the start and end points.

2. Velocity and Speed

Velocity is the rate of change of position with respect to time.

Velocity Vector

The velocity vector \(\mathbf{v}(t)\) (also denoted as \(\dot{\mathbf{r}}(t)\)) is found by differentiating the position vector:
\$\(\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \frac{dx}{dt}\mathbf{i} + \frac{dy}{dt}\mathbf{j} + \frac{dz}{dt}\mathbf{k}\)\$
* Direction: The velocity vector is always tangent to the path of motion at any point \(t\).
* Unit Tangent Vector: A unit vector in the direction of motion is given by \(\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}\).

Speed

Speed is a scalar quantity representing the magnitude of the velocity vector.
\$\(v = |\mathbf{v}(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}\)\$

EXAM TIP: If a question asks for the “direction of motion,” provide the velocity vector or the angle it makes with a positive axis. If it asks for “speed,” you must calculate the magnitude of the velocity vector.

3. Acceleration

Acceleration is the rate of change of velocity with respect to time.

Acceleration Vector

The acceleration vector \(\mathbf{a}(t)\) (also denoted as \(\ddot{\mathbf{r}}(t)\) or \(\dot{\mathbf{v}}(t)\)) is the derivative of the velocity vector:
\$\(\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d^2\mathbf{r}}{dt^2} = \frac{d^2x}{dt^2}\mathbf{i} + \frac{d^2y}{dt^2}\mathbf{j} + \frac{d^2z}{dt^2}\mathbf{k}\)\$

Summary of Vector Calculus Relationships

COMMON MISTAKE: When integrating acceleration to find velocity, or velocity to find position, do not forget the constant vector of integration \(\mathbf{c} = c_1\mathbf{i} + c_2\mathbf{j} + c_3\mathbf{k}\). You must use initial conditions (e.g., \(\mathbf{v}(0)\) or \(\mathbf{r}(0)\)) to solve for \(\mathbf{c}\).

4. Distance Travelled

While displacement is the change in position, distance travelled is the total length of the path taken by the particle.

For a particle moving from \(t=a\) to \(t=b\), the distance \(s\) is the integral of speed:
\$\(s = \int_{a}^{b} |\mathbf{v}(t)| \, dt = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt\)\$

VCAA FOCUS: Distance calculations often lead to integrals that require tech-active solutions (CAS) or specific trigonometric substitutions in tech-free exams. Always check if the question asks for “distance” (scalar) or “displacement” (vector).

5. Path of Motion (Cartesian Equation)

The path of a particle is often described by a vector function, but VCAA frequently requires converting this to a Cartesian equation (an equation involving only \(x, y\) and sometimes \(z\), with the parameter \(t\) eliminated).

Steps to find the Cartesian Equation:

1. Identify the parametric equations: \(x = f(t)\) and \(y = g(t)\).
2. Rearrange one equation to isolate \(t\) (if possible) and substitute into the other.
3. Use trigonometric identities if \(x\) and \(y\) involve circular functions:
   • \(\sin^2(t) + \cos^2(t) = 1\) (for circles and ellipses)
   • \(\sec^2(t) - \tan^2(t) = 1\) (for hyperbolas)
4. Define the Domain: The domain of the Cartesian equation is determined by the range of the parametric function \(x(t)\) for the given interval of \(t\).

STUDY HINT: Practice sketching paths from vector functions. Remember that the direction of motion is indicated by arrows on the Cartesian path as \(t\) increases.

6. Constant Acceleration Formulas (Vector Form)

If the acceleration vector \(\mathbf{a}\) is constant, the following vector equations (analogous to linear SUVAT equations) can be used:

1. \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\)
2. \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\)
3. \(\mathbf{r} = \mathbf{r}_0 + \frac{1}{2}(\mathbf{u} + \mathbf{v})t\)

Where \(\mathbf{u}\) is initial velocity, \(\mathbf{v}\) is final velocity, and \(\mathbf{r}_0\) is initial position.

REMEMBER: These formulas only apply when \(\mathbf{a}\) is a constant vector (i.e., its components do not contain \(t\)). If \(\mathbf{a}\) depends on \(t\), you must use calculus (integration).