Use vector functions of time and differentiate them to find velocity, speed and acceleration along a path

What this dot point is asking

SCSA wants you to model motion with a vector function, differentiate it to obtain velocity and acceleration, find speed, and describe the path.

Position as a vector function of time

A path is described by $\mathbf{r}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j} + z(t)\,\mathbf{k}$, giving the particle's position at each time $t$. Evaluating at particular times gives points on the path; eliminating the parameter $t$ between the component equations gives the cartesian relation of the curve.

Differentiation is componentwise

Each component is differentiated separately using ordinary calculus. The velocity vector is tangent to the path, pointing in the direction of motion; the acceleration vector records how velocity changes.

Speed is a scalar

Using the calculus on motion

To find when a particle is momentarily at rest, set $\mathbf{v}(t) = \mathbf{0}$ (all components zero). To find when velocity is perpendicular to acceleration, set $\mathbf{v}\cdot\mathbf{a} = 0$. Position, displacement and distance follow by evaluating or integrating as needed.

Distance travelled along the path

Displacement and distance are different quantities. The displacement over a time interval is the vector $\mathbf{r}(t_2) - \mathbf{r}(t_1)$, a straight-line change in position. The distance travelled is the length of the actual path, found by integrating the speed: $\text{distance} = \displaystyle\int_{t_1}^{t_2} |\mathbf{v}(t)|\, dt$. For motion around a closed curve the displacement can be zero while the distance is large, so a question asking for distance requires the speed integral, not just the endpoints.

Circular motion and centre-seeking acceleration

A particle moving on a circle of radius $a$ at constant angular rate $\omega$ has position $\mathbf{r}(t) = a\cos\omega t\,\mathbf{i} + a\sin\omega t\,\mathbf{j}$. Differentiating twice gives $\mathbf{a}(t) = -\omega^2\mathbf{r}(t)$, an acceleration of constant magnitude $a\omega^2$ directed straight back toward the centre. This is the vector-calculus statement of centripetal acceleration, and SCSA frequently asks students to derive it by differentiating the position vector and recognising $\mathbf{a} = -\omega^2\mathbf{r}$.

The tangent and normal directions

Because velocity is always tangent to the path, the unit vector $\hat{\mathbf{v}} = \dfrac{\mathbf{v}}{|\mathbf{v}|}$ gives the direction of motion at any instant. The acceleration can be split into a component along $\hat{\mathbf{v}}$, which changes the speed, and a component perpendicular to it, which changes the direction. When the speed is constant the acceleration is entirely perpendicular to the velocity, which is exactly why $\mathbf{v}\cdot\mathbf{a} = 0$ for uniform circular motion.