Describe motion in the plane using a position vector r(t), differentiate to get velocity v(t)=r'(t) and acceleration a(t)=r''(t), and use the dot product to analyse the motion (speed, perpendicular velocities, angle between v and a)

What this dot point is asking

NESA wants you to describe a particle moving in the plane by a single position vector $\mathbf{r}(t)$ whose components are functions of time, and then to do calculus on it: differentiate once for the velocity vector, again for the acceleration vector, and read off the speed as a magnitude. On top of that, you must use the dot product as an analytical tool, to find the angle between velocity and acceleration, to decide whether the particle is speeding up or slowing down, to test when two particles' velocities are perpendicular, and to find when the distance from a fixed point is increasing.

This is the general version of the idea behind two earlier topics. Parametric vector equations of lines, $\mathbf{r} = \mathbf{a} + t\mathbf{d}$, describe constant-velocity motion in a straight line, the special case where the components are linear in $t$. Projectile motion is the special case $\ddot x = 0$, $\ddot y = -g$. Here the components $x(t)$ and $y(t)$ can be any functions of time, so the path can curve, the speed can change, and you find velocity and acceleration by differentiating rather than reading them off.

The answer

The position vector and its derivatives

A particle moving in the plane is located at time $t$ by its position vector

As $t$ varies, the head of $\mathbf{r}(t)$ traces out the path (or trajectory) of the particle. To differentiate a vector function, you simply differentiate each component, because $\mathbf{i}$ and $\mathbf{j}$ are fixed:

The velocity $\mathbf{v}(t)$ is a vector: its direction is the direction of motion, tangent to the path, and its magnitude is the speed

Speed is a scalar and is never negative. The acceleration $\mathbf{a}(t)$ is the rate of change of the velocity vector; it points to the side the path is turning towards, and it need not be along the velocity.

Picturing position, velocity and acceleration together

At any instant the three vectors live in the one diagram. The position vector $\mathbf{r}$ reaches from the origin to where the particle is; the velocity $\mathbf{v}$ is tangent to the path, pointing the way the particle is heading; and the acceleration $\mathbf{a}$ points to the concave side of the curve. When $\mathbf{a}$ leans forwards (an acute angle to $\mathbf{v}$) the particle is speeding up; when it leans backwards (an obtuse angle) it is slowing down.

The path (trajectory)

The path is the curve in the plane that the particle travels along, with time stripped away. To find its Cartesian equation, eliminate $t$ between $x(t)$ and $y(t)$, exactly as for any parametric curve. For instance, if $x = 2t$ and $y = t^2$, then $t = \tfrac{x}{2}$ and $y = \tfrac{x^2}{4}$, a parabola. The path tells you the shape of the journey; the velocity and acceleration tell you the timing and dynamics along it. Two particles can share a path yet move along it at completely different rates.

The dot product as the analytical tool

The dot product turns geometric questions about these vectors into arithmetic. Recall the two formulas for vectors $\mathbf{p} = (p_1, p_2)$ and $\mathbf{q} = (q_1, q_2)$:

where $\theta$ is the angle between them. Two consequences drive almost every analysis question.

Angle between two vectors. Rearranging the geometric formula,

Apply this with $\mathbf{p} = \mathbf{v}$ and $\mathbf{q} = \mathbf{a}$ to find the angle between velocity and acceleration.

Perpendicular means zero dot product. Two non-zero vectors are perpendicular exactly when $\mathbf{p} \cdot \mathbf{q} = 0$ (because $\cos 90^\circ = 0$). This single fact answers "when are the velocities perpendicular?" and "when is $\angle AOB$ a right angle?", and it underlies the closest-approach result below.

Speeding up or slowing down: the sign of $\mathbf{v}\cdot\mathbf{a}$

Whether a particle is gaining or losing speed is not about the size of the acceleration, it is about its direction relative to the velocity. The rate of change of speed is governed by the dot product $\mathbf{v}\cdot\mathbf{a}$:

• $\mathbf{v}\cdot\mathbf{a} > 0$ (angle acute): acceleration has a forward component, so the speed is increasing.
• $\mathbf{v}\cdot\mathbf{a} < 0$ (angle obtuse): acceleration has a backward component, so the speed is decreasing.
• $\mathbf{v}\cdot\mathbf{a} = 0$ (perpendicular): the speed is momentarily stationary (a turning point of the speed, or purely "centripetal" turning with no change of speed).

The cleanest way to see this is to differentiate the square of the speed. Since $|\mathbf{v}|^2 = \mathbf{v}\cdot\mathbf{v}$, the product rule for the dot product gives

The speed $|\mathbf{v}|$ rises and falls together with $|\mathbf{v}|^2$, so the sign of $\mathbf{v}\cdot\mathbf{a}$ is the sign of the rate of change of speed. (This is also why the speed of a projectile is least at the top of its flight: there the velocity is horizontal and the acceleration is straight down, so $\mathbf{v}\cdot\mathbf{a} = 0$.)

When is the distance from a point increasing?

Let $D(t) = |\mathbf{r}(t)|$ be the distance of the particle from the origin (or, for a fixed point $C$, take $\mathbf{r}(t) - \mathbf{c}$). Differentiating $D$ directly drags in a square root, so instead differentiate $D^2 = |\mathbf{r}|^2 = \mathbf{r}\cdot\mathbf{r}$:

Because $D \ge 0$, the distance $D$ increases, decreases or is stationary exactly as $D^2$ does, so the sign of $\mathbf{r}\cdot\mathbf{v}$ decides it:

• $\mathbf{r}\cdot\mathbf{v} > 0$: the particle is moving away from $O$ (distance increasing).
• $\mathbf{r}\cdot\mathbf{v} < 0$: the particle is moving towards $O$ (distance decreasing).
• $\mathbf{r}\cdot\mathbf{v} = 0$: the distance is stationary, the instant of closest (or farthest) approach, where $\mathbf{r}$ is perpendicular to the velocity.

The last point has a clean geometric meaning: the closest approach to $O$ happens when the position vector is at right angles to the direction of motion, which for straight-line motion is the foot of the perpendicular from $O$ to the line.

Distinguishing from constant-velocity motion

It is worth being deliberate about how this differs from the parametric-line work. If $\mathbf{r}(t) = \mathbf{a} + t\mathbf{d}$, then $\mathbf{v}(t) = \mathbf{d}$ is constant and $\mathbf{a}(t) = \mathbf{0}$: the particle moves in a straight line at constant speed $|\mathbf{d}|$, and "speeding up or slowing down" never arises. The moment a component is quadratic or higher (or trigonometric, or exponential) in $t$, the velocity changes, the acceleration is non-zero, the path curves, and the dot-product analysis above comes into play. A quick diagnostic: differentiate; if $\mathbf{a} = \mathbf{0}$ you are back in the constant-velocity world, and if not you are in genuine vector-calculus motion.

Time-parameterised points and $\overrightarrow{OA}\cdot\overrightarrow{OB} = 0$

Some questions give two moving points $A$ and $B$ with position vectors $\overrightarrow{OA}(t)$ and $\overrightarrow{OB}(t)$ and ask when $\angle AOB = 90^\circ$, or when one displacement is perpendicular to another. The recipe is always the same: write the dot product as a function of $t$, set it to zero, and solve. Because the components are functions of $t$, the dot product is typically a quadratic in $t$, so there can be two instants (or none), not just one, a point students routinely miss by stopping at the first root.

How exam questions ask about motion as a vector function

• "Find the velocity / acceleration / speed at time $t$." Differentiate $\mathbf{r}$ once for $\mathbf{v}$, again for $\mathbf{a}$; take $|\mathbf{v}|$ for the speed. Keep speeds as exact surds unless a decimal is asked for.
• "Find the angle between the velocity and acceleration." Use $\cos\theta = \dfrac{\mathbf{v}\cdot\mathbf{a}}{|\mathbf{v}|\,|\mathbf{a}|}$.
• "Is the particle speeding up or slowing down (at this instant)?" Compute $\mathbf{v}\cdot\mathbf{a}$ and read its sign; positive means speeding up.
• "Find when the speed is least / greatest." Solve $\mathbf{v}\cdot\mathbf{a} = 0$ (or minimise $|\mathbf{v}|^2$ directly), then check the sign change.
• "Find when the particle is closest to $O$ / to the point $C$." Solve $\mathbf{r}\cdot\mathbf{v} = 0$ (using $\mathbf{r} - \mathbf{c}$ for a general point), confirm a minimum, then evaluate the distance.
• "Show that the distance from $O$ is increasing." Show $\mathbf{r}\cdot\mathbf{v} > 0$ (via $\frac{d}{dt}|\mathbf{r}|^2 = 2\,\mathbf{r}\cdot\mathbf{v}$).
• "When are the two velocities (or $\overrightarrow{OA}$ and $\overrightarrow{OB}$) perpendicular?" Set the relevant dot product to zero and solve for $t$; expect a quadratic, so look for two answers.
• "Do the particles collide? If so, at what angle do they meet?" Equate the two position vectors with the same $t$, require both components to match (collision), then find the angle between $\mathbf{v}_P$ and $\mathbf{v}_Q$ at that time.