Position vectors don't just sit still — let them depend on time, and every idea from the last three lessons becomes a way to describe moving particles and combined forces.
10 min read4 worked examples
Modelling motion with position vectors
A particle moving with constant velocity $\mathbf{v}$ has a position vector that changes steadily with time $t$. If the particle starts at position vector $\mathbf{r_0}$ when $t=0$, then after time $t$ it has moved a displacement $t\mathbf{v}$ from its start — since velocity multiplied by time gives displacement. This gives the position vector at any time $t$:
$\mathbf{r}(t) = \mathbf{r_0}+t\mathbf{v}$
This is the vector equivalent of the familiar straight-line graph $y=mx+c$: a fixed starting value, plus a constant rate of change multiplied by time. Every component of $\mathbf{r}(t)$ moves at a constant rate — the $\mathbf{i}$-component increases (or decreases) at the rate given by the $\mathbf{i}$-component of $\mathbf{v}$, and independently for $\mathbf{j}$ (and $\mathbf{k}$, in 3D).
Velocity and speed
Velocity is a vector — it has both a size and a direction. Speed is the corresponding scalar: the magnitude of the velocity vector, $|\mathbf{v}|$, with the direction stripped away. Two particles can have the same speed but different velocities, if they're moving at the same rate in different directions.
Worked example 1
Position and speed at a given time
A particle $P$ has initial position vector $(2\mathbf{i}+3\mathbf{j})$ m, and moves with constant velocity $(4\mathbf{i}-\mathbf{j})$ m/s. Find (a) the position vector of $P$ after 3 seconds, and (b) the speed of $P$.
MethodSubstitute into r(t) = r₀ + tv, then take the magnitude of v separately for speed
Speed is the magnitude of the (constant) velocity vector — it doesn't depend on t at all.
▪ solved
Worked example 2
Finding when a particle crosses an axis
A particle has position vector $(2\mathbf{i}+6\mathbf{j})$ m at time $t=0$, and moves with constant velocity $(3\mathbf{i}-2\mathbf{j})$ m/s. Find the time at which the particle crosses the $x$-axis, and its position at that moment.
MethodSet the j-component of r(t) to zero and solve for t
Write the general position vector at time t, componentwise.
2
On the x-axis, the j-component is zero: $6-2t=0 \Rightarrow t=3$.
Every point on the x-axis has a y-coordinate (here, j-component) of 0.
3
i-component at $t=3$: $2+3(3) = 11$.
Substitute the value of t back into the i-component.
4
The particle is at $(11,0)$ after 3 seconds.
State the full position.
▪ solved
Two particles: meeting
When two particles $P$ and $Q$ move independently, each has its own position vector as a function of time, $\mathbf{r_P}(t)$ and $\mathbf{r_Q}(t)$. The particles meet (are at the same point at the same instant) exactly when $\mathbf{r_P}(t) = \mathbf{r_Q}(t)$ for some value of $t$ — which, written out componentwise, means both the $\mathbf{i}$-components and the $\mathbf{j}$-components must agree, at the same value of $t$.
Worked example 3
Do two particles meet?
At time $t=0$, particle $P$ has position vector $(\mathbf{i}+2\mathbf{j})$ m and velocity $(3\mathbf{i}+\mathbf{j})$ m/s; particle $Q$ has position vector $(13\mathbf{i}+8\mathbf{j})$ m and velocity $(-\mathbf{i}-\mathbf{j})$ m/s. Show that $P$ and $Q$ meet, and find the time and position at which this happens.
MethodEquate both components of the two position vectors, and check the resulting t-values agree
Solve the j-component equation independently — it must give the same t for the particles to genuinely meet.
4
Both equations give $t=3$, so $P$ and $Q$ really do meet, at $t=3$ seconds.
If the two t-values had disagreed, the particles' paths would cross without them ever being there at the same time.
5
Position at $t=3$: $\mathbf{r_P}(3) = (1+9)\mathbf{i}+(2+3)\mathbf{j} = 10\mathbf{i}+5\mathbf{j}$, i.e. $(10,5)$.
Substitute t=3 into either position vector — both give the same point, as a check.
▪ solved
Meeting vs. crossing paths
Two particles' paths can cross as geometric lines without the particles ever meeting — that only requires the i- and j-equations to give the same t. If solving the two component equations separately gives two different values of t, the paths cross but the particles pass through that point at different times, and never actually meet.
Forces as vectors
Forces acting on a particle are also vectors, each with a magnitude and a direction. When several forces act at once, the resultant force is simply their vector sum, $\mathbf{F_1}+\mathbf{F_2}+\cdots$, found by adding componentwise exactly as with any other vectors. A particle is in equilibrium when the forces acting on it balance perfectly — the resultant force is the zero vector:
This single vector equation is equivalent to one scalar equation per component — in 2D, the $\mathbf{i}$-components must sum to zero, and the $\mathbf{j}$-components must sum to zero, independently.
Worked example 4
Finding a force for equilibrium · Challenge
Three forces, $\mathbf{F_1}=(2\mathbf{i}+5\mathbf{j})$ N, $\mathbf{F_2}=(-4\mathbf{i}+\mathbf{j})$ N, and $\mathbf{F_3}$, act on a particle in equilibrium. Find $\mathbf{F_3}$, and its magnitude to 3 significant figures.
MethodUse the equilibrium condition to make F₃ the subject, then apply the magnitude formula
1
Equilibrium requires $\mathbf{F_1}+\mathbf{F_2}+\mathbf{F_3} = \mathbf{0}$, so $\mathbf{F_3} = -(\mathbf{F_1}+\mathbf{F_2})$.
Rearrange the equilibrium condition to isolate the unknown force.
Apply the magnitude formula to find the size of the balancing force.
▪ solved
Common mistake
Solving only one of the two component equations when checking whether particles meet or forces balance — and assuming the answer carries over to the other component automatically. Each component gives an independent equation; both must be checked (and, for a meeting problem, must produce the same t) before drawing a conclusion.
Check yourself
A particle has position vector $(5\mathbf{i}-2\mathbf{j})$ m at time $t=0$, and moves with constant velocity $(-\mathbf{i}+3\mathbf{j})$ m/s. Find the position vector of the particle after 4 seconds, and its distance from the origin at that time, to 3 significant figures. (Hint: use r(t) = r₀ + tv, then take the magnitude of the result.)