Displacement, velocity and force are vectors, not plain numbers. This lesson puts the i–j notation from Pure Mathematics to work modelling real mechanics problems — combining forces, tracking a particle's position, and reading off magnitude and direction.
8 min read4 worked examples
Vector quantities in mechanics
The first lesson of this topic drew the line between scalar quantities (magnitude only) and vector quantities (magnitude and direction). The test never changes: if a direction is needed to pin the quantity down completely, it's a vector; if a single number already says everything there is to say, it's a scalar. Some concrete examples of each, as they show up in mechanics:
Scalars
Distance — $500$ m
Speed — $30$ m s$^{-1}$
Mass — $5$ kg
Time — $4$ s
Energy — e.g. kinetic energy, in J
Vectors
Displacement — $300$ m north-east
Velocity — $30$ m s$^{-1}$ due south
Weight — $5g$ N vertically downwards
Acceleration — $2$ m s$^{-2}$ upwards
Force — $10$ N along a rope
Notice that a car doing $30$ m s$^{-1}$ due south and a car doing $30$ m s$^{-1}$ due north have the same speed but different velocities — the direction is real information, not decoration. Displacement, velocity, acceleration and force are the vectors this topic uses most, and every one of them can be written in exactly the same $\mathbf{i}$–$\mathbf{j}$ component form you met in the Vectors topic. Here, $\mathbf{i}$ and $\mathbf{j}$ are simply a fixed pair of perpendicular directions — typically "east" and "north", or "horizontal" and "vertical" — against which a mechanics quantity is measured.
Worked example 1
Classifying scalars and vectors
State whether each of the following is a scalar or a vector, giving a reason: (a) a cyclist's distance travelled, $12$ km; (b) the cyclist's displacement, $12$ km due north; (c) the cyclist's speed, $4$ m s$^{-1}$; (d) the cyclist's velocity, $4$ m s$^{-1}$ on a bearing of $065°$; (e) the force needed to hold a gate shut, $30$ N.
MethodAsk: does a direction change what the quantity means?
a
Distance is a scalar: $12$ km fully describes it, and "$12$ km, due north" would be a strange thing to say — how far you've travelled doesn't care which way you faced.
A size on its own, complete.
b
Displacement is a vector: $12$ km due north is a genuinely different outcome from $12$ km due south, even though both cover the same distance.
Size and direction both matter.
c
Speed is a scalar: $4$ m s$^{-1}$ tells you how quickly the cyclist covers ground, with no reference to which way they're heading.
A rate with no direction attached.
d
Velocity is a vector: it's the same $4$ m s$^{-1}$ as the speed, but stating the bearing $065°$ turns it into a quantity with a direction as well as a size.
Same size as (c), but now with a direction.
e
Force is a vector: $30$ N pushing the gate shut and $30$ N pulling it open are opposite in effect, so the direction the force acts in is essential information, not an optional extra.
The direction changes the physical outcome.
▪ solved
Same word, two meanings
Notice (a)–(d): distance and speed are the scalar "sizes" that sit underneath displacement and velocity, which are the corresponding vectors. This pairing — a scalar magnitude with a vector built from it by adding a direction — comes up throughout mechanics, so it's worth being able to name both halves of each pair.
Writing mechanics quantities in i–j form
A velocity of $\mathbf{v} = 3\mathbf{i} + 4\mathbf{j}$ m s$^{-1}$ means "$3$ m s$^{-1}$ in the $\mathbf{i}$ direction, combined with $4$ m s$^{-1}$ in the $\mathbf{j}$ direction" — exactly the same idea as a column vector $\begin{pmatrix}3\\4\end{pmatrix}$, just written on one line. The two tools you already have from Pure Mathematics carry over unchanged:
Magnitude — for $\mathbf{v}=x\mathbf{i}+y\mathbf{j}$, the size of the quantity (its speed, if $\mathbf{v}$ is a velocity, or the size of the force, if $\mathbf{v}$ is a force) is $|\mathbf{v}| = \sqrt{x^2+y^2}$, by Pythagoras.
Direction — the angle $\theta$ the quantity makes with the $\mathbf{i}$ direction satisfies $\tan\theta = \dfrac{y}{x}$, found (and adjusted for quadrant, after a sketch) exactly as before.
Units travel with the vector
Unlike a pure-maths vector, a mechanics vector always carries units — m for a displacement, m s$^{-1}$ for a velocity, N for a force. State them on the vector itself (e.g. $\mathbf{F} = (2\mathbf{i}+6\mathbf{j})$ N) and again on any magnitude or direction you calculate from it.
Resultant force and resultant velocity
When more than one force acts on a particle, the single force that has the same overall effect is the resultant force, found by adding the individual force vectors component by component — add all the $\mathbf{i}$ parts, then add all the $\mathbf{j}$ parts. The same idea gives a resultant velocity when a particle's motion is the combination of two or more velocity contributions acting at once, such as a drone's own powered motion together with a steady wind.
Worked example 2
Resultant force, magnitude and direction
Two ropes pull a crate across smooth ice: rope $A$ exerts a force $\mathbf{F}_A = (3\mathbf{i}+4\mathbf{j})$ N and rope $B$ exerts $\mathbf{F}_B = (-\mathbf{i}+2\mathbf{j})$ N. Find the resultant force $\mathbf{R}$ on the crate, and hence its magnitude and direction (as an angle from $\mathbf{i}$), to 1 decimal place.
MethodAdd components, then Pythagoras and tan⁻¹
1
$\mathbf{R} = \mathbf{F}_A+\mathbf{F}_B = (3-1)\mathbf{i} + (4+2)\mathbf{j} = (2\mathbf{i}+6\mathbf{j})$ N.
Add $\mathbf{i}$ parts and $\mathbf{j}$ parts separately.
2
$|\mathbf{R}| = \sqrt{2^2+6^2} = \sqrt{40} = 2\sqrt{10} \approx 6.3$ N.
Pythagoras on the components.
3
Both components are positive, so $\mathbf{R}$ lies in the first quadrant and no adjustment is needed: $\theta = \tan^{-1}\!\left(\dfrac{6}{2}\right) = \tan^{-1}(3) \approx 71.6°$.
Sketch first to confirm the quadrant.
▪ solved
Worked example 3
Resultant velocity
A drone's rotors give it a velocity of $(5\mathbf{i}+2\mathbf{j})$ m s$^{-1}$ through still air. A steady wind adds a further $(-2\mathbf{i}+3\mathbf{j})$ m s$^{-1}$. Find the drone's resultant velocity $\mathbf{R}$, and hence its speed and direction (as an angle from $\mathbf{i}$), to 1 decimal place.
MethodSame idea as a resultant force — add the components
1
$\mathbf{R} = (5-2)\mathbf{i} + (2+3)\mathbf{j} = (3\mathbf{i}+5\mathbf{j})$ m s$^{-1}$.
The drone's own velocity plus the wind's velocity, added component by component.
Both components are positive, so $\theta = \tan^{-1}\!\left(\dfrac{5}{3}\right) \approx 59.0°$ from $\mathbf{i}$.
First quadrant, no adjustment needed.
▪ solved
The wind changes the drone's actual path
The drone's rotors are still driving it along $(5\mathbf{i}+2\mathbf{j})$ — that hasn't changed. What actually happens to the drone is the resultant $\mathbf{R}$: a different speed and a different direction from what the rotors alone would produce. This is exactly why resultants matter — they tell you what really happens when several vector effects act at once.
Worked example 4
Equilibrium with unknown componentsChallenge
A particle in equilibrium is acted on by three forces: $\mathbf{F}_1 = (5\mathbf{i}-2\mathbf{j})$ N, $\mathbf{F}_2 = (-3\mathbf{i}+a\mathbf{j})$ N and $\mathbf{F}_3 = (b\mathbf{i}+6\mathbf{j})$ N. Find the values of $a$ and $b$.
MethodEquilibrium means the resultant is the zero vector
1
Equilibrium means $\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3 = \mathbf{0}$, so the $\mathbf{i}$ components sum to $0$ and the $\mathbf{j}$ components sum to $0$ independently.
Never add or compare vectors as if they were single numbers — $(3\mathbf{i}+4\mathbf{j}) + (-\mathbf{i}+2\mathbf{j})$ is not "found" by adding the magnitudes $5$ and $\sqrt5$. Combine components first, and only take a magnitude (or angle) right at the end, from the combined vector.
Check yourself
A force is given by $\mathbf{F} = (6\mathbf{i}-8\mathbf{j})$ N. Find (a) the magnitude of $\mathbf{F}$, (b) the unit vector in the direction of $\mathbf{F}$, and (c) a force of magnitude $20$ N acting in the same direction as $\mathbf{F}$.
(a) $|\mathbf{F}| = \sqrt{6^2+(-8)^2} = \sqrt{100} = 10$ N.
(b) Unit vector $= \dfrac{\mathbf{F}}{|\mathbf{F}|} = \dfrac{6\mathbf{i}-8\mathbf{j}}{10} = 0.6\mathbf{i}-0.8\mathbf{j}$.
(c) Scale the unit vector to length $20$: $20(0.6\mathbf{i}-0.8\mathbf{j}) = (12\mathbf{i}-16\mathbf{j})$ N.