Vectors in Kinematics and Coordinate Systems

Ask a spacecraft navigator where a probe is and you will get six numbers: three of position, three of velocity — the state vector, exactly the $\vec r$ and $\vec v$ of this article, logged against time. JPL's Horizons system serves these vectors for every tracked body in the Solar System, and mission design is the craft of steering them with small, precisely timed contributions to $\vec a$. Closer to the ground, your phone's GPS chip does 3D kinematics dozens of times per second, fusing satellite ranging and Doppler velocities with accelerometer components $(a_x, a_y, a_z)$ through filters whose internal model is literally equation (7) applied over millisecond steps — and when the screen flips from portrait to landscape, the software has just noticed gravity's components migrating from one axis to another. The tangential/normal split earns its keep wherever machines move fast. Autonomous vehicles plan trajectories under an $a_\perp = v^2/R$ ceiling set by tyre grip — which is why route planners slow the car before the bend — and highway engineers design clothoid curves whose curvature grows gradually so that $a_\perp$ ramps instead of jumping. And every drone flight controller runs the moving-basis mathematics of this article in real time, converting body-frame sensor readings into world-frame motion by differentiating rotating unit vectors — the $-r\dot\theta^2$ and $2\dot r\dot\theta$ story, industrialized.

Stand a metre stick slanted against sunlight coming straight down: its shadow on the floor is shorter than the stick. Shine light horizontally from the side: the shadow on the wall is shorter too. Neither shadow is the stick — but the two shadows together pin it down completely. That is a vector and its components: the floor shadow is $V_x$, the wall shadow is $V_y$, and the stick is $\vec V$, hypotenuse of the right triangle whose legs are its own shadows. Tilt the stick flatter and the floor shadow grows while the wall shadow shrinks; the trade is governed by nothing more than the two trig ratios you already own.

Two things make this more than a geometry exercise. First, the components carry signs: aim the arrow up-and-to-the-left and its $x$-shadow points backwards — $V_x$ comes out negative, and the algebra tracks direction for you. Second, and this is the payoff, the two shadows are independent: a crosswind blowing due north does not slow a plane flying due east by a single km/h — it only adds a northward drift. And an arrow's home address is not part of its identity: "5 m, northeast" is the same displacement wherever on the page you draw it, which is exactly the licence that lets arrows slide head-to-tail when you add them. Now set the arrows in motion. Track a drone with a laser rangefinder and log the arrow from you to it at every instant: that stream of arrows is $\vec r(t)$. Over a short interval the drone slides from $\vec r(t)$ to $\vec r(t+\Delta t)$; the little connecting arrow $\Delta\vec r$ is a chord of the flight path, and as $\Delta t$ shrinks the chord tips into the tangent: velocity $\vec v = d\vec r/dt$ points exactly along the path, always, no exceptions. Acceleration is where 2D breaks your 1D instincts. In one dimension it could only stretch or shrink the velocity; in the plane it gains a second power — it can rotate the velocity without touching its length. Watch a car take a bend at a steady 60 km/h: the speedometer never flickers, yet the velocity vector is swinging around, so the car is accelerating the entire time. Every acceleration vector splits into these two jobs — the part along the velocity works the throttle; the part perpendicular to it turns the wheel.

The last piece of the picture is the choice of basis. Nail your axes to the laboratory floor and they never rotate — the simplest possible bookkeeping. But you may instead strap your axes to the particle, pointing one unit vector "outward from the origin" and the other "sideways": the polar basis. Those directions swing around as the particle moves, and the swinging itself generates terms in the acceleration that Cartesian axes never show — nothing physical changes, but the components reorganize, and two of the reorganized pieces are famous enough to have names: centripetal and Coriolis. Pick coordinates to match the problem's symmetry — projectiles love Cartesian, orbits love polar, helices love cylindrical — and a coupled mess can collapse into separable, exactly solvable pieces.

Put the tail of $\vec V$ at the origin and measure $\theta$ counter-clockwise from the positive $x$-axis. Drop a perpendicular from the tip to the $x$-axis and you have a right triangle with $\vec V$ as hypotenuse: the horizontal leg hugs the angle (adjacent — cosine), the vertical leg faces it (opposite — sine), and the two shadows read off directly:

where: $V$ — the magnitude, always positive; $\theta$ — the angle from the positive $x$-axis, counter-clockwise positive; $V_x, V_y$ — signed numbers in the same unit as $V$. Read the cosine as a dimmer switch: at $\theta = 0°$ it passes the full length, fades to nothing at $90°$, and turns negative beyond — the shadow then points backwards along $x$, and the algebra has tracked the direction with no extra effort from you. At $45°$ each component is $0.707\,V$, not $V/2$: perpendicular lengths combine by squares, not by halves. Just as often you travel the other way — instruments report an eastward speed and a northward speed, and the pilot wants one airspeed and heading. Pythagoras rebuilds the length from the legs, and the direction hides in the ratio of the shadows:

where: the rebuilt $V$ is always positive because the squares erase any minus signs. The quadrant patch is not optional: an arctangent only ever answers between $-90°$ and $+90°$, and it cannot tell $(-3, +3)$ from $(+3, -3)$ — both hand it the same ratio. The ratio fixes the tilt of the line; the signs of the pair choose which end of the line the arrow occupies. The real reason to resolve is that it turns vector addition from a drawing problem into bookkeeping. Think of a long trek with many legs: however tangled the route, each leg deposits its east-shadow into an east ledger and its north-shadow into a north ledger, and the ledgers never interact. At trek's end the two totals are the components of the net displacement:

where: $R_x, R_y$ — components of the resultant $\vec R = \vec A_1 + \vec A_2 + \cdots$; this step is ordinary signed arithmetic, the whole reason the method scales to any number of vectors. It agrees exactly with the geometric picture you may have met first — sliding arrows head-to-tail and drawing the arrow that closes the polygon. For just two vectors at mutual angle $\theta$, that geometry compresses into a single magnitude formula, which the derivation below shows is hiding inside the components:

where: $A, B$ — the magnitudes; $\theta$ — the angle between the vectors placed tail-to-tail. To see that (4) is hiding inside (3), expand $R^2 = (A_x+B_x)^2 + (A_y+B_y)^2 = A^2 + B^2 + 2AB(\cos\alpha\cos\beta + \sin\alpha\sin\beta)$ and apply the cosine difference identity — the ledger and the parallelogram can never disagree. The cross term $2AB\cos\theta$ is the interaction — positive when the vectors help each other, zero at right angles, negative when they fight — and it sweeps the resultant through its entire allowed window, from $A + B$ (parallel) down to $|A - B|$ (opposed): a 3 N and a 4 N force can combine to anything from 1 N to 7 N, but never 8. Two smaller tools complete the kit. Multiplying a vector by a scalar $k$ stretches every component by the same factor — $k\vec V = (kV_x,\ kV_y)$, magnitude $|k|\,V$ — keeping the direction for $k>0$ and reversing it exactly $180°$ for $k<0$; no scalar ever rotates a vector. And dividing a vector by its own magnitude distils a pure direction of length one, the unit vector $\hat V = \vec V/|\vec V|$, so that every vector factorises into "how big" times "which way": $\vec V = |\vec V|\,\hat V$. The Cartesian $\hat\imath, \hat\jmath, \hat k$ are exactly such pure directions — and the polar $\hat r, \hat\theta$ we meet below are unit vectors too, just less obedient ones. There is one more question the toolkit must answer before it is trustworthy: what makes a set of numbers a vector at all? A single arrow in space is one object; its components are merely shadows on a chosen set of axes. Rotate the axes and the shadows change — but the arrow, its length and its direction relative to physical landmarks, does not. That invariance is the defining property that distinguishes a true vector from an arbitrary list of numbers, and it can be proven in four lines.

Fix an origin and Cartesian unit vectors $\hat\imath, \hat\jmath, \hat k$, nailed to the laboratory floor. To say where a particle is in space you need three numbers that change with time, and the position vector packages them into a single object — the arrow drawn from the origin to the particle at each instant:

where: $x, y, z$ — the coordinate functions, each an ordinary function of time; $\hat\imath, \hat\jmath, \hat k$ — mutually perpendicular unit vectors, constant in time. That italicized phrase is the load-bearing one: the arrow $\vec r$ swings and stretches as the particle moves, but every bit of that motion is carried by the three coordinate functions while the basis stands perfectly still. How does the arrow change? Differentiate it. The product rule wants to act on both the coordinates and the basis vectors — but $\dot{\hat\imath} = \dot{\hat\jmath} = \dot{\hat k} = 0$ for a fixed basis, so the derivatives land only on the coordinates, and the same happens again at the second derivative:

where: the overdot means $d/dt$; $\vec v$ points along the instantaneous tangent to the trajectory and $|\vec v|$ is the speed; $\vec a$ is unrestricted in direction. Read (6) as three private conversations: $\dot x$ is computed from $x(t)$ alone and never consults $y$ or $z$. This is the decoupling that lets one curved problem split into parallel straight ones — the entire simplification hangs on those three zeros, and it is the cleanest illustration of why basis choice matters. It is exactly why projectile motion under gravity separates into $\ddot x = 0$ and $\ddot y = -g$, two independent equations sharing only a clock. When the acceleration vector is constant — free flight near the ground, a charged particle between parallel plates, a rover under steady thrust — both integrations can be done once and for all, and the entire trajectory collapses to two lines:

where: $\vec r_0, \vec v_0$ — initial position and velocity; each Cartesian component of (7) is exactly the 1D constant-acceleration kinematics you already know. Geometrically, (7) confines the motion to the flat plane spanned by $\vec v_0$ and $\vec a$, and the path inside that plane is always a parabola — three dimensions of freedom, yet the trajectory never leaves one sheet. The split of $\vec a$ into speed-changing and path-bending parts is one dot product away. Differentiate the squared speed: $\frac{d}{dt}(v^2) = \frac{d}{dt}(\vec v\cdot\vec v) = 2\,\vec v\cdot\vec a$ — so only the component of $\vec a$ along $\vec v$ can touch the speedometer. If $\vec a \perp \vec v$, the speed cannot change at all; the perpendicular remainder can only swing the velocity around, and for a path with local radius of curvature $R$ the swing rate fixes its size:

where: $a_\parallel$ — the throttle, the projection $\vec a\cdot\vec v/v$; $a_\perp$ — the wheel, with $R$ the radius of the circle that locally hugs the path. In a short time $dt$ the velocity turns by $d\phi = (v\,dt)/R$ as the particle rides the local circle, so the sideways velocity change is $v\,d\phi = (v^2/R)\,dt$. This decomposition is coordinate-free — it belongs to the path itself, not to any axis choice: uniform circular motion is all $a_\perp$, straight-line braking is all $a_\parallel$, everything else is a mixture.

Which formula answers which question

Describe planar motion by distance-from-origin $r$ and bearing $\theta$, with unit vectors $\hat r$ (outward) and $\hat\theta$ (sideways, toward increasing $\theta$). Projected onto the fixed Cartesian pair they read:

where: both are unit vectors anchored to the particle's current bearing — as the particle moves, they swing. Differentiating them with the chain rule (the fixed $\hat\imath, \hat\jmath$ ride along inertly) gives the two relations that generate everything else in this section:

where: $\dot\theta$ — the angular velocity, the rate at which the bearing changes. Each basis vector rotates into (minus) the other at exactly that rate — the mathematical heart of every rotating-frame effect, from this article's Coriolis term to the fictitious forces of non-inertial frames later in the track.

Add a straight axis to the polar plane and you have cylindrical coordinates $(\rho, \phi, z)$ — the natural home of helices, such as a charged particle spiralling along a magnetic field:

where: $\rho$ — perpendicular distance from the $z$-axis; $\phi$ — azimuthal angle in the $xy$-plane. The $(\rho, \phi)$ sector is plane polar verbatim, while the $z$ part decouples like a Cartesian coordinate — the helix separates into uniform circular motion plus free drift along the axis. Wrap the polar idea around a sphere and you reach spherical coordinates $(r, \theta, \phi)$ — distance from the origin, colatitude from the $z$-axis, azimuth around it. The velocity components can be read off with almost no calculus at all, which is the point of the final derivation:

where: $r\sin\theta$ — the radius of the horizontal circle traced by varying $\phi$ at fixed $r, \theta$; a point near the pole sweeps a small circle, a point on the equator the largest one. The corresponding radial acceleration picks up two centripetal contributions, one from each angular motion: $a_r = \ddot r - r\dot\theta^2 - r\sin^2\theta\,\dot\phi^2$.

1 · The kicked soccer ball (confidence builder)

A soccer ball leaves the boot at $20.0\ \mathrm{m/s}$, angled $35.0°$ above the level pitch. Split this velocity into horizontal and vertical components, and check the split.

Step 1. Set axes: $x$ downfield along the grass, $y$ straight up. The velocity arrow sits in the first quadrant, so both components will be positive.

Step 2. Horizontal component from (1): $V_x = 20.0 \times \cos 35.0° = 20.0 \times 0.819 = 16.4\ \mathrm{m/s}$.

Step 3. Vertical component from (1): $V_y = 20.0 \times \sin 35.0° = 20.0 \times 0.574 = 11.5\ \mathrm{m/s}$.

Step 4. Rebuild to check with (2): $\sqrt{16.4^2 + 11.5^2} = \sqrt{268.9 + 132.2} = \sqrt{401.1} = 20.0\ \mathrm{m/s}$ — the original speed returns exactly. Angle check: $\arctan(11.5/16.4) = \arctan(0.701) = 35.0°$ ✓.

Answer: $V_x = 16.4\ \mathrm{m/s}$ downfield, $V_y = 11.5\ \mathrm{m/s}$ upward. Read it physically: 16.4 m/s of the kick is chasing the goal while 11.5 m/s is fighting gravity — and from here on, those two missions proceed independently.

2 · Throttle and wheel, separated (exam level)

A warehouse rover starts at $\vec r_0 = (2, 1)\ \mathrm{m}$ with velocity $\vec v_0 = 3\,\hat\imath\ \mathrm{m/s}$ and holds constant acceleration $\vec a = 1.5\,\hat\jmath\ \mathrm{m/s^2}$. At $t = 4\ \mathrm{s}$, find its position, velocity, and speed — then split $\vec a$ into the part changing the speed and the part bending the path, and find the local radius of curvature.

Step 1. Position from (7), component by component: $x = 2 + 3(4) = 14\ \mathrm{m}$; $y = 1 + \tfrac12(1.5)(16) = 13\ \mathrm{m}$. So $\vec r(4) = (14, 13)\ \mathrm{m}$.

Step 2. Velocity from (7): $\vec v(4) = 3\,\hat\imath + (1.5)(4)\,\hat\jmath = (3, 6)\ \mathrm{m/s}$; speed $v = \sqrt{9 + 36} = \sqrt{45} = 6.71\ \mathrm{m/s}$.

Step 3. Throttle part — project $\vec a$ on the velocity direction: $a_\parallel = \dfrac{\vec a \cdot \vec v}{v} = \dfrac{(0)(3) + (1.5)(5)}{6.71} = \dfrac{9}{6.71} = 1.34\ \mathrm{m/s^2}$. Positive: the rover is speeding up.

Step 4. Wheel part — the perpendicular remainder: $a_\perp = \sqrt{|\vec a|^2 - a_\parallel^2} = \sqrt{2.25 - 1.80} = \sqrt{0.45} = 0.67\ \mathrm{m/s^2}$.

Step 5. Radius of curvature from (8): $R = \dfrac{v^2}{a_\perp} = \dfrac{45}{0.67} = 67\ \mathrm{m}$. The path — a parabola opening in $+y$ — is locally as gentle as a 67-metre circle at this point.

Step 6. Cross-check: early in the run ($t \to 0$) the velocity was $(3, 0)$ and $\vec a = (0, 1.5)$ was fully perpendicular — all wheel, no throttle, $R = 9/1.5 = 6\ \mathrm{m}$. As the rover gains $y$-speed, $\vec a$ tilts ever more parallel to $\vec v$: same acceleration vector, jobs continuously renegotiated — the exact 2D analogue of "zero velocity does not mean zero acceleration."

Answer: $\vec r(4) = (14, 13)\ \mathrm{m}$, $\vec v(4) = (3, 6)\ \mathrm{m/s}$, $v = 6.71\ \mathrm{m/s}$; $a_\parallel = 1.34\ \mathrm{m/s^2}$, $a_\perp = 0.67\ \mathrm{m/s^2}$, $R = 67\ \mathrm{m}$.

Texts. Morin, Introduction to Classical Mechanics, Ch. 1 & 10 (polar coordinates, accelerating frames) · Goldstein, Classical Mechanics, 3rd ed., Ch. 1 & 4 (vectors, orthogonal transformations) · Landau & Lifshitz, Mechanics (Vol 1), §§1–4 · Taylor, Classical Mechanics, Ch. 1 (vectors) & §1.7 (polar coordinates) · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 1.

Provenance. All curvilinear acceleration formulas verified against Morin (Ch. 10), Taylor (§1.7), Goldstein (Ch. 1, 4), and Landau-Lifshitz (Vol 1, §§1–4). A verifier should recompute: (1) the polar acceleration by twice differentiating $\mathbf{r}=r\hat{\mathbf{r}}$ using $\dot{\hat{\mathbf{r}}}=\dot\theta\hat{\boldsymbol{\theta}}$; (2) magnitude invariance via orthogonality of the 2D rotation matrix; (3) spherical scale factors $1,r^2,r^2\sin^2\theta$ from the line element. The numeric checks — centripetal magnitude $v^2/r$ and angular-momentum conservation $r^2\dot\theta=\text{const}$ — are standard textbook results. Worked-example numbers on this page ($V_x = 16.4$, $V_y = 11.5\ \mathrm{m/s}$ rebuilding to $20.0$; $\vec r(4) = (14,13)$ m, $a_\parallel = 1.34$, $a_\perp = 0.67\ \mathrm{m/s^2}$, $R = 67$ m) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Position, Displacement, and Reference Frames · Velocity and Acceleration · One-Dimensional Kinematics. Leads to: Projectile Motion · Circular Motion and Centripetal Acceleration · Relative Motion and Galilean Transformations · Newton's Second Law.