Motion in a Plane: 2D Kinematics

Put two coins on the edge of your desk. Flick one hard so it shoots off sideways; at the same instant, nudge the other so it simply drops. One coin sails a metre through the air on a graceful curve, the other falls straight down — yet listen: a single click. They land together, every time. The sideways rush buys exactly zero extra airtime. Something about motion in two dimensions is profoundly simpler than it looks, and that click is your first evidence. This module is where kinematics grows its second dimension. Position becomes an arrow; velocity becomes an arrow that always points along the path; acceleration becomes an arrow free to point anywhere — and what it does depends entirely on its angle to the velocity. Best of all, the apparatus comes with a gift: pick two perpendicular axes, and the motion along one runs completely independently of the motion along the other. One curved journey splits into two straight-line problems you already know how to solve, sharing nothing but the clock. Every projectile you will ever launch, every circular loop you will ever analyse, and a good fraction of JEE-Main's kinematics section run on the ideas of this page. Master the split — axes for bookkeeping, one shared time, Pythagoras to reassemble — and 2D motion reads like two easy problems stapled together, because that is precisely what it is.

Key takeaways

In a plane, position is the vector $\vec r = x\,\hat\imath + y\,\hat\jmath$ . Velocity always points along the tangent to the path; its magnitude is the speed $v = \sqrt{v_x^2 + v_y^2}$ — Pythagoras, never $v_x + v_y$ .
Perpendicular axes are independent: each obeys its own 1D kinematics, and the ONLY thing they share is time $t$ . The dropped coin and the launched coin land together.
Acceleration in a plane has two jobs: its component along $\vec v$ changes the speed; its component perpendicular to $\vec v$ turns the direction, leaving the speedometer untouched.
For constant acceleration, each axis runs its own suvat set with the same $t$ : $x$ -data never enters a $y$ -equation.
Key idea: one 2D motion = two independent 1D motions on a shared clock — resolve, solve each axis alone, recombine with Pythagoras and trigonometry.

The core insight

Motion in a plane splits along any two perpendicular axes into independent one-dimensional motions that share only the time — and because velocity is a vector, acceleration along the velocity changes how fast you move while acceleration perpendicular to it changes only which way you move.

Where this lives today

Every drone hovering steadily runs this module's split in silicon: its flight controller keeps separate control loops per axis, each nudging one velocity component toward its target on the shared clock. The same architecture drives a 3D printer's head and a CNC mill's cutter — a curved contour decomposed into $x(t)$ and $y(t)$ commands executed by two motors that never consult each other: the two shadows, mechanised. Game physics engines make the deepest point unmissable: to fly a football across the screen, the engine stores $(x, y)$ , $(v_x, v_y)$ , $(a_x, a_y)$ and updates each component independently every frame — equation (4) evaluated on a shared clock, billions of times per second worldwide. And when a programmer forgets Pythagoras and limits speed by $v_x + v_y$ , characters visibly run faster on diagonals — a bug so common the industry named it ("diagonal speed boost"), a pure specimen of this module's central trap.

The physical picture

Watch a drone fly a smooth curve, and track it with two spotlights: one overhead, casting a shadow on the ground, one from the side, casting a shadow on a wall. The drone performs a single elegant 2D motion — but each shadow performs a plain 1D motion, and neither shadow knows the other exists. Film only the shadows and you can reconstruct the whole flight: the drone is its two shadows, reassembled. That is the entire strategy of this module — the curved path is one motion, the axes are our bookkeeping, and every vector of the motion splits into an $x$ -part and a $y$ -part that evolve separately, welded together only by the shared stopwatch.

Figure

The anatomy of motion in a plane — position arrow, tangent velocity, and the two jobs of acceleration

Two features of this picture break 1D habits, and both are exam gold. First, the velocity arrow is chained to the path — always along the tangent — while the position arrow $\vec r$ points from the origin to the particle, generally a different direction entirely. Second, acceleration is chained to nothing: along the velocity, against it, straight across it, or anywhere between. In one dimension acceleration could only stretch or shrink your velocity; in a plane it learns to turn it — the upgrade that makes circles, orbits and every curved trajectory possible.

Arrows that move: position, velocity, acceleration

Fix an origin and two perpendicular unit vectors: $\hat\imath$ pointing along $x$ and $\hat\jmath$ along $y$ , both nailed down once and never moved. A particle's location at any instant is then the arrow

\vec r(t) = x(t)\,\hat\imath + y(t)\,\hat\jmath \qquad (1)

where: $x(t), y(t)$ — the two coordinate functions of time; $\hat\imath, \hat\jmath$ — fixed unit vectors. Equation (1) is a contract: however curved the flight, it is fully described by two ordinary functions of one variable — the two shadows. A drone at $\vec r = 30\,\hat\imath + 40\,\hat\jmath$ (metres) is 30 m east and 40 m up, a straight-line $\sqrt{30^2 + 40^2} = 50$ m from the origin. Velocity asks how the position arrow changes. Over an interval $\Delta t$ the particle slides from $\vec r$ to $\vec r + \Delta\vec r$ , and the average velocity is the displacement divided by the time — but now it is an arrow:

\vec v_{\text{avg}} = \frac{\Delta \vec r}{\Delta t} = \frac{\Delta x}{\Delta t}\,\hat\imath + \frac{\Delta y}{\Delta t}\,\hat\jmath \qquad (2)

where: $\Delta \vec r$ — the chord joining two positions on the path; $\Delta x, \Delta y$ — its components, each built from its own axis's data alone. Shrink the window and the chord tips into the tangent (the first derivation watches it happen): the instantaneous velocity $\vec v = v_x\,\hat\imath + v_y\,\hat\jmath$ has components that are the tangent-slopes of each coordinate's own graph against time — position–time-graph reading, done twice. Magnitude and direction come from right-triangle geometry:

v = |\vec v| = \sqrt{v_x^2 + v_y^2}, \qquad \tan\theta = \frac{v_y}{v_x} \qquad (3)

where: $v$ — the speed, what a speedometer reads; $\theta$ — the angle of motion above the $x$ -axis. A walker doing 1 m/s across a travelator carrying her along at 2 m/s moves at $\sqrt{4+1} = \sqrt 5 \approx 2.24$ m/s — not 3. Components combine like the sides of a right triangle, never by plain addition. Acceleration repeats the construction one level up: $\vec a = a_x\,\hat\imath + a_y\,\hat\jmath$ , each component the rate of change of its own velocity component — each axis again minding its own business. (Writing these rates as $dx/dt$ and $dv/dt$ is the calculus shorthand; Calculus in Kinematics makes it precise, and nothing here needs more than the tangent-slope idea.)

Derivation AAverage velocity is the chord and instantaneous velocity is the tangent

We want to see, with actual numbers, that the average-velocity arrow lies along the chord of the path, and that as the time window shrinks it settles onto the tangent — which is why instantaneous velocity always points along the path. The strategy: take the motion $x = 3t$ , $y = 4t - t^2$ (metres, seconds), sit at $t = 2$ s, and shrink the window, watching the chord's direction converge.

Step 1. Locate the particle: $\vec r(2) = (6, 4)$ m. Over a window $\Delta t$ , the chord components are $\Delta x = 3\Delta t$ and $\Delta y = \big[4(2{+}\Delta t) - (2{+}\Delta t)^2\big] - 4 = -\Delta t^2$ . Why the algebra matters: the linear terms in $\Delta y$ cancel exactly — the $y$ -motion is momentarily poised at its turning point, so its change is second-order small.

Step 2. Take windows. $\Delta t = 1$ : chord $(3, -1)$ , direction $\tan^{-1}(1/3) = 18.4°$ below the $x$ -axis, average speed $\sqrt{10} = 3.16$ m/s. $\Delta t = 0.1$ : chord $(0.3, -0.01)$ , $1.9°$ below. $\Delta t = 0.01$ : chord $(0.03, -0.0001)$ , $0.19°$ below — collapsing onto the horizontal.

Step 3. Read the limit. The average velocity $(3, -\Delta t)$ settles to $(3, 0)$ : instantaneous velocity 3 m/s, exactly horizontal — the tangent direction there, since $y$ has just stopped rising. The tangent-slope rules confirm it: $v_x = 3$ throughout, $v_y = 4 - 2t = 0$ at $t = 2$ s. Verified: $\vec r(3) = (9, 3)$ gives the 1-second chord $(3, -1)$ ✓; $|(3,-1)| = \sqrt{10} = 3.16$ ✓; the shrinking-window limit matches $v_y = 4-2t$ at $t=2$ .

The independence principle: two problems, one clock

Here is the deepest fact in this module: perpendicular components of a motion evolve independently. The acceleration's $x$ -component changes only $v_x$ ; its $y$ -component changes only $v_y$ . Gravity, pulling straight down, has no vote on horizontal velocity — so the flicked coin's sideways rush does nothing to delay its fall. Both coins' vertical motions are identical twins: same start ( $v_y = 0$ ), same $g$ downward, same 1.25 m drop, same $t = \sqrt{2h/g} = 0.5$ s of airtime (taking $g = 10\ \mathrm{m/s^2}$ , the usual CBSE/JEE/NEET numerical choice; with $g = 9.8$ , as AP, IB and A-Level prefer, $t = 0.505$ s). The launched coin simply also covers ground sideways while its vertical twin-story plays out.

Figure

Strobe photograph of the independence principle — dropped and launched, level with each other at every flash

Independence is a licence to compute. Under any constant acceleration, each axis runs the full 1D suvat machinery on its own private data, stapled to the other axis by a single shared quantity:

v_x = u_x + a_x t, \quad x = u_x t + \tfrac12 a_x t^2 \qquad\Big|\qquad v_y = u_y + a_y t, \quad y = u_y t + \tfrac12 a_y t^2 \qquad (4)

where: $u_x, u_y$ — initial-velocity components ( $u_x = u\cos\theta$ , $u_y = u\sin\theta$ from magnitude-and-angle); $a_x, a_y$ — the constant acceleration's components; $t$ — the same number in every slot. That last clause is the whole discipline of 2D problem-solving: time is not an $x$ -thing or a $y$ -thing but the one universal bridge, and most lost marks in this chapter come from quietly using different times on the two axes.

Go deeper · Derivation B · 1 minWhy the axes cannot talk to each other

Derivation BWhy the axes cannot talk to each other

We want to justify the split (4): that a 2D constant-acceleration motion is exactly two 1D suvat problems running in parallel. The strategy is bookkeeping honesty — write the velocity update in components and watch that no equation ever contains the other axis's symbols. The step to watch is the first: it works only because the axes never move.

Step 1. Because $\hat\imath$ and $\hat\jmath$ are fixed, a change in the velocity arrow can come only from changes in its components: $\Delta \vec v = \Delta v_x\,\hat\imath + \Delta v_y\,\hat\jmath$ .

Step 2. Match components with the definition of acceleration, $\Delta \vec v = \vec a\,\Delta t$ : so $\Delta v_x = a_x\,\Delta t$ and $\Delta v_y = a_y\,\Delta t$ — two scalar statements, each mentioning one axis only. Why this is the crux: the $x$ -statement contains no $y$ -symbol. Whatever $v_y$ does — grow, shrink, reverse — the rule updating $v_x$ never hears about it.

Step 3. Run each statement through the 1D playbook: constant $a_x$ with initial $u_x$ gives $v_x = u_x + a_x t$ and $x = u_x t + \tfrac12 a_x t^2$ , exactly as in Kinematic Equations, using only $x$ -data; identically for $y$ . Equation (4) assembled.

Step 4. Reconnect through time: evaluate both axis-stories at the same $t$ , recombine with Pythagoras and $\tan\theta = v_y/v_x$ . Verified: for $u = (3, 4)$ m/s, $a = (0, -2)\ \mathrm{m/s^2}$ , equation (4) at $t = 2$ s gives $(6, 4)$ m — reproducing the path of the first derivation ✓. Velocity-dependent forces (air drag) make $a_x$ depend on $v_y$ and break the split — see Limiting cases.

Fine print: the split works whenever each acceleration component minds only its own axis. Air resistance, acting against the total velocity, entangles the axes — hence "neglect air resistance" in every exam problem.

What acceleration does in a plane: throttle and steering

In 1D, acceleration had one job: change the speed. In a plane it has two, set by its direction relative to the velocity: the part along $\vec v$ is the throttle (or brake), the part across $\vec v$ is the steering wheel. A car holding 60 km/h through a bend, needle frozen, is accelerating the entire time — all steering, no throttle.

Go deeper · Derivation C · 1 minPerpendicular acceleration turns the velocity without changing the speed

Derivation CPerpendicular acceleration turns the velocity without changing the speed

We want the claim most suspicious to 1D-trained instincts: an acceleration at right angles to the velocity changes the direction of motion but leaves the speed alone. The strategy: update the velocity arrow over a short interval and compute its new length with Pythagoras — the perpendicular geometry will make the speed change vanish to first order.

Step 1. Over a short time $\Delta t$ , the velocity updates by $\vec v\,' = \vec v + \vec a\,\Delta t$ . Take $\vec v = (v, 0)$ and a perpendicular $\vec a = (0, a)$ : then $\vec v\,' = (v, \,a\Delta t)$ .

Step 2. New speed by Pythagoras: $|\vec v\,'| = \sqrt{v^2 + a^2\Delta t^2}$ — the change involves only $\Delta t^2$ . Halve the interval and the speed change drops to a quarter; in the instant-by-instant limit it contributes nothing. Why first order is what matters: every genuine rate is read from the part proportional to $\Delta t$ . A perpendicular push has zero projection along the motion, so that part is exactly zero.

Step 3. The direction changes in proportion to $\Delta t$ : the arrow tilts by roughly $a\Delta t/v$ . Turning is first-order; speeding is second-order. Perpendicular acceleration is pure steering.

Step 4. In general, only the component of $\vec a$ along $\vec v$ changes the speed; only the component across $\vec v$ changes the direction. Held perpendicular forever with constant magnitude, $\vec a$ turns the velocity steadily through a full circle — the doorway to Circular Motion. Verified: $v = 4$ m/s, $a = 2\ \mathrm{m/s^2}$ , $\Delta t = 0.01$ s: $\vec v\,' = (4, 0.02)$ , speed $\sqrt{16.0004} = 4.00005$ m/s (second-order small), tilt $\tan^{-1}(0.02/4) = 0.29°$ (first-order). General cross-check: $|\vec v\,'|^2 = v^2 + 2(\vec a \cdot \vec v)\Delta t + a^2 \Delta t^2$ , and the first-order term vanishes exactly when $\vec a \perp \vec v$ .

This settles favourite exam questions instantly. Can a particle accelerate at constant speed? Yes — keep $\vec a \perp \vec v$ : uniform circular motion. At minimum speed, what is the angle between $\vec a$ and $\vec v$ ? Ninety degrees — the throttle-component has hit zero, so all of $\vec a$ is momentarily steering. The worked examples cash this in.

Limiting cases

case	result	meaning
$\vec a = 0$	straight line, constant speed	both shadows run uniformly; the path cannot bend
$\vec a$ always parallel to $\vec v$	straight line, changing speed	pure throttle: 1D motion wearing 2D clothes
$\vec a$ always perpendicular to $\vec v$ , constant size	circle at constant speed	pure steering — the preview of circular motion
one component of motion zero throughout	1D kinematics recovered	the plane contains the line: 2D is a strict upgrade
$v_y = 0$ at an instant, $a_y \neq 0$	path momentarily level, still curving	zero velocity component ≠ zero acceleration component
acceleration depends on total velocity (air drag)	axes couple, independence fails	the split is a gift of constant, axis-clean acceleration

The fifth row deserves a pause: in the first derivation, $v_y$ passed through zero while $a_y$ stayed at $-2\ \mathrm{m/s^2}$ — level for one instant, already bending down. The 2D twin of "zero velocity does not mean zero acceleration."

Worked examples

Board level — reading a motion from its components (CBSE/ISC pattern)

A particle moves in the x–y plane with position $x = 3t$ m and $y = (4t - t^2)$ m, where $t$ is in seconds. Find (a) the velocity components and the acceleration, (b) the instant at which the particle moves horizontally, and (c) the speed and direction of motion at $t = 3$ s.

Step 1. Component rates via the tangent-slope rules (origin: Calculus in Kinematics): $v_x = 3$ m/s constant; $v_y = 4 - 2t$ m/s; hence $a_x = 0$ , $a_y = -2\ \mathrm{m/s^2}$ : $\vec a = -2\,\hat\jmath\ \mathrm{m/s^2}$ , constant.

Step 2. "Moving horizontally" means no vertical velocity: $4 - 2t = 0$ , so $t = 2$ s, where the speed is just $|v_x| = 3$ m/s. (Note $a_y$ is still $-2\ \mathrm{m/s^2}$ there — level path, nonzero acceleration.)

Step 3. At $t = 3$ s: $v_x = 3$ , $v_y = 4 - 6 = -2$ m/s. Speed $= \sqrt{3^2 + (-2)^2} = \sqrt{13} = 3.61$ m/s.

Step 4. Direction: $\tan\theta = |v_y|/v_x = 2/3$ , so $\theta = 33.7°$ below the $+x$ -axis (the minus sign on $v_y$ says descending). State the direction — a magnitude alone loses the direction mark.

Answer: $\vec v = 3\,\hat\imath + (4-2t)\,\hat\jmath$ m/s, $\vec a = -2\,\hat\jmath\ \mathrm{m/s^2}$ ; horizontal at $t = 2$ s; at $t = 3$ s the speed is $3.61$ m/s at $33.7°$ below the x-axis.

Competitive level — minimum speed and the perpendicularity payoff (JEE-Main pattern)

A particle starts with velocity $\vec u = 4\,\hat\imath - 3\,\hat\jmath$ m/s and moves with constant acceleration $\vec a = 2\,\hat\jmath\ \mathrm{m/s^2}$ . Find (a) the minimum speed during the motion and (b) the angle between $\vec v$ and $\vec a$ at that instant.

Step 1. Split by axes: $v_x = 4$ m/s forever (no $x$ -acceleration); $v_y = -3 + 2t$ m/s; so $v = \sqrt{16 + (2t-3)^2}$ .

Step 2. Minimise by inspection — no calculus: $16$ is fixed, and the square $(2t-3)^2$ is smallest when zero, at $t = 1.5$ s. Minimum speed $= \boxed{4\ \text{m/s}}$ .

Step 3. At $t = 1.5$ s the velocity is $(4, 0)$ , purely along $x$ , while $\vec a = (0, 2)$ is purely along $y$ : the angle is $\boxed{90°}$ .

Step 4. The reusable lesson: at minimum speed the changeable velocity component has hit zero, leaving $\vec a$ entirely perpendicular to $\vec v$ . "Minimum speed ⇔ $\vec a \perp \vec v$ " is a standing JEE shortcut. Timing note: 60–90 seconds. Write $v^2$ as a fixed number plus a square, kill the square, and quote the perpendicularity — differentiating $v^2$ works too but is slower. The distractor grid will offer $\sqrt{16+9} = 5$ m/s (the initial speed) for students who never track the components.

Competitive level — velocity angle vs position angle (NEET/JEE-Main pattern)

A body starts from rest-in-y at the origin with $v_x = 6$ m/s constant, and gains vertical velocity under $a_y = 4\ \mathrm{m/s^2}$ . Find (a) when its velocity makes $45°$ with the x-axis, (b) its speed then, and (c) the angle of its position vector at that instant.

Step 1. Components: $v_x = 6$ always; $v_y = 4t$ . A $45°$ velocity direction needs $v_y = v_x$ : $4t = 6$ , so $t = 1.5$ s.

Step 2. Speed there: $\sqrt{6^2 + 6^2} = 6\sqrt 2 = 8.49$ m/s.

Step 3. Position by equation (4): $x = 6(1.5) = 9$ m; $y = \tfrac12(4)(1.5)^2 = 4.5$ m. Position angle: $\tan^{-1}(4.5/9) = \tan^{-1}(0.5) = 26.6°$ .

Step 4. The punchline: the velocity points at $45°$ but the position vector at only $26.6°$ — different arrows answering different questions (where you are going now, versus where you are seen from the origin). Options equating the two angles are planted precisely for this moment. Timing note: about 90 seconds — two component reads, one suvat line per axis, two arctangents. Every answer comes from components at one shared $t$ ; never reconstruct the path shape.

Exam traps

Adding component magnitudes. The speed of $(6, 8)$ m/s is 10 m/s, not 14. Components recombine by Pythagoras and $\tan\theta$ — the most-farmed error in this chapter, and the distractor grid always contains the plain sum.

Two clocks. Using $t$ from a $y$ -calculation in an $x$ -equation at a different moment. There is one stopwatch: write $t$ once, use it everywhere; if two events differ, name them $t_1$ and $t_2$ .

Velocity direction confused with position direction. $\vec v$ is tangent to the path; $\vec r$ points from the origin. The third worked example lives on the difference — expect one option built from each angle.

Average velocity aimed wrongly. It points along the chord $\Delta \vec r$ — not along the final velocity. On a curved path it generally points where the particle never actually headed at any instant.

Sine–cosine swap when resolving. $u_x = u\cos\theta$ when $\theta$ is measured from the $x$ -axis — the cosine hugs the axis the angle touches. Angles given from the vertical flip the pair; re-derive from the triangle instead of quoting.

One perpendicular instant sold as a circle. $\vec a \perp \vec v$ at an instant means the speed is momentarily steady — nothing more. Circular motion needs perpendicularity at every instant with constant magnitude. Assertion–reason setters love the gap.

Degrees–radians slips. Direction answers are arctangents; a calculator left in radian mode turns $33.7°$ into an alien number. Check the mode before the first arctan.

Origins · 1 minChristiaan Huygens and how this was discovered

Christiaan Huygens

1629–1695 · The Hague and Paris

Huygens was the first person to compute what it costs to change a velocity's direction. In 1659, analysing a stone whirled on a string, he derived the exact measure of the turning acceleration — proportional to the speed squared, inversely to the radius — decades before Newton's Principia printed the equivalent result. His Horologium Oscillatorium (1673), written after he built the first pendulum clock, treats motion with full two-dimensional thinking: velocities as directed quantities, compounded and resolved along perpendicular directions. The throttle-and-steering split of this module is the idea Huygens quantified first, and every banked road and orbiting satellite still runs on his number, $v^2/r$ .

History

1637

— Descartes' La Géométrie welds algebra to geometry: coordinates make a curved path a pair of functions.

1638

— Galileo's Two New Sciences composes horizontal and vertical motions independently — the two-shadow idea before vectors existed.

1673

— Huygens' Horologium Oscillatorium quantifies the acceleration hidden in a turning velocity.

1687

— Newton's Principia, Corollary I: motions from simultaneous "forces" compound by the parallelogram — independence made law.

1881

— Gibbs prints his Elements of Vector Analysis; position, velocity and acceleration take their modern arrow notation.

Going further

Texts. NCERT Physics XI Part 1, Ch. 4 (Motion in a Plane) — the CBSE/NEET canonical treatment · H.C. Verma, Concepts of Physics Vol. 1, Ch. 3–4 · Halliday, Resnick & Walker, Ch. 4. The full vector-calculus treatment (rotating axes, the origin of centripetal and Coriolis terms) is the UG deep dive: Kinematics in 2D & 3D (aq-mech tier).

Provenance. All results are standard kinematics per NCERT Ch. 4. A verifier should recompute: coin drop $t = \sqrt{2(1.25)/10} = 0.5$ s ( $0.505$ s with $g{=}9.8$ ); travelator $\sqrt 5 = 2.24$ m/s; chord table for $x=3t, y=4t-t^2$ at $t=2$ : $(3,-1), (0.3,-0.01), (0.03,-0.0001)$ , directions $18.4°, 1.9°, 0.19°$ below horizontal, $v_y = 4-2t = 0$ ; perpendicular update $(4, 0.02)$ , speed $4.00005$ m/s, tilt $0.29°$ ; board example: horizontal at $t=2$ s, $\sqrt{13} = 3.61$ m/s at $33.7°$ below axis; minimum speed $4$ m/s at $t = 1.5$ s, angle $90°$ ; angle example $6\sqrt2 = 8.49$ m/s, position angle $\tan^{-1}(0.5) = 26.6°$ .

Prerequisites. Components & Resolution of Vectors — the splitting machinery every problem starts with · Acceleration — rates of change of velocity, signs and all. Leads to. Projectile Motion — the constant- $g$ special case with its trajectory, range and height formulas · Circular Motion — what permanently perpendicular acceleration builds.

How can the x-motion truly not know about the y-motion? They belong to one object!

Because "the x-motion" is the shadow of the object on the x-axis, and that shadow's changes are driven only by the x-component of the acceleration. The object is one thing, but the bookkeeping splits perfectly: the update rule for $v_x$ is $\Delta v_x = a_x \Delta t$ , and no $y$ -symbol appears in it. The two coins are the experimental proof — the launched coin's sideways speed buys it zero extra airtime, because gravity is entirely a $y$ -agent. The split fails only when the acceleration itself mixes axes, as air drag does, which is why exam problems switch drag off.

Is speed the sum of the component speeds?

Never. Components are the perpendicular legs of a right triangle whose hypotenuse is the velocity, so speed is $\sqrt{v_x^2 + v_y^2}$ : 1 m/s across a 2 m/s travelator gives $\sqrt 5 \approx 2.24$ m/s, not 3. The plain sum overcounts because the components point in different directions — metres east and metres north are not the same kind of progress. Addition works only when both vectors share one line, which is 1D kinematics again.

Why does turning count as accelerating if the speedometer never moves?

Because velocity is an arrow, and acceleration is the rate of change of that arrow — length or direction. A car rounding a bend at a steady 60 km/h has a velocity swinging from "60 east" to "60 north": a real change, produced by a real sideways force from the tyres, which passengers feel. The speedometer reports only the arrow's length, so it is blind to the turning. Physics keeps the vector definition because forces change the vector: the same push that speeds you up when aimed along your motion turns you when aimed across it.

Do x and y really share the same t, or is that just a convention?

Physical fact, not convention. There is one particle and one clock; " $t = 1.5$ s" names a single event in its life, at which it has one position (two coordinates) and one velocity (two components). Solving the y-story for an event's time and feeding it into the x-story uses the only honest bridge between the axes. What is convention is the axes themselves — rotate them at will; a good choice makes the acceleration components simplest.

Does the independence principle survive air resistance?

No — and seeing why sharpens the principle. Drag acts against the total velocity with a size set by the total speed, so its x-component depends on $v_y$ too. The axes then exchange information through the force and the two-shadows split breaks. That is why exams state "neglect air resistance": it is the condition under which the constant-acceleration split (4) is exact. Real ballistics — and game engines with serious aerodynamics — handle the coupling numerically.

Which way does average velocity point on a curved path?

Along the chord — the displacement arrow $\Delta\vec r$ from start to end, divided by the elapsed time. It ignores the route entirely, and on a curved path it usually matches the instantaneous velocity at no moment of the journey: in the first derivation the 1-second chord pointed 18.4° below horizontal while the instantaneous velocity at the window's start was exactly horizontal. Exams ask for both in one question — compute them separately, from endpoints and from tangent-slopes respectively.