Module 3: Kinematics · Theory · Prerequisites: K10, Foundation 4
Lifting the one-dimensional restriction adds a direction to every quantity — and reveals that speed and acceleration can be utterly unrelated.
Mahavakya
Two-dimensional motion is not two separate motions happening at the same time. It is one motion described in two coordinates.
On this page
1. Position, velocity and acceleration as vectors
In one dimension position was a signed number and velocity was its derivative. Both steps carry directly to two dimensions — the only change is that both become vectors.
r(t) = x(t) i + y(t) j
v(t) = dr/dt = vx i + vy j
a(t) = dv/dt = ax i + ay j
Coordinates specify the position; the position vector represents it relative to the chosen origin. The speed is the magnitude |v|, not the vector — Foundation 2’s distinction between a vector and its magnitude, appearing with physical content for the first time.
One immediate consequence: position is a point in the plane, not a point on a line. Moving the origin changes all the position values. Moving to a different frame changes all the velocities. Accelerations survive both — the same invariance K1 established, now with two components instead of one.
2. Why the components are independent
The x and y components of Newton’s second law are separate equations. Whatever happens along x does not appear in the y equation, and vice versa. So as long as the forces separate — as long as nothing in the x direction depends on y or on vy — the components evolve independently.
This is Foundation 4’s linearity of differentiation applied to a vector. The derivative of a sum is the sum of the derivatives, so differentiating r = x i + y j gives dx/dt i + dy/dt j — each component differentiates on its own.
The independence belongs to the equations, not to the motion. The particle is not doing two things simultaneously; it is doing one thing, and we have chosen to describe it in two coordinates. Nothing physical changes if we rotate the axes.
This matters for projectiles: the ball is not “moving horizontally” and “falling” at the same time. It is following one curved path that happens to separate neatly in Cartesian coordinates under gravity.
3. Differentiating component by component
Worked example
A particle moves with r(t) = (3t) i + (4t − 5t²) j metres. Find the velocity, speed and acceleration at t = 0.4 s.
Step 1 — compute components first:
vx = 3 m/s vy = 4 − 10t
ax = 0 ay = −10 m/s²
Step 2 — evaluate at t = 0.4 s:
vy(0.4) = 4 − 4 = 0
v = 3 i + 0 j = 3 i m/s
Step 3 — take the magnitude last: speed = |v| = 3 m/s
This is the apex of the trajectory. The vertical velocity has vanished, but the horizontal velocity is intact and the acceleration is still 10 m/s² downward. Speed is 3 m/s; acceleration magnitude is 10 m/s². They have nothing to do with each other at this instant.
v and a are perpendicular here. In two dimensions the angle between them can be anything from 0 to 180 degrees.
Always compute components before magnitudes. Taking |v| at the start destroys the directional information you need for the next step. The magnitude is the very last operation.
4. The angle between velocity and acceleration
In one dimension, v and a can only be parallel or antiparallel — the same direction or opposite. That is the whole of K4’s speeding-up / slowing-down analysis.
In two dimensions the angle between them can be anything from 0° to 180°, and that angle is what determines the shape of the path.
- Parallel (θ = 0°) — speed increasing, path straight.
- Antiparallel (θ = 180°) — speed decreasing, path straight.
- Perpendicular (θ = 90°) — speed instantaneously constant, path curving. The apex of a projectile, and uniform circular motion throughout.
- Any other angle — both speed and direction changing. The general case.
K4 said acceleration is the rate of change of a vector, not a magnitude. Here is the case that proves why that matters. At the apex, the speed is momentarily constant — and the acceleration is 10 m/s² downward. Non-zero acceleration, zero rate of change of speed. That is only possible because the velocity still has a direction that the acceleration can change.
5. What zero velocity looks like in two dimensions
In one dimension v = 0 is one condition. In two dimensions v = 0 requires both components to vanish simultaneously:
vx = 0 and vy = 0
This is the two-dimensional form of K4’s warning. vx = 0 alone means the particle is not moving horizontally — it may still be moving fast vertically. And |v| = 0 requires the entire vector to vanish, which is a far stronger condition than any one component being zero.
6. Uniform circular motion derived
Consider a particle moving on a circle of radius A at constant angular speed ω. Its position is:
r(t) = A cos ωt i + A sin ωt j
Differentiate once:
v = −Aω sin ωt i + Aω cos ωt j
The speed is |v| = Aω, which is constant. And the dot product v · r = A²ω(−sin ωt cos ωt + sin ωt cos ωt) = 0, so v is always perpendicular to r — that is, tangent to the circle.
Differentiate again:
a = −Aω² cos ωt i − Aω² sin ωt j = −ω² r
The acceleration points opposite to the position vector — toward the centre — with magnitude Aω² = v²/A.
Speed is constant. Velocity direction changes every instant. So acceleration is not zero.
Three results from one parametrisation:
- |v| = Aω — constant, derived, not assumed.
- v ⊥ r — velocity is tangent to the circle.
- a = −ω² r — acceleration points to the centre, magnitude v²/A.
This is the same result Foundation 8 derived using rotating basis vectors. Two routes, one result — which is why it is a consequence rather than a formula.
7. Three independent quantities, now six
K5 noted that position, velocity and acceleration are three independent quantities — knowing one tells you nothing about the others. In two dimensions there are now six: x, y, vx, vy, ax, ay.
A particle far from the origin in the x-direction may have zero x-velocity. A particle with a large y-velocity may have zero y-acceleration. None of the six implies any of the others.
The one new trap: vx = 0 is not a turning point in the x-direction unless the sign of vx also changes. It might simply be a momentary stop in one coordinate while the particle continues moving in the other.
Contemplation
The circular-motion calculation derived three results simultaneously from one parametrisation, without assuming any of them. That is the Mahavakya’s point in quantitative form: it is one motion, and choosing to describe it in Cartesian coordinates is a tool, not a fact about the motion.
Rotate the axes and the components change. The speed, the direction of the acceleration and the shape of the path do not. The coordinate system is ours; the motion is not.
That is also why the dot product v · r = 0 is a cleaner statement than “velocity is perpendicular to position” — the dot product is a number, unchanged by axis rotation, and it says the same thing in every frame.
Common misconceptions
1. Two-dimensional motion is two separate motions
It is one motion described in two coordinates. The components are independent in the equations, not in the physics.
2. Take the magnitude of the velocity first
Compute components first, then differentiate or integrate, then take the magnitude. Taking |v| early destroys direction information.
3. Zero speed means zero acceleration
Speed and acceleration are unrelated. Uniform circular motion has constant non-zero speed and non-zero acceleration throughout.
4. vx = 0 means the particle is at rest
Rest requires both components to vanish. The apex of a projectile has vy = 0 and vx = u cos θ ≠ 0.
5. Centripetal acceleration is a force
It is an acceleration that circular motion requires. Something physical — tension, friction, gravity — must supply the corresponding force. The name describes what the acceleration does, not what causes it.
Reflection
- A particle moves in a circle at constant speed. In one dimension it would be considered not accelerating. What changes in two dimensions, and why?
- The parametric derivation gave v · r = 0 without drawing anything. What does that dot product mean geometrically, and why is the algebraic route cleaner?
- At what point on a projectile’s path are v and a parallel? Antiparallel? At what angle at a general point?
Key takeaways
- Position, velocity and acceleration are vectors; differentiate component by component.
- Component independence belongs to the equations, not to the motion. The particle does one thing.
- Compute components first, then evaluate, then take magnitudes. Never the other way round.
- The angle between v and a can be anything in two dimensions. Only when it is 0° or 180° does the one-dimensional intuition hold.
- Zero velocity requires both components to vanish simultaneously.
- Uniform circular motion: |v| constant, v ⊥ r, a = −ω²r toward the centre.
What comes next
This part showed that the x and y motions are independent when the forces separate. Projectile motion under gravity is exactly that case: gravity acts only vertically, the horizontal component feels nothing, and the trajectory follows from combining two one-dimensional motions on a shared clock. K12 builds it out.
Prerequisites: K10 and Foundation 4 · Reading: 16 min · Practice: 30 min · Difficulty: Intermediate