University Physics I · 2D & 3D Kinematics · 3.6
Motion in three dimensions
Position, velocity and acceleration in space carry three components. The calculus barely changes; the work is deciding when the third component can be dropped, and when dropping it destroys the physics.
Build the model
Connect the measurement to the mechanism.
Motion in space is one vector function of one variable: r(t) = x(t)î + y(t)ĵ + z(t)k̂. Because the Cartesian basis vectors never change, differentiation acts on the components one at a time, so v = dr/dt and a = dv/dt are three copies of one-dimensional kinematics tied together by a single clock. Nothing about the calculus is harder in three dimensions.
The judgement is geometric: a path is two-dimensional only when every displacement from the start can be built from two fixed vectors. Constant acceleration guarantees that, and so does any central acceleration, which is why projectiles and orbits are planar problems. A velocity component along a magnetic field, crosswind drag, or a Coriolis term does not, and forcing those into a plane deletes real physics.
Reduce when the geometry earns it, and choose the third axis along whatever stays fixed.
- Simple definition
- Three-dimensional kinematics describes a particle with one position vector of three time-dependent components, differentiated and integrated component by component in a fixed Cartesian basis.
- Example
- A drone at r(t) = (2.0 m s⁻¹)t î + (3.0 m) ĵ + (1.5 m s⁻²)t² k̂ has v = (2.0 m s⁻¹) î + (3.0 m s⁻²)t k̂, so its acceleration is a steady 3.0 m s⁻² along k̂.
î, ĵ and k̂ never change, so space motion is three 1D problems on one clock.
a = ẍ î + ÿ ĵ + z̈ k̂ · m, m s⁻¹, m s⁻² · differentiation passes to the components
Speed is the magnitude of the tangent vector; arc length accumulates it over time.
m s⁻¹ and m · path length s is never less than |Δr|
Each component integrates on its own, and each carries its own initial value.
constant a: r(t) = r₀ + v₀t + ½at² · six scalar constants in all
The path lies in the plane through r₀ spanned by v₀ and a; a straight line if v₀ × a = 0.
m · every displacement is a combination of the fixed vectors v₀ and a
The triple product vanishes exactly when the plane of the curve stops turning.
m³ s⁻⁶ · zero torsion, required at every instant where v × a ≠ 0
Circling in x–y while advancing in z: the standard motion that no plane can hold.
m · pitch = 2π v∥ / ω · R and ω fix the circle, v∥ the steady drift
One vector function, three component functions
A particle in space is located by r(t) = x(t) î + y(t) ĵ + z(t) k̂, measured from a chosen origin in a chosen frame. The three coordinate functions share a single parameter, t, and the points they trace form the trajectory — a curve in space. Displacement between two instants is the vector difference Δr = r₂ − r₁, and its magnitude is the straight-line separation, never larger than the distance travelled along the curve. Nothing in this description privileges an axis. Which coordinates happen to stay constant is a statement about the axes you picked, not about the motion.
Differentiate component by component
In a fixed Cartesian frame, î, ĵ and k̂ have constant magnitude and constant direction, so their derivatives vanish and d/dt passes straight through to the coefficients: v = ẋ î + ẏ ĵ + ż k̂ and a = ẍ î + ÿ ĵ + z̈ k̂. Velocity is tangent to the path at every point, and speed is |v| = √(ẋ² + ẏ² + ż²) in m s⁻¹. This is the whole reason three-dimensional kinematics is three one-dimensional problems sharing one clock. The property belongs to the fixed basis alone: in cylindrical or spherical coordinates the unit vectors turn as the particle moves, and differentiating them produces the extra terms that carry the centripetal and Coriolis pieces.
Going backwards costs six numbers
Given a(t), integrate componentwise: v(t) = v₀ + ∫a dt′ and r(t) = r₀ + ∫v dt′. Each integration introduces one constant per component, so a complete solution needs six scalars — three for r₀ and three for v₀. For constant a this collapses to r(t) = r₀ + v₀t + ½at², the same statement as in one dimension with vectors in place of signed numbers. Arc length is the one quantity that refuses to decompose: s = ∫|v| dt must be built from the magnitude, because √(ẋ² + ẏ² + ż²) is not the sum of the separate pieces.
Constant acceleration is always a plane problem
Read r(t) − r₀ = v₀t + ½at² as a recipe: every displacement from the launch point is v₀ times a number plus a times a number. Two fixed vectors span a plane, so the whole trajectory lies in the plane through r₀ containing v₀ and a — at every t, for any launch angle. If v₀ × a = 0 the two are parallel and the path degenerates to a straight line. A ball thrown with v₀ = (3.0 î + 4.0 ĵ + 12.0 k̂) m s⁻¹ under a = −9.81 k̂ m s⁻² is therefore not a three-dimensional problem. Rotate the horizontal axes to put one along 3î + 4ĵ and the launch reads 5.0 m s⁻¹ horizontally with 12.0 m s⁻¹ vertically: ordinary projectile motion, rising 12.0²/(2 × 9.81) = 7.34 m and returning to launch height after 2 × 12.0/9.81 = 2.45 s.
When the plane is not guaranteed
Two conditions cover most planar cases. Constant acceleration, as above. And central acceleration, with a always along the line to one fixed point: then d(r × v)/dt = v × v + r × a = 0, so r × v is fixed and r stays perpendicular to it. Orbits are planar for that reason, not by assumption. For a general path the exact test is zero torsion, (v × a)·(da/dt) = 0 at every instant where v × a ≠ 0; fail it and the trajectory leaves every plane. A charge moving partly along a uniform magnetic field is the standard failure: the perpendicular velocity circles while the parallel velocity is untouched, giving a helix. A proton in a 0.50 T field turns at ω = |q|B/m = 4.8 × 10⁷ rad s⁻¹, so v⊥ = 2.0 × 10⁶ m s⁻¹ and v∥ = 1.0 × 10⁶ m s⁻¹ give a radius v⊥/ω = 4.2 cm and a pitch 2πv∥/ω = 13 cm.
Choose the axes to do the reducing
When a reduction exists, make it explicit rather than assumed. If the motion is planar, the plane normal is n̂ = (v₀ × a)/|v₀ × a| for constant acceleration, or the direction of r × v for a central force; align k̂ with n̂, put the origin on the plane, and the z equation becomes z = 0. If the motion is genuinely three-dimensional, look instead for a direction along which the equations decouple: put k̂ along B for the helix and z(t) = z₀ + v∥t separates completely from the circular x–y motion, leaving one 2D problem plus one trivial 1D problem. Dropping a dimension is a claim about geometry, so state the claim — the trajectory lies in the plane containing v₀ and a — rather than quietly deleting a coordinate.
Change one variable at a time
Make the relationship visible.
Pull the advance down to zero: the coil closes onto the dashed starting plane and the torsion readout reaches zero. Give it any advance at all and no plane can hold the path.
ARC LENGTH PER TURN1.39 m
|Δr| AFTER ONE TURN0.60 m
TORSION τ1.94 m⁻¹
CURVATURE κ4.07 m⁻¹
Live interpretationARC LENGTH PER TURN: 1.39 m. |Δr| AFTER ONE TURN: 0.60 m. TORSION τ: 1.94 m⁻¹. CURVATURE κ: 4.07 m⁻¹
Catch the common trap
Explain before calculating.
A ball is launched with v₀ = (6.0 î − 8.0 ĵ + 15.0 k̂) m s⁻¹, k̂ vertical and drag negligible, so a = −9.81 k̂ m s⁻². Which statement is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA drone's position, in metres with t in seconds and k̂ vertical, is r(t) = (9.0t) î + (12.0t) ĵ + (8.0t − 4.9t²) k̂. Find v(t) and a(t), and the speed at t = 2.0 s.
- î, ĵ and k̂ never change, so d/dt passes to the coefficients one at a time: v = dr/dt = (9.0 m s⁻¹) î + (12.0 m s⁻¹) ĵ + (8.0 − 9.8t) m s⁻¹ k̂.
- Differentiate again: a = dv/dt = −(9.8 m s⁻²) k̂ — constant and vertical, so only the k̂ component of v ever changes.
- At t = 2.0 s the vertical component is 8.0 − 9.8 × 2.0 = −11.6 m s⁻¹, so v = (9.0 î + 12.0 ĵ − 11.6 k̂) m s⁻¹: already falling, two seconds in.
- Speed is the one quantity that will not decompose — take the magnitude: |v| = √(9.0² + 12.0² + 11.6²) = √(81 + 144 + 134.56) = √359.56 = 19.0 m s⁻¹.
Answerv = 9.0 î + 12.0 ĵ + (8.0 − 9.8t) k̂ m s⁻¹; a = −9.8 k̂ m s⁻²; |v| = 19.0 m s⁻¹ at t = 2.0 s
MediumTake the same drone flight. Show that it is a plane problem, find the unit normal of that plane, and give the two-dimensional launch it reduces to.
- a is constant, so r(t) − r₀ = v₀t + ½at²: every displacement from the launch point is a number times v₀ = (9.0 î + 12.0 ĵ + 8.0 k̂) m s⁻¹ plus a number times a = −(9.8 m s⁻²) k̂. Two fixed vectors span a plane, so the whole flight lies in it.
- The normal runs along v₀ × a = (12.0 × (−9.8) − 0) î + (0 − 9.0 × (−9.8)) ĵ + 0 k̂ = (−117.6 î + 88.2 ĵ) m² s⁻³.
- |v₀ × a| = √(117.6² + 88.2²) = √21609 = 147.0, so n̂ = −0.80 î + 0.60 ĵ. The normal is horizontal, so the plane is vertical — it has to be, since the plane contains a.
- Rotate the horizontal axes to put one along the horizontal part of v₀, which has magnitude √(9.0² + 12.0²) = 15.0 m s⁻¹; the vertical part is 8.0 m s⁻¹ and the coordinate out of the plane stays zero for every t. That is a launch at √(15.0² + 8.0²) = 17.0 m s⁻¹, arctan(8.0/15.0) = 28° above the horizontal.
- From there it is ordinary projectile work, with g = 9.8 m s⁻² as the 4.9t² implies: apex 8.0²/(2 × 9.8) = 3.27 m above launch, back to launch height at 2 × 8.0/9.8 = 1.63 s, by which point it has covered 15.0 × 1.63 = 24.5 m across.
AnswerPlanar, with n̂ = −0.80 î + 0.60 ĵ; the reduced launch is 17.0 m s⁻¹ at 28° above the horizontal (15.0 m s⁻¹ across, 8.0 m s⁻¹ up)
HardA charged particle follows r(t) = (0.25 m) cos ωt î + (0.25 m) sin ωt ĵ + (1.2 m s⁻¹)t k̂ with ω = 8.0 rad s⁻¹. Find its speed, the pitch, the arc length of one turn, and test whether any plane holds the path.
- Componentwise again: v = −Rω sin ωt î + Rω cos ωt ĵ + v∥ k̂, with Rω = 0.25 × 8.0 = 2.0 m s⁻¹ for the circling part and v∥ = 1.2 m s⁻¹ for the drift.
- Speed: |v| = √((Rω)² + v∥²) = √(2.0² + 1.2²) = √5.44 = 2.33 m s⁻¹, and it never changes, because both contributions are constant.
- One turn takes T = 2π/ω = 2π/8.0 = 0.785 s, in which k̂ advances by the pitch v∥T = 1.2 × 0.785 = 0.942 m.
- Arc length over that turn is s = ∫|v| dt = |v|T = 2.33 × 0.785 = 1.83 m, against a straight-line |Δr| of only 0.942 m — the circling returns x and y to their starting values, so the whole displacement is the pitch.
- Planarity: a = −Rω²(cos ωt î + sin ωt ĵ), so v × a = v∥Rω²(sin ωt î − cos ωt ĵ) + R²ω³ k̂ and da/dt = Rω³(sin ωt î − cos ωt ĵ). Their dot product collapses to v∥R²ω⁵ = 1.2 × 0.25² × 8.0⁵ = 2.46 × 10³ m³ s⁻⁶.
- That is never zero while v∥ ≠ 0, so no plane holds this path. Set v∥ = 0 and it vanishes — along with the drift, leaving a circle of radius 0.25 m in the x–y plane.
Answer|v| = 2.33 m s⁻¹; pitch = 0.942 m; arc length = 1.83 m per turn against |Δr| = 0.942 m; (v × a)·(da/dt) = 2.46 × 10³ m³ s⁻⁶ ≠ 0, so the motion is genuinely three-dimensional