Relative Position & Velocity

Write v_PA for the velocity of P measured by an observer at rest in A. The second label is not decoration: one trolley has a different velocity for every observer. Positions add tip to tail at each instant, r_PB = r_PA + r_AB, and when the frames share one clock and do not rotate relative to each other, differentiating term by term gives v_PB = v_PA + v_AB. Inner labels match and cancel, so longer chains work the same way, and reversing a pair flips the vector, v_AB = −v_BA, because r_AB = −r_BA. An attendant pushes a trolley toward the tail at 1.2 m s⁻¹ relative to the cabin of an airliner flying at 230 m s⁻¹ over the ground. With forward positive, v_TG = v_TC + v_CG = −1.2 + 230 = 228.8 m s⁻¹. From a second airliner overtaking on a parallel track at 250 m s⁻¹, v_TO = v_TG + v_GO = 228.8 − 250 = −21.2 m s⁻¹: the trolley slides backwards past its windows.

When two velocities are known in the same frame, the relative velocity is their difference, v_BA = v_BG − v_AG. Subtract component by component, then rebuild the magnitude and direction. Rain falling straight down at 8.0 m s⁻¹ in still air, seen from a car driving at 14.0 m s⁻¹, has v_RC = (0, −8.0) − (14.0, 0) = (−14.0, −8.0) m s⁻¹. Its speed relative to the car is √(14.0² + 8.0²) = 16.1 m s⁻¹, neither 22.0 nor 6.0, and the streaks on a side window slant back at arctan(14.0/8.0) = 60.3° from the vertical. Run the argument backwards and the slant measures the rain's speed. In general the relative speed is √(v_A² + v_B² − 2v_A v_B cos θ), where θ is the angle between the two velocities. It equals the difference of the speeds only when they point the same way, and their sum only when they point opposite ways.

An aircraft's engines and wings set its velocity relative to the air: airspeed and heading. The wind is the air's velocity relative to the ground, and the sum v_PG = v_PA + v_AG is the track and groundspeed that carry the aircraft across the map. To hold a chosen track, turn the nose into the wind until the cross-track part of the air velocity cancels the crosswind, v sin α = w_⊥; what remains along the track sets the progress. A light aircraft with 62 m s⁻¹ airspeed must track due north while the wind blows toward the east at 12 m s⁻¹. It heads α = arcsin(12/62) = 11.2° west of north and makes good √(62² − 12²) = 60.8 m s⁻¹; a headwind would then subtract directly from that. Two consequences follow. The nose does not point where the aircraft goes, and once the crosswind reaches the airspeed no heading holds the track at all.

Interception is easiest in the target's frame. A target on a steady velocity is at rest there, so the pursuer reaches it by moving straight down the line of sight: v_PT must point from P to T. Resolve along and across that line. Across it, the pursuer must match the target, v sin β = u sin γ; along it, the closing speed v cos β − u cos γ must be positive. Navigators use the rule in reverse: a vessel whose bearing holds steady while its range falls is on a collision course. A patrol boat making 12 m s⁻¹ spots a runabout 4.0 km away, crossing the line of sight at 7.0 m s⁻¹. Leading it by β = arcsin(7.0/12) = 35.7°, the boat closes at √(12² − 7.0²) = 9.75 m s⁻¹ and intercepts in 410 s. Pointing at the runabout throughout instead traces a curve, and that pure pursuit takes vD/(v² − u²) = 505 s, because it always aims where the target is rather than where it will be.

Two ships on steady courses are awkward to follow on a chart and simple in either ship's frame. Relative to A, B starts at r₀ and moves in a straight line, r(t) = r₀ + vᵣₑₗ t. The squared separation D² = r₀² + 2(r₀ ⋅ vᵣₑₗ)t + vᵣₑₗ² t² is a parabola in t, so d(D²)/dt = 0 gives t* = −(r₀ ⋅ vᵣₑₗ)/vᵣₑₗ². Substituting back, dₘᵢₙ is the perpendicular distance from A to B's relative track, |x₀ v_y − y₀ vₓ|/vᵣₑₗ. Take two airliners at one altitude, B 30 km east and 40 km north of A, with v_BA = (−150, −150) m s⁻¹. Then r₀ ⋅ vᵣₑₗ = −1.05 × 10⁷ m² s⁻¹ and vᵣₑₗ² = 4.5 × 10⁴ m² s⁻², so t* = 233 s and dₘᵢₙ = (1.5 × 10⁶)/212.1 m = 7.07 km. Where the two ground tracks cross is beside the point; only the relative track matters. A negative t* means the closest point is already past.

The chain v_PB = v_PA + v_AB rests on two assumptions: the frames do not rotate relative to each other, and every observer shares one clock. Differentiate once more and a_PB = a_PA + a_AB. Frames in constant relative velocity therefore agree on every acceleration, which is why the laws of motion can hold in all of them at once. A frame that accelerates does not: when a bus brakes at 3.0 m s⁻², a ball rolling freely on its floor keeps its velocity over the road yet accelerates forward at 3.0 m s⁻² relative to the bus, with nothing pushing it. A rotating frame, a turntable or the spinning Earth, adds velocity terms the simple sum does not contain. And the shared clock fails near light speed: two velocities of 0.60c along one line compound to 1.20c by this rule but are measured at (u + v)/(1 + uv/c²) = 0.88c. For a 300 m s⁻¹ projectile fired forward from a 250 m s⁻¹ airliner, uv/c² is about 8 × 10⁻¹³.