---
title: Relative Motion
module: Kinematics
moduleNumber: 1
lessonNumber: 4
order: 104
summary: >
  A velocity is only ever measured relative to some observer, so a boat's speed
  through the water, over the ground, and as seen from another boat are three
  different vectors. Choosing the right frame — and subtracting one motion from
  another — collapses river crossings, crosswind headings, pursuit, and
  closest-approach problems into a single vector equation. We build the
  relative-velocity and relative-position relations for uniformly moving frames, show
  why acceleration is the one quantity all such observers agree on, and note where
  rotating frames break the simple subtraction.
topics: [Kinematics]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 3 — Motion in Two and Three Dimensions; §3-1"
---

## Relative position and velocity

Relative motion compares positions measured in two reference frames. For particle
$P$, frame $A$, and frame $B$,

$$
\vec r_{P/B}=\vec r_{P/A}+\vec r_{A/B}.
$$

Differentiation gives the Galilean velocity transformation,

$$
\vec v_{P/B}=\vec v_{P/A}+\vec v_{A/B}.
$$

Subscripts read “of the first object relative to the second.” If frame $A$ moves
at constant velocity relative to $B$, acceleration has the same value in both
frames:

$$
\vec a_{P/B}=\vec a_{P/A}.
$$

$$
% caption: The position triangle relates a particle and two frames. Following the
% P-to-A arrow and then the A-to-B arrow gives the same directed displacement as
% the direct P-to-B arrow, which is the geometric basis of the velocity relation.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[fill=white, draw=black, thick] (0.4,0.4) circle (2pt) node[below] {B};
  \draw[fill=white, draw=black, thick] (3.0,0.9) circle (2pt) node[below] {A};
  \draw[fill=acc!12, draw=acc, thick] (4.4,3.0) circle (2pt) node[above] {P};
  \draw[->, black, thick] (0.4,0.4) -- (3.0,0.9) node[midway, below] {A/B};
  \draw[->, acc, thick] (3.0,0.9) -- (4.4,3.0) node[midway, right] {P/A};
  \draw[->, acc, very thick] (0.4,0.4) -- (4.4,3.0) node[midway, above left] {P/B};
\end{tikzpicture}
$$

**Intercept tests and closest approach.**

Under constant relative velocity, the intercept condition

$$
\vec r_0+\vec v_{\rm rel}t=\vec0
$$

has a physical solution only if both component equations produce the same
nonnegative time. Dividing one component equation by another is unsafe when a
component of relative velocity is zero. A more robust geometric test uses the
two-dimensional scalar cross product: $\vec r_0\times\vec v_{\rm rel}=0$
is required for the two vectors to be parallel. The dot product must also be
negative for a future intercept, since the relative velocity must have a component
toward the origin.

When this parallel condition fails, closest approach still has a definite time. The
squared separation is

$$
R^2(t)=(\vec r_0+\vec v_{\rm rel}t)\cdot
(\vec r_0+\vec v_{\rm rel}t).
$$

Its derivative is zero at

$$
t_{\rm c}=-\frac{\vec r_0\cdot\vec v_{\rm rel}}
{|\vec v_{\rm rel}|^2}.
$$

The closest-approach time is relevant only when nonnegative. If it is negative, the
objects are already separating and their closest approach over the future interval
is their current separation. The formula works without inventing a pursuit force or
assuming either object turns toward the other.

> **Worked example.** A drone lies $\vec r_0=(300,400)\ \mathrm m$ from a fixed receiver and moves at $\vec v_{\rm rel}=(-20,10)\ \mathrm{m\,s^{-1}}$ relative to it. Find the time and distance of closest approach.
>
> With $\vec r_0\cdot\vec v_{\rm rel}=-2000\ \mathrm{m^2\,s^{-1}}$ and $|\vec v_{\rm rel}|^2=500\ \mathrm{m^2\,s^{-2}}$,
> $$
> t_{\rm c}=-\frac{\vec r_0\cdot\vec v_{\rm rel}}{|\vec v_{\rm rel}|^2}=-\frac{-2000}{500}=4.0\ \mathrm s.
> $$
> The relative position is then $(220,440)\ \mathrm m$, of magnitude $492\ \mathrm m$. The cross product $|300(10)-400(-20)|=11\,000\ \mathrm{m^2\,s^{-1}}$ is nonzero, so the paths never intersect. The initially negative range rate means the objects close at first but do not collide.

**Components in a common basis.**

Vector addition requires a single coordinate basis. With east as $x$ and north as
$y$, a boat's ground velocity is

$$
v_{B/G,x}=v_{B/W,x}+v_{W/G,x},
\qquad
v_{B/G,y}=v_{B/W,y}+v_{W/G,y}.
$$

Speeds add directly only for collinear vectors. A current or wind changes the
ground track through vector components, so a travel time and a landing position
require separate calculations.

$$
% caption: Ground velocity is the diagonal of a velocity addition. Through-water
% boat velocity and river current belong to different reference frames; their
% component sum fixes the observed ground track.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (5.8,0) node[right] {east};
  \draw[->, black] (0,0) -- (0,3.8) node[above] {north};
  \draw[->, black, thick] (0,0) -- (2.4,0) node[midway, below] {current};
  \draw[->, acc, thick] (2.4,0) -- (2.4,2.8) node[midway, right] {boat water};
  \draw[->, acc, very thick] (0,0) -- (2.4,2.8) node[above left] {boat ground};
\end{tikzpicture}
$$

> **Worked example.** A $240\ \mathrm m$-wide river flows east at $2.0\ \mathrm{m\,s^{-1}}$; a boat aimed due north moves at $4.0\ \mathrm{m\,s^{-1}}$ relative to the water. Find the crossing time, downstream drift, and ground speed.
>
> Only the north component crosses the river, so
> $$
> t=\frac{240}{4.0}=60\ \mathrm s.
> $$
> The current carries the boat $(2.0)(60)=120\ \mathrm m$ downstream, and the ground speed is $\sqrt{4.0^2+2.0^2}=4.47\ \mathrm{m\,s^{-1}}$. Time, landing point, and ground speed are three separate answers.

## Moving media and frame transformations

Aircraft navigation names three velocities:

- **Airspeed** — velocity of the aircraft relative to the air.
- **Groundspeed** — velocity of the aircraft relative to the ground.
- **Wind** — velocity of the air relative to the ground.

They combine by the frame chain

$$
\vec v_{A/G}=\vec v_{A/{\rm air}}+\vec v_{{\rm air}/G}.
$$

A prescribed ground track fixes the direction of $\vec v_{A/G}$, not the
heading direction of $\vec v_{A/{\rm air}}$. A crosswind requires a heading
offset whose lateral component cancels the drift. The velocity triangle fixes
the required airspeed direction before the ground-track components are evaluated.

$$
% caption: An aircraft heading is offset into the crosswind so that its lateral
% airspeed component cancels wind. The resulting ground-velocity arrow is aligned
% with the desired northward track; its length determines groundspeed and travel time.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (4.4,0) node[right] {east};
  \draw[->, black] (0,0) -- (0,4.0) node[above] {north};
  \draw[->, acc, very thick] (2.5,0) -- (2.5,3.3) node[midway, right] {ground track};
  \draw[->, acc, thick] (2.5,0) -- (0.9,3.3) node[midway, left] {heading};
  \draw[->, black, thick] (0.9,3.3) -- (2.5,3.3) node[midway, above] {wind};
\end{tikzpicture}
$$

> **Worked example.** An aircraft with airspeed $180\ \mathrm{km\,h^{-1}}$ meets an eastward wind of $50\ \mathrm{km\,h^{-1}}$. Find the heading and groundspeed for a due-north ground track.
>
> The heading must supply a westward airspeed component of $50\ \mathrm{km\,h^{-1}}$ to cancel the wind, leaving a northward groundspeed
> $$
> v_G=\sqrt{180^2-50^2}=173\ \mathrm{km\,h^{-1}}.
> $$
> The heading offset is $\sin^{-1}(50/180)=16.1^\circ$ west of north.

**Galilean frame transformation.**

Frames moving at constant relative velocity use the coordinate transformation

$$
\vec r'=\vec r-\vec Vt,
\qquad \vec v'=\vec v-\vec V,
\qquad \vec a'=\vec a.
$$

The same event has different position and velocity coordinates in the two frames,
but acceleration is identical. This is the Newtonian form of the relativity
principle for inertial frames.

The constant vertical separation in velocity is the frame-speed difference; equal
slopes give the common acceleration.

$$
% caption: A passenger walks forward inside a train while the train moves over
% ground. The passenger-ground velocity is the sum of passenger-train and train-
% ground arrows; switching to the train frame subtracts the train velocity.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, acc, thick] (1.0,2.5) -- (2.15,2.5) node[midway, above] {walk};
  \draw[->, black, thick] (2.15,2.5) -- (3.9,2.5) node[midway, above] {train};
  \draw[->, acc, very thick] (1.0,1.7) -- (3.9,1.7) node[midway, below, align=center] {passenger\\over ground};
  \draw[black, dotted] (1.0,1.6) -- (1.0,2.6);
  \draw[black, dotted] (3.9,1.6) -- (3.9,2.6);
  \draw[black, thick] (0.5,-0.15) rectangle (4.4,0.7);
  \node at (2.45,0.275) {train frame};
\end{tikzpicture}
$$

## Pursuit, interception, and range rate

For two particles,

$$
\vec r_{A/B}=\vec r_A-\vec r_B,
\qquad
\vec v_{A/B}=\vec v_A-\vec v_B.
$$

With constant relative velocity, interception requires

$$
\vec r_{A/B}(0)+\vec v_{A/B}t=\vec0.
$$

The $x$ and $y$ components must produce the same positive time.

**Rotating-frame boundary.**

The Galilean transformation applies only between frames with constant relative
velocity and parallel fixed axes. A rotating platform has axes whose directions
change with time. Velocities measured there include additional rotation-dependent
terms, so the simple subtraction $\vec v'=\vec v-\vec V$ is incomplete.

**Range rate.**

Radar and navigation systems often measure range $R=|\vec r_{A/B}|$ and its
time derivative rather than full relative velocity. Differentiating the magnitude
gives radial range rate,

$$
\dot R=\hat R\cdot\vec v_{A/B}.
$$

Only the component of relative velocity along the line of sight changes range.
A transverse relative velocity can be large while instantaneous range rate is zero.

$$
% caption: Relative velocity is resolved into line-of-sight and transverse parts.
% The radial component changes radar range, whereas the perpendicular component
% changes bearing but gives zero instantaneous range rate.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[fill=white, draw=black, thick] (0,0) circle (2pt) node[below] {observer};
  \draw[fill=acc!12, draw=acc, thick] (4.2,2.1) circle (2pt) node[above right,xshift=2pt] {object};
  \draw[acc, thick] (0,0) -- (4.2,2.1) node[midway,below] {range};
  \draw[->, black, thick] (4.2,2.1) -- (2.8,1.4) node[midway,below] {radial};
  \draw[->, acc, thick] (4.2,2.1) -- (3.5,3.5) node[right] {cross};
\end{tikzpicture}
$$

## Relative acceleration and time histories

Relative acceleration follows from component subtraction:

$$
\vec a_{A/B}=\vec a_A-\vec a_B.
$$

It describes curvature of one trajectory as viewed from the other. Two vehicles
with equal acceleration have zero relative acceleration even while their individual
velocities change. A frame attached to an accelerating vehicle is not inertial,
although its relative-position coordinates can still be used with the appropriate
inertial-force correction.

**Component trace.**

Each component calculation carries its frame label.

| Quantity | East component | North component | Meaning |
| --- | ---: | ---: | --- |
| boat relative water | $-2.0$ | $3.46$ | chosen heading |
| water relative ground | $+2.0$ | $0$ | current |
| boat relative ground | $0$ | $3.46$ | direct crossing |

The direct-crossing geometry and component construction represent the same
frame-labelled sum.

The component diagram resolves the eastward cancellation shown by the river track.

**Frame labels and closure checks.**

Relative-motion notation prevents a common category error: subtracting a position
from a velocity or combining two velocities that refer to incompatible pairs of
objects. The ordered label $P/A$ means “particle P measured from frame A.” In the
sum $\vec v_{P/B}=\vec v_{P/A}+\vec v_{A/B}$, the intermediate label A
closes: the first arrow ends at A and the second begins at A. The result connects P
to B. This is the same cancellation rule used when adding directed displacements.

Reversing a relation changes its sign. Thus $\vec v_{A/B}=-\vec v_{B/A}$,
and a stationary passenger in a train has $\vec v_{P/T}=\vec0$ even though
$\vec v_{P/G}=\vec v_{T/G}$ in the ground frame. Zero velocity is never
complete without the reference object or frame. A boat “at rest” relative to water
can drift rapidly relative to shore; a spacecraft “at rest” relative to a docking
port can move quickly relative to a planet.

The same physical event is described at one shared Newtonian time in all Galilean
frames. Positions at that event differ because their origins differ. Comparing the
position of a train passenger at one ground-clock reading with the position of that
same passenger at a different train-clock reading is not a transformation; it is a
comparison of two different events.

> **Worked example.** A train moves east at $18\ \mathrm{m\,s^{-1}}$ relative to the platform. A
> passenger walks east along the carriage at $1.4\ \mathrm{m\,s^{-1}}$ relative to
> the train. The label-closing sum gives
>
> $$
> v_{P/G}=v_{P/T}+v_{T/G}=1.4+18=19.4\ \mathrm{m\,s^{-1}}.
> $$
>
> A passenger walking west at $1.4\ \mathrm{m\,s^{-1}}$ has ground-frame velocity
> $16.6\ \mathrm{m\,s^{-1}}$ east, rather than $-1.4\ \mathrm{m\,s^{-1}}$ east. The latter
> is the velocity relative to the train. Stating “west” without the reference frame
> would make the two descriptions appear contradictory when they are not.

**Galilean transformations in components.**

Let frame $S'$ move with constant velocity $\vec V=(V_x,V_y)$ relative to S,
with parallel axes and coincident origins at $t=0$. The transformation is applied
component by component:

$$
x'=x-V_xt,\qquad y'=y-V_yt,
\qquad v_x'=v_x-V_x,\qquad v_y'=v_y-V_y.
$$

The inverse transformation adds the same frame velocity. This symmetry provides a
quick sign check: transforming from ground to train subtracts train-over-ground;
transforming back from train to ground adds it. A moving observer can describe an
object as moving west while a ground observer describes it as moving east. Both use
the same acceleration when their frames have constant relative velocity.

> **Worked example.** A drone has ground-frame velocity $(12,5)\ \mathrm{m\,s^{-1}}$. A truck-mounted
> observer moves at $(8,0)\ \mathrm{m\,s^{-1}}$ relative to ground. The drone
> velocity measured from the truck is
>
> $$
> \vec v_{D/T}=(12,5)-(8,0)=(4,5)\ \mathrm{m\,s^{-1}}.
> $$
>
> Its speed in the truck frame is $\sqrt{4^2+5^2}=6.4\ \mathrm{m\,s^{-1}}$, whereas
> its ground-frame speed is $13\ \mathrm{m\,s^{-1}}$. The north component is
> unchanged because the frame translation has no north component. Neither speed is
> more “real”; each is a measurement relative to a different inertial observer.

**Relative position and separation histories.**

The relative-position vector $\vec r_{A/B}=\vec r_A-\vec r_B$ points
from B to A. Its magnitude is separation, but the vector also retains bearing. A
constant relative velocity produces a straight relative-position trajectory. The
origin of this relative coordinate system is attached to B, so B remains fixed at
zero and A appears to move with $\vec v_{A/B}$. This change of description can
turn two moving ground-frame trajectories into one moving point and one fixed
origin.

Separation alone is not enough to determine an intercept. The range may decrease
at first and then increase if the relative-velocity vector does not point exactly
toward the origin. A genuine intercept requires the full vector equation
$\vec r_{A/B}(t)=\vec0$. In two dimensions, the starting separation and
relative velocity must be parallel and oppositely directed for a constant-velocity
collision to occur. A nonzero sideways component produces a closest approach rather
than a meeting.

**Fixed-speed pursuit and feasibility.**

An interceptor that can select its heading but has a fixed speed relative to a
medium faces a different problem from passive relative motion. Let the target have
known ground velocity $\vec v_T$, and let the interceptor have fixed ground-speed
magnitude $u$. An intercept at time $t$ requires its constant ground velocity to be

$$
\vec v_I=\frac{\vec r_T(0)-\vec r_I(0)}{t}+\vec v_T.
$$

The heading is feasible only if $|\vec v_I|=u$. Solving that magnitude condition
for positive $t$ finds possible intercept times. A solution can fail to exist when
the interceptor is too slow in the relevant direction. Being faster in speed alone
is not always enough when a target begins far ahead and the allowed time or region
is restricted.

The geometry can be viewed in velocity space. For a selected time, the displacement
needed to reach the target fixes a required average ground velocity. As time grows,
the displacement-per-time contribution shrinks and the required velocity approaches
the target velocity. Very short intercept times demand very large speed; very long
times may be physically possible but tactically irrelevant. The mathematical root
must be checked against any specified time window and against the assumption that
both velocities remain constant.

> **Worked example.** A rescue craft starts $600\ \mathrm m$ south of a buoy drifting east at $2.0\ \mathrm{m\,s^{-1}}$; the craft holds a ground speed of $10\ \mathrm{m\,s^{-1}}$. Find the intercept time and heading.
>
> With the craft at the origin, an intercept at time $t$ needs required velocity $(2.0,600/t)\ \mathrm{m\,s^{-1}}$, whose magnitude equals the fixed ground speed:
> $$
> 2.0^2+\left(\frac{600}{t}\right)^2=10^2.
> $$
> The positive root is $t=61.2\ \mathrm s$, giving heading components $(2.0,9.80)\ \mathrm{m\,s^{-1}}$, or $11.5^\circ$ east of north. The east component matches the buoy's drift rather than adding a detour.

**Relative acceleration and noninertial observers.**

Relative acceleration is always obtained by subtracting accelerations measured in a
common inertial frame: $\vec a_{A/B}=\vec a_A-\vec a_B$. Equal
accelerations give zero relative acceleration, so separation can change linearly
even while both objects accelerate in the ground frame. The same conclusion covers
two cars braking equally and two objects dropped in the same local gravitational
field.

An observer attached to one of those accelerating objects is not an inertial frame.
The relative-coordinate equations still describe geometry, but Newton's second law
in that accelerating observer's frame requires an additional inertial-force term.
The ordinary Galilean statement that all observers agree on acceleration applies
only to frames moving at constant relative velocity. A rotating observer adds still
more terms because its axes change direction as well as its origin's motion.

> **Worked example.** Two vehicles on a straight road start $30\ \mathrm m$ apart with relative velocity $-4.0\ \mathrm{m\,s^{-1}}$ and brake at the same rate. When do they meet?
>
> Equal braking gives zero relative acceleration, so the separation stays linear:
> $$
> s(t)=30-4.0t\ \mathrm{m}.
> $$
> It reaches zero at $t=7.5\ \mathrm s$, provided the braking stage lasts that long. The shared deceleration cancels from the closing calculation.

**Range, bearing, and collision assessment.**

Range rate measures only the radial part of relative velocity. A negative range
rate means the distance between two objects is decreasing at that instant; a positive
range rate means it is increasing. Neither quantity alone determines collision risk.
An object can have a strongly negative range rate and still pass at a safe lateral
distance, while an object with zero range rate may be at closest approach before
moving away. The relative-position vector and the transverse component of relative
velocity supply the missing geometric information.

Bearing is the direction of the relative-position vector. In planar motion, a
constant bearing together with decreasing range is a warning sign: the relative
velocity is directed along the line of sight, so the relative-position vector can
reach zero. A changing bearing usually signals a transverse component and hence a
miss, though a changing reference orientation or noisy angle measurement must not
be mistaken for physical transverse motion. Navigation systems often use both range
and bearing histories because neither record alone identifies the complete relative
velocity vector.

The radial-transverse split is found by projecting relative velocity onto the unit
line-of-sight vector. The radial projection changes range; the remainder changes
bearing. At closest approach the radial projection is zero, so relative velocity is
perpendicular to the separation vector. This perpendicular condition is the
geometric version of setting the derivative of squared range to zero.

**Bearing-based interpretation.**

A ship observes another vessel at a bearing that remains fixed while range falls
from $4.0\ \mathrm{km}$ to $2.0\ \mathrm{km}$ over ten minutes. The average radial
range rate is $-0.20\ \mathrm{km\,min^{-1}}$. Under the stated constant-bearing
and constant-relative-velocity model, the relative path lies along the line of
sight and collision would follow ten minutes later if neither vessel changes course.
If the bearing changes by several degrees over the same interval, the conclusion is
no longer valid without a full relative-vector calculation.

## Solution methods and coordinate changes

Start by naming the objects and selecting one inertial frame as the common basis.
Write every supplied velocity with its two labels, then translate all vectors into
components in that same basis. A diagram may suggest the direction of a result, but
component equations determine its signs and allow a numerical check. Keep speed
separate from velocity: a speed is a magnitude and cannot be inserted into a vector
sum until a direction or component decomposition is supplied.

For crossing and wind problems, identify which velocity is constrained by the
question. A specified heading constrains the vehicle-medium velocity; a specified
ground track constrains the vehicle-ground velocity. For interception, form relative
position and relative velocity before solving time. For a frame conversion, state
the frame velocity and use subtraction in the requested direction. These choices
prevent an otherwise correct triangle from answering a different physical question.

After solving, test limiting cases. With zero current or wind, the ground and
medium velocities should coincide. With two objects sharing the same velocity,
relative velocity should vanish. With equal accelerations, relative acceleration
should vanish. A proposed direct river crossing becomes impossible if current speed
exceeds the available through-water speed, because no heading can supply a lateral
component large enough to cancel the current. Such checks expose sign mistakes and
impossible geometric assumptions before numerical rounding hides them.

**Reporting an intercept result.**

An intercept answer needs more than a time. A complete statement identifies the
reference frame, the intercept location or required heading when relevant, and the
model interval over which velocities were assumed constant. “Interception occurs
after $45\ \mathrm s$” is incomplete if the target is moving through a current or if
the pursuer has a finite speed constraint. “In the ground frame, the craft reaches
the target after $45\ \mathrm s$ by holding the stated heading” makes the relation
between the calculation and the physical claim reproducible.

**Common failure modes.**

Three mistakes recur in relative-motion work:

- **Adding speeds, not vectors** — combining two speeds rather than two labeled velocity vectors loses direction and frame information.
- **Wrong reference medium** — using a ground speed where the problem supplied a speed relative to air or water changes the physical triangle before any calculation begins.
- **Accepting a negative time** — a negative intercept time ignores the direction of relative motion; it usually places the intersection in the past.

A brief label audit and a sign check on the final time prevent all three.

Relative coordinates also clarify apparently paradoxical statements. Two objects
can move rapidly in a ground frame while remaining fixed relative to each other.
Conversely, an object stationary in one frame can move in every other frame whose
origin has nonzero relative velocity. The frame label resolves the apparent
contradiction without changing any physical event.

**Relative acceleration as a vector time history.**

Relative acceleration describes how one object's velocity changes as seen from the
other. In a common inertial frame,

$$
\vec a_{A/B}(t)=\vec a_A(t)-\vec a_B(t).
$$

The equation remains valid when either acceleration changes with time. Integrating
it over an interval gives the change in relative velocity, and integrating again
gives the change in relative position. Initial relative position and initial
relative velocity are still required: acceleration history alone determines neither
the initial separation nor the current closing speed.

Constant relative acceleration gives the relative trajectory the same form as
ordinary constant-acceleration motion,

$$
\vec r_{A/B}(t)=\vec r_0+\vec v_0t+
\frac12\vec a_{A/B}t^2.
$$

It gives the difference between two independently measured motions. A chaser can
have a larger ground-frame acceleration than a target but
still move away initially if its relative velocity points outward. The acceleration
controls curvature of the relative path; the initial relative velocity controls its
initial tangent.

When acceleration is supplied as a sampled vector history, componentwise updates
are safer than using changing magnitudes. Over a short interval, add
$\vec a_{A/B}\Delta t$ to relative velocity, then use the appropriate average
relative velocity to update relative position. A changing direction of acceleration
can rotate the relative velocity even when its magnitude is nearly constant. A
single scalar “closing acceleration” is adequate only when all relevant vectors are
collinear with the line of sight.

> **Worked example.** At $t=0$, object A is at $(80,20)\ \mathrm m$ relative to B and has relative
> velocity $(-6,4)\ \mathrm{m\,s^{-1}}$. During the next $3.0\ \mathrm s$, its
> relative acceleration is $(-1,2)\ \mathrm{m\,s^{-2}}$. The ending relative
> velocity is
>
> $$
> \vec v_{A/B}(3)=(-6,4)+(-1,2)(3)=(-9,10)\ \mathrm{m\,s^{-1}}.
> $$
>
> The relative-position change is not the ending velocity multiplied by time. For
> constant acceleration it is the initial velocity term plus the acceleration term,
>
> $$
> \Delta\vec r=(-6,4)(3)+\frac12(-1,2)(3)^2=(-22.5,21)\ \mathrm m.
> $$
>
> Thus the new relative position is $(57.5,41)\ \mathrm m$. The northward separation
> has increased even though the eastward separation has decreased. A scalar distance
> calculation would hide this mixed component behavior.

**Changing origins without changing the events.**

A frame transformation has two parts: a translation of the origin and, possibly, a
relative velocity of the origins. If the primed origin has ground-frame position
$\vec R(t)$, then coordinates of the same event obey

$$
\vec r'=\vec r-\vec R(t).
$$

Differentiation gives $\vec v'=\vec v-\dot{\vec R}$ and
$\vec a'=\vec a-\ddot{\vec R}$. The familiar Galilean form is the special
case $\vec R(t)=\vec R_0+\vec Vt$, for which $\ddot{\vec R}=0$.
Writing the origin trajectory explicitly makes clear why a frame attached to an
accelerating car does not share the ground-frame acceleration values.

At constant frame velocity, the subtracted displacement is $\vec Vt$ over the
same elapsed time.

An origin shift at one fixed time changes every displayed position by the same
vector. It does not alter separations: for two objects A and B,

$$
(\vec r_A-\vec R)-(\vec r_B-\vec R)=\vec r_A-\vec r_B.
$$

The cancellation leaves relative position independent of the arbitrary coordinate
origin. Two observers can therefore disagree about each object's map coordinate
while agreeing exactly on the displacement from B to A. The same cancellation holds
for a common constant-velocity frame translation in relative velocity, but fails if
one observer mistakenly subtracts a velocity measured in a different direction or
at a different time.

The same origin shift applies to positions recorded at each common clock time; it
does not pair positions from different events.

> **Worked example.** At $t=4.0\ \mathrm s$, a delivery robot has ground position
> $(30,12)\ \mathrm m$. A van-mounted origin began at $(4,0)\ \mathrm m$ and moves
> at $(5,1)\ \mathrm{m\,s^{-1}}$. Its origin position at the same event is
>
> $$
> \vec R=(4,0)+(5,1)(4)=(24,4)\ \mathrm m.
> $$
>
> The robot coordinate in the van frame is $(6,8)\ \mathrm m$. If the robot ground
> velocity is $(7,3)\ \mathrm{m\,s^{-1}}$, the van-frame velocity is $(2,2)\
> \mathrm{m\,s^{-1}}$. The calculation uses a common time for robot and origin;
> subtracting the van position from a different time would produce a coordinate
> difference with no physical event attached to it.

**Vector time histories and integration order.**

Velocity and acceleration histories are vector functions, not sequences of unrelated
arrows. A velocity arrow at a later time begins at the same velocity-space origin as
an earlier arrow; its difference is the accumulated acceleration area. A position
arrow at a later time differs from an earlier position arrow by the accumulated
velocity area. Drawing these histories tail-to-tail prevents the common mistake of
connecting successive velocity tips as though they were locations in physical space.

A piecewise-constant relative-acceleration table should record the interval, relative
acceleration, beginning relative velocity, ending relative velocity, and relative
displacement. The ending values carry forward. This procedure also records
when an assumed model changes: a turn, engine burn, or control input belongs at an
interval boundary rather than being smeared across the entire history.

## Measurement, uncertainty, and feasibility

Closest approach is a relative-motion calculation with a geometric condition rather
than an assumption that the objects meet. Place object B at the origin of the
relative coordinate system at the initial time. The position of A is then
$\vec r_0$, and its straight-line relative path is

$$
\vec r(t)=\vec r_0+\vec v_{\rm rel}t.
$$

At the minimum separation, the vector from B to A is perpendicular to the
relative-velocity vector. Otherwise a component of velocity would still be reducing
or increasing the separation. The condition is

$$
\vec r(t_{\rm c})\cdot\vec v_{\rm rel}=0,
$$

which yields the closest-approach time already obtained by differentiating squared
range. Squared range is preferred over range itself because it avoids a square root
and retains the same minimum for nonnegative distance.

The calculation must distinguish three outcomes. A positive closest-approach time
and zero closest distance gives an intercept. A positive time and nonzero distance
gives a future miss. A negative closest-approach time means the mathematical
minimum occurred before the selected initial time; from now on, the objects separate
under the constant-velocity model. These statements are about the relative path,
not about which object is considered the pursuer.

$$
% caption: The full closest-approach construction uses a relative-position vector
% from B to A and a constant relative-velocity direction. At the marked point the
% separation is perpendicular to the path, so its dot product with relative velocity
% vanishes; this yields the future miss distance without assuming an intercept.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0.5,2.4) -- (5.4,2.4);
  \node[black, anchor=west] at (5.4,2.4) {path of A};
  \draw[fill=white, draw=black, thick] (2.4,0) circle (2pt) node[below] {B};
  \draw[->, acc, very thick] (2.4,0) -- (4.6,2.4);
  \node[acc, anchor=north west] at (3.5,1.15) {separation};
  \draw[fill=acc!14, draw=acc, thick] (4.6,2.4) circle (1.8pt) node[above right] {A};
  \draw[->, black, thick] (4.4,2.4) -- (3.1,2.4) node[midway, above] {rel. velocity};
  \draw[black, dashed] (2.4,0) -- (2.4,2.4);
  \node[black, anchor=east] at (2.35,1.2) {miss};
  \draw[fill=acc!14, draw=acc, thick] (2.4,2.4) circle (1.8pt);
  \node[acc, anchor=south east] at (2.3,2.5) {closest};
  \draw[black] (2.25,2.4) -- (2.25,2.25) -- (2.4,2.25);
\end{tikzpicture}
$$

> **Worked example.** At the initial observation, aircraft A is $\vec r_0=(8.0,6.0)\ \mathrm{km}$
> relative to aircraft B. Their ground velocities are
>
> $$
> \vec v_A=(0.20,-0.10)\ \mathrm{km\,min^{-1}},
> \qquad
> \vec v_B=(-0.05,0.15)\ \mathrm{km\,min^{-1}}.
> $$
>
> The relative velocity of A with respect to B is therefore
>
> $$
> \vec v_{A/B}=\vec v_A-\vec v_B=(0.25,-0.25)
> \ \mathrm{km\,min^{-1}}.
> $$
>
> The initial dot product is
>
> $$
> \vec r_0\cdot\vec v_{A/B}=8.0(0.25)+6.0(-0.25)
> =0.50\ \mathrm{km^2\,min^{-1}},
> $$
>
> and the relative-speed square is
>
> $$
> |\vec v_{A/B}|^2=0.25^2+(-0.25)^2=0.125
> \ \mathrm{km^2\,min^{-2}}.
> $$
>
> The closest-approach time is $t_{\rm c}=-0.50/0.125=-4.0\ \mathrm{min}$. It is
> negative. The two aircraft were closest four minutes before the observation; no
> future closest point along the unaltered straight-line paths exists. The appropriate
> future minimum is the present separation,
>
> $$
> R(0)=\sqrt{8.0^2+6.0^2}=10\ \mathrm{km}.
> $$
>
> Changing only the sign of the relative velocity produces a different physical
> case. If $\vec v_{A/B}=(-0.25,0.25)\ \mathrm{km\,min^{-1}}$, then the dot
> product is $-0.50\ \mathrm{km^2\,min^{-1}}$ and $t_{\rm c}=+4.0\ \mathrm{min}$.
> The relative position at that time is
>
> $$
> (8.0,6.0)+(-0.25,0.25)(4.0)=(7.0,7.0)\ \mathrm{km}.
> $$
>
> The closest distance is $\sqrt{7.0^2+7.0^2}=9.90\ \mathrm{km}$, so this is a
> future miss rather than a collision. The result can be checked geometrically:
> $(7.0,7.0)\cdot(-0.25,0.25)=0$. The perpendicular dot-product check is valuable
> because it detects a sign error in the time calculation without redoing the entire
> derivative.
>
> The closest-approach result has separate geometric and temporal tests. Reporting
> both prevents a positive closest-time calculation from being mistaken for an
> intercept condition.
>
> | Quantity | Relation | Interpretation |
> | --- | --- | --- |
> | Candidate time | $t_*=-\vec r_0\cdot\vec v_{\rm rel}/v_{\rm rel}^2$ | closest approach only when $t_*>0$ |
> | Miss vector | $\vec r(t_*)=\vec r_0+\vec v_{\rm rel}t_*$ | zero vector is required for collision |
> | Geometric closure | $\vec r(t_*)\cdot\vec v_{\rm rel}=0$ | closest separation is perpendicular to the path |

**Model limits in encounter prediction.**

Closest-approach predictions assume that both ground velocities remain constant over
the computed interval. A turn, a wind change, a control response, or a different
altitude invalidates the straight relative path after the change. The calculation is
still applicable as a short-horizon prediction, but its conclusion should be reported
with the time window and the velocity source. It predicts geometric proximity, not
intent, communication, or collision avoidance behavior.

**Sensitivity and uncertainty in relative predictions.**

Relative calculations combine measurements from two objects, so their uncertainty
often exceeds the uncertainty of either individual measurement. A small error in
each velocity component changes the relative velocity by their difference. Over a
long prediction time, that velocity error becomes a growing relative-position error.
For constant relative velocity, an uncertainty $\delta\vec v$ contributes roughly
$t\,\delta\vec v$ to the predicted separation after time $t$. A reported
intercept time should therefore not carry more precision than the position and
velocity measurements allow.

The closest-approach time is especially sensitive when the dot product
$\vec r_0\cdot\vec v_{\rm rel}$ is small. In that situation the objects are
already close to perpendicular relative motion, and a small change in a measured
velocity component can shift the predicted closest time from slightly future to
slightly past. The minimum separation itself can be more robust than the time, or
the reverse, depending on the geometry. Reporting both the input time interval and
the measurement precision gives the calculation its proper scope.

Sensitivity can be recorded by separating the inputs that perturb event time from
those that perturb miss distance. The same speed uncertainty does not affect both
outputs equally for every encounter geometry.

| Input perturbation | Primary effect | Diagnostic |
| --- | --- | --- |
| along-track velocity error | shifts $t_*$ | repeat range measurement over a longer baseline |
| transverse velocity error | changes minimum separation | bearing history or second sensor direction |
| common clock offset | biases both objects' state times | synchronized timestamp check |
| position-origin error | shifts $\vec r_0$ | independent reference mark |

Relative velocity can also be inferred from repeated range-and-bearing observations,
but numerical differentiation amplifies angle and range noise. Fitting a straight
relative trajectory to several observations is often more reliable than subtracting
two nearly equal position readings from a single pair of frames. The fitted model
should still be checked against residuals: systematic curvature indicates turning,
changing wind, or a noninertial reference issue rather than random measurement
scatter.

> **Worked example.** If a relative-speed estimate is $0.30\pm0.02\ \mathrm{km\,min^{-1}}$ and the
> forecast interval is $20\ \mathrm{min}$, the velocity uncertainty alone corresponds
> to roughly $0.4\ \mathrm{km}$ of along-track position uncertainty. Reporting a
> predicted miss distance of $0.37\ \mathrm{km}$ without that qualification would
> suggest a false level of certainty. The appropriate conclusion is that the model
> predicts a close approach within the stated measurement resolution, followed by a
> request for updated observations or a more complete motion model.

**Assumption audit.**

Every relative-motion result rests on an observation time, a coordinate convention,
and a model for how velocities evolve afterward. Verify that both velocity vectors
were measured in the same frame, their timestamps were synchronized, and constant
velocity or constant relative acceleration remained credible over the forecast
interval. A numerical answer can be algebraically correct and physically unusable
when any of those conditions is missing. Naming the reference frame and forecast
horizon turns a vector result into a testable statement about a real encounter.

The encounter record should expose the calculation in the same order as the physical
inference. A scalar range or speed is insufficient when the transverse component
controls the miss geometry.

| Record | Relation | Check against observation |
| --- | --- | --- |
| Initial separation | $\vec r_{A/B}(t_0)$ | common timestamp and coordinate basis |
| Relative velocity | $\vec v_{A/B}=\vec v_A-\vec v_B$ | frame labels on both input vectors |
| Closest approach | $t_*=-\vec r_0\cdot\vec v_{\rm rel}/v_{\rm rel}^2$ | $t_*$ inside the forecast interval |
| Predicted miss | $\vec r(t_*)$ | updated range-and-bearing residuals |

[^tipler31]: Tipler and Mosca, _Physics for Scientists and Engineers_, 6th ed., §3-1.

**Derivatives of relative coordinates.**

Relative coordinates form an ordinary vector-valued function of time. Once
$\vec r_{A/B}(t)=\vec r_A(t)-\vec r_B(t)$ has been defined at common
times, differentiation produces relative velocity and relative acceleration without
any new kinematic rule. The important condition is synchronization: positions from
different time stamps cannot be subtracted to estimate an instantaneous separation,
and velocities taken from different instants cannot be subtracted to estimate an
instantaneous relative velocity.

Repeated differentiation clarifies what each graph or sensor record can support.
Position differences yield relative position. Their time derivative yields relative
velocity. A further derivative yields relative acceleration. Each derivative removes
one constant of integration, so the reverse process needs an initial condition at
the appropriate level. Starting from relative acceleration and integrating twice
without an initial separation and relative velocity produces a family of possible
encounter paths, not one prediction.

The magnitude of relative position, $R=|\vec r|$, has a derivative that depends
on direction as well as magnitude. The range rate is the projection of relative
velocity onto the current line of sight. A large transverse velocity leaves range
unchanged at one instant but rotates the line of sight. In planar motion, the rate
of change of bearing is proportional to the transverse component divided by range.
Thus the same sideways speed produces a faster bearing sweep when the objects are
near than when they are far apart.

Tracking requires this derivative structure. A radar that reports range but not
bearing cannot recover a full Cartesian relative-velocity vector from one instant.
Conversely, a camera that reports bearing but no range cannot distinguish a nearby
slow target from a distant fast one using a single frame. Multiple time samples and
a stated motion model turn these partial observations into a relative trajectory.

> **Worked example.** At one instant, an object is $2.0\ \mathrm{km}$ east of an observer and moves
> $(0.30,0.40)\ \mathrm{km\,min^{-1}}$ relative to the observer. The radial direction
> is east, so the range rate is $+0.30\ \mathrm{km\,min^{-1}}$. The north component
> is transverse. It changes bearing but does not contribute to the instantaneous range
> change. Replacing the relative velocity by its speed, $0.50\ \mathrm{km\,min^{-1}}$,
> would incorrectly treat the whole motion as radial.

**Feasibility before solving an intercept time.**

An algebraic intercept time is meaningful only after the geometry and motion
constraints have been checked. For two constant-velocity objects, the relative
position must be driven to zero by a single scalar time. In two dimensions, both
component equations must agree. Different east and north component solutions describe
different hypothetical meeting events. No common intercept time exists.

Fixed-speed interception adds a second constraint. The required interceptor velocity
must have the allowed magnitude and must satisfy any heading, acceleration, or
operating-region limits. A boat cannot exceed its through-water speed, and an
aircraft cannot turn instantaneously under the stated model. Such a solution is
infeasible under the constraints; report that outcome instead of a rounded time with
an impossible velocity attached.

The sign of time is a separate feasibility test. A negative root identifies an
intersection of the straight paths before the chosen initial observation. It may be
appropriate for reconstructing a past encounter, but it does not predict a future
intercept. A zero root means the objects are already coincident at the selected
initial time. Positive roots should still be checked against the interval over which
the velocities are credible.

When relative acceleration is present, the relative-position equation can yield more
than one positive root. The objects may meet, separate, and meet again after a turn
or reversal. Each root must be tested against the piecewise motion stage that
produced it. A quadratic solver cannot decide whether a motor was still running or
a braking stage had already ended.

> **Worked example.** At $t=0$, target T is at $(120,80)\ \mathrm m$ relative to pursuer P. Their
> constant velocities are $(2,1)\ \mathrm{m\,s^{-1}}$ and
> $(5,0)\ \mathrm{m\,s^{-1}}$, respectively. The target relative to pursuer has
> velocity $(-3,1)\ \mathrm{m\,s^{-1}}$. The east equation would give $t=40\ \mathrm
> s$, while the north equation would require $t=-80\ \mathrm s$. The paths do not
> intersect in the future. The decreasing east separation does not compensate for the
> increasing north separation.

**Radar observations and Cartesian relative data.**

Radar observations are naturally polar: range $R$ and bearing $\theta$. Cartesian
relative coordinates are obtained only after a direction convention is stated. With
east as the positive horizontal axis and bearing measured counterclockwise from east,

$$
x=R\cos\theta,\qquad y=R\sin\theta.
$$

Navigation displays often measure bearing clockwise from north instead. Using the
same sine and cosine pair without converting the angle rotates or reflects the
relative position. The safest method is to sketch the axes, identify the component
adjacent to the stated bearing, and test a cardinal direction: a bearing due north
must produce zero east component and positive north component in the chosen basis.

Successive polar observations also require care during differentiation. A changing
range and a changing bearing both contribute to Cartesian relative velocity. The
radial term points along the line of sight, while the angular term is perpendicular
to it. A tracker that reports a constant range and rapidly changing bearing is not
observing a stationary target; it is observing predominantly transverse motion.
Likewise, a steady bearing with decreasing range is consistent with radial closing,
but only the full vector history establishes whether the track stays exactly radial.

Cartesian conversion puts measurements from separate sensors into one component system.
Once all observations use the same origin, time base, and axes, position differences
and fitted velocity components can be compared directly. It also exposes a frequent
mistake in hand sketches: range is a scalar length, while the Cartesian pair retains
the direction needed for addition and subtraction.

> **Worked example.** A radar reports range $R=5.0\ \mathrm{km}$ at a bearing of $53.1^\circ$ north of
> east. The relative Cartesian position is $(3.0,4.0)\ \mathrm{km}$. A second report
> one minute later gives $(3.2,3.6)\ \mathrm{km}$ after the same conversion. The
> estimated relative velocity is $(0.20,-0.40)\ \mathrm{km\,min^{-1}}$. Taking a
> simple difference of the two range values would retain only radial change and would
> miss the west-east and north-south components that determine the path.

**Uncertainty propagation in relative predictions.**

Uncertainty in relative position comes from both position measurements and from the
coordinate transformation used to compare them. When independent Cartesian position
components have comparable random uncertainty, subtracting two positions increases
the uncertainty in each relative component by a root-sum-square combination. A
small uncertainty in bearing can dominate the transverse coordinate at long range,
because an angular error corresponds to a sideways distance approximately equal to
range times the angle error in radians.

Velocity uncertainty grows when it is inferred from short time intervals. Dividing a
position difference by a smaller interval improves temporal resolution but magnifies
the same position noise in the velocity estimate. For encounter forecasting, a
fitted relative trajectory through several measurements often gives a more stable
velocity estimate than one pair of observations, provided the residuals support the
constant-velocity model.

Prediction uncertainty expands with forecast time. A component error that is modest
at the current observation can shift a predicted intercept or closest approach by a
large distance later. The report should distinguish uncertainty in range, uncertainty
in bearing, uncertainty in inferred velocity, and uncertainty caused by model change.
Combining them into a single unexplained error bar hides which new measurement would
most improve the forecast.

> **Worked example.** At a range of $12\ \mathrm{km}$, a bearing uncertainty of $1.0^\circ$ corresponds
> to a transverse position uncertainty of about $12(\pi/180)=0.21\ \mathrm{km}$.
> If the inferred transverse relative speed has an additional uncertainty of
> $0.03\ \mathrm{km\,min^{-1}}$, a ten-minute extrapolation adds roughly
> $0.30\ \mathrm{km}$ of transverse uncertainty. A predicted miss distance of
> $0.10\ \mathrm{km}$ is therefore not resolved by these observations. The defensible
> statement is that the current data cannot distinguish a very close pass from an
> intercept under the assumed straight-line model.

**Measurement-priority decision.**

The next measurement should target the component that limits the decision. When
bearing uncertainty dominates a long-range forecast, a better angular observation is
more valuable than another precise range reading. When a predicted closest approach
is highly sensitive to relative speed, a longer time baseline between synchronized
position observations can improve the velocity estimate. The relative-coordinate
model therefore identifies the current prediction and the observation most likely to
reduce its uncertainty.