Relative Motion — Relate relative motion between points/frames in 2D (intro) - Statics and Dynamics

SECTION 1

Historical Context & Motivation

The notion that motion depends on who is watching is far older than any formal dynamics course. From the earliest astronomical debates to modern vehicle-crash analysis, the ability to translate one observer's measurements into another's has driven some of the most consequential advances in mechanics. In engineering practice, relative motion analysis underpins everything from linkage design in mechanisms to the guidance of satellites rendezvousing in orbit. Understanding the historical arc of this idea clarifies why the vector-addition framework taught in this lesson is both powerful and surprisingly general.

1632 • Galileo's Relativity Principle

In Dialogue Concerning the Two Chief World Systems, Galileo argued that a passenger in the hold of a uniformly moving ship cannot distinguish the ship's motion from rest. This was the first explicit statement that the laws of mechanics are the same in all inertial frames—a precursor to relative-motion vector equations.

1687 • Newton's Principia

Newton formalized the laws of motion with respect to an absolute frame, but his Corollary V acknowledged that relative velocities obey vector addition. The equation v_B = v_A + v_B/A is essentially embedded in Newtonian mechanics from the start.

1835 • Coriolis & Rotating Frames

Gaspard-Gustave de Coriolis derived the additional acceleration terms that arise when the reference frame itself rotates. His work extended the relative-motion formalism beyond translation, setting the stage for the rotating-frame analyses covered in advanced dynamics.

1900s • Mechanism & Machine Design

The kinematic analysis of four-bar linkages, slider-cranks, and cam mechanisms depended heavily on relative velocity and acceleration diagrams. These graphical methods, developed by Reuleaux and later refined by engineering educators, made 2-D relative motion a core competency in mechanical engineering curricula.

1960s–present • Orbital Rendezvous & Robotics

NASA's Gemini program required precise relative-motion calculations for spacecraft docking. Today, the same vector-addition principles extend to multi-link robot arms and autonomous vehicles operating in dynamic environments.

The central question this lesson addresses is deceptively simple: if observer A measures a particle's position, velocity, and acceleration, and observer B is located somewhere else—possibly moving—how do we systematically convert A's measurements into B's? Answering this question in two dimensions requires nothing more than careful vector addition, yet the framework scales to every mechanism and multi-body system you will encounter in your engineering career.

SECTION 2

Core Principles & Definitions

Before writing any equations, it is essential to lock down the vocabulary and conceptual building blocks of relative motion in a plane. In this introductory treatment we restrict attention to translating reference frames—frames whose axes remain parallel to a fixed frame at all times but whose origins may be moving. Rotating frames introduce additional Coriolis and centripetal terms and will be treated in a subsequent lesson. Throughout, we adopt the standard dynamics notation where a subscript such as B/A is read as "B relative to A" or "B as observed from A."

Reference Frame

A coordinate system attached to an observer or body, comprising an origin and a set of orthogonal axes. The fixed (inertial) frame is typically labeled O-xy, and a moving frame is attached to a point A.

Absolute vs. Relative Quantities

An absolute position, velocity, or acceleration is measured with respect to a fixed inertial frame. A relative quantity is the difference vector observed from a (possibly moving) point.

Vector Addition Rule

The position of B in the fixed frame equals the position of A in the fixed frame plus the position of B relative to A: r_B = r_A + r_B/A. Differentiating once yields velocities; twice yields accelerations.

Translating Frame (No Rotation)

When the moving frame's axes stay parallel to the fixed frame at all instants, differentiation of the position equation is straightforward. No Coriolis or centripetal corrections appear.

Subscript Notation Convention

The subscript B/A always means "of B as seen by A." Note that v_B/A = −v_A/B; reversing the subscripts flips the sign.

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KEY TAKEAWAY

Think of relative motion like GPS navigation between two moving cars on a highway. Your navigation display shows the vector from your car (frame A) to the other car (point B). Even though both cars have absolute velocities measured from the road (inertial frame), what matters to you is the difference vector—the velocity of B as seen from A. The vector-addition equation simply decomposes B's absolute motion into A's absolute motion plus B's motion relative to A. Master that decomposition, and you can analyze any two-body or multi-body problem in a plane.

SECTION 3

Visual Explanation — The Position Triangle

The entire framework of 2-D relative motion for translating frames can be captured in a single geometric picture: the position triangle. Consider a fixed origin O, a moving point A, and a second point B whose motion we wish to describe from A's perspective. The three position vectors form a triangle, and the vector equation r_B = r_A + r_B/A is nothing more than the tip-to-tail rule of vector addition. The diagram below makes this relationship explicit.

The position triangle. The violet arrow is the absolute position of A, the blue arrow is the absolute position of B, and the dashed cyan arrow is the relative position of B as seen from A. The vector equation r_B = r_A + r_B/A is simply the tip-to-tail rule.

This triangle persists at every instant in time, even though A and B may each be moving along complicated paths. At the next instant the triangle simply reshapes. By differentiating the position relationship with respect to time, we obtain the analogous velocity and acceleration triangles—each governed by the same tip-to-tail addition. The geometry is always the same; only the physical meaning of the vectors changes (position, velocity, or acceleration).

SECTION 4

Mathematical Framework

With the position triangle established, deriving the velocity and acceleration equations is a matter of straightforward calculus. Because the translating frame's axes remain parallel to the fixed frame, the unit vectors î and ĵ do not change with time, so time-differentiation passes through to the scalar components without generating extra terms. This simplicity is what distinguishes the translating-frame case from the rotating-frame case.

RELATIVE POSITION

r_B = r_A + r_{B/A}

r_B = absolute position of B (from fixed origin O); r_A = absolute position of A; r_B/A = position of B relative to A.

RELATIVE VELOCITY

v_B = v_A + v_{B/A}

Differentiating the position equation once with respect to time. Here v_B/A = dr_B/A/dt, the time rate of change of the relative position vector.

RELATIVE ACCELERATION

a_B = a_A + a_{B/A}

Differentiating a second time. Because the frame only translates (no rotation), there are no Coriolis or centripetal terms. This equation is the complete acceleration relation for translating frames.

Each vector equation decomposes into two scalar equations when resolved onto the fixed x- and y-axes. For velocity, for instance, the scalar forms are:

COMPONENT FORM — VELOCITY

v_{Bx} = v_{Ax} + v_{(B/A)x} ; v_{By} = v_{Ay} + v_{(B/A)y}

Subscript x and y refer to projections along the fixed-frame axes. Because the moving frame's axes are parallel to the fixed axes, we may project directly.

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Sign Convention Reminder

Always define a positive direction for each axis before resolving vectors into components. A common source of error in relative-motion problems is an inconsistent sign between the velocity of A and the relative velocity of B/A. Drawing the velocity triangle before plugging in numbers catches most sign errors.

SECTION 5

The Velocity Triangle — Detailed Breakdown

Just as the position equation creates a triangle in space, the velocity equation v_B = v_A + v_B/A creates a velocity triangle in velocity space. Drawing this triangle to scale (or at least sketching it with approximately correct directions) is the single most valuable habit you can develop for solving relative-motion problems. The triangle visually enforces the vector equation and immediately reveals unknown directions or magnitudes. In mechanism analysis the velocity triangle (or "velocity polygon") is the classical graphical tool for finding link velocities.

Left: physical space showing points A and B with their absolute velocity vectors. Right: the velocity triangle constructed tip-to-tail. The pink arrow is v_A, the amber arrow is v_B, and the dashed emerald arrow from the tip of v_A to the tip of v_B is v_B/A.

Notice the construction in the right panel: both absolute velocity vectors share a common tail (the origin of velocity space). The relative velocity v_B/A is the vector that must be added to v_A to reach v_B. Equivalently, v_B/A = v_B − v_A. This subtraction form is often the quickest route in component-based calculations. The same triangle logic applies identically to acceleration vectors. In problems where two of the three vectors are fully known (magnitude and direction), the triangle is determinate and the third vector can be solved by simple trigonometry or component algebra.

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Common Pitfall

Students sometimes sketch the relative-velocity arrow from B's tip to A's tip instead of from A's tip to B's tip. Remember: v_B/A starts at the tip of v_A and ends at the tip of v_B. If you reverse it, you get v_A/B = −v_B/A, which has the wrong sign.

SECTION 6

Worked Example — Two Boats on a River

Boat A travels due east at 5 m/s relative to the ground. Boat B travels at 8 m/s in a direction 60° north of east, also relative to the ground. Find the velocity of B as observed from A (i.e., v_B/A), giving both magnitude and direction.

Relative Velocity of Two Boats

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Step 1 — Establish the Coordinate System

Let the positive x-axis point east and the positive y-axis point north. Both velocities are given relative to the ground, which serves as the fixed (inertial) frame.
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Step 2 — Resolve Absolute Velocities into Components

Boat A travels due east: v_Ax = 5 m/s, v_Ay = 0. Boat B travels at 60° north of east: v_Bx = 8 cos 60° = 4 m/s, v_By = 8 sin 60° = 6.928 m/s.

v_A = (5, 0) m/s ; v_B = (4, 6.928) m/s
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Step 3 — Apply the Relative Velocity Equation

From v_B/A = v_B − v_A: component-wise, v_(B/A)x = 4 − 5 = −1 m/s, and v_(B/A)y = 6.928 − 0 = 6.928 m/s.

v_B/A = (−1, 6.928) m/s
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Step 4 — Compute Magnitude

|v_B/A| = √((−1)² + (6.928)²) = √(1 + 48) = √49 = 7 m/s.

| $v_{B/A}$ | = 7 m/s
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Step 5 — Determine Direction

θ = arctan(v_(B/A)y / v_(B/A)x) = arctan(6.928 / (−1)) = arctan(−6.928). Since the x-component is negative and the y-component is positive, the vector lies in the second quadrant. The reference angle is arctan(6.928/1) ≈ 81.8°, so the direction is 180° − 81.8° = 98.2° measured counter-clockwise from the positive x-axis (i.e., about 8.2° west of due north).

$v_{B/A}$ = 7 m/s at 98.2° from east (≈ 8.2° west of north)
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Step 6 — Interpret the Result

To a passenger on Boat A, Boat B appears to be moving almost due north at 7 m/s, with a slight westward drift. This makes physical sense: A is heading east, which "subtracts" an eastward component from B's already partially eastward motion, leaving B's apparent velocity dominated by the northward component.

SECTION 7

Strengths, Limitations & When to Use Each Form

The translating-frame relative-motion equations are powerful for a wide class of engineering problems, but they have clear boundaries. The table below summarizes where this introductory framework excels and where more advanced tools are needed.

Aspect	Translating Frame (This Lesson)	Rotating Frame (Advanced)
Frame axes	Stay parallel to the fixed frame at all times	Rotate with angular velocity ω relative to the fixed frame
Velocity equation	$v_B$ = $v_A$ + $v_{B/A}$ (simple addition)	$v_B$ = $v_A$ + Ω × $r_{B/A}$ + ( $v_{B/A}$ )_rel
Acceleration equation	$a_B$ = $a_A$ + $a_{B/A}$ (no extra terms)	Includes Coriolis (2Ω × v_rel) and centripetal (Ω × (Ω × r)) terms
Typical applications	Vehicle interception, projectile-from-moving-platform, two-particle problems	Rigid-body linkages, gear trains, rotating machinery, planetary motion
Complexity	Low — scalar component algebra suffices	High — cross products and angular-velocity tracking required

Comparison of translating vs. rotating frame relative-motion analyses

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KEY TAKEAWAY

Think of the translating-frame equations as the base case of relative motion analysis—simple, direct, and valid whenever neither the observer nor the object is rigidly connected to a rotating body. Just as a software engineer starts with a base recursive case before handling the general case, mastering v_B = v_A + v_B/A gives you the foundation on which all rotating-frame and rigid-body kinematics are built.

SECTION 8

Connection to Rotating Frames & Rigid-Body Kinematics

The translating-frame equations form the scaffolding upon which the full rotating-frame transport theorem is built. When the moving frame rotates with angular velocity Ω, the time derivative of any vector Q observed in the fixed frame relates to its derivative in the rotating frame by (dQ/dt)_fixed = (dQ/dt)_rot + Ω × Q. When Ω = 0, this reduces exactly to the translating-frame result. The table below sketches the progression you will encounter through a typical dynamics course.

Topic	New Element Introduced	Key Equation Added
Translating frames (this lesson)	Moving origin, parallel axes	$v_B$ = $v_A$ + $v_{B/A}$
Rigid-body rotation (planar)	Points on same rigid body → $r_{B/A}$ has constant length	$v_B$ = $v_A$ + ω × $r_{B/A}$
Rotating frames (general)	Axes rotate, point slides within frame	$a_B$ = $a_A$ + α × r + ω × (ω × r) + 2ω × v_rel + a_rel
3-D kinematics	Euler angles, rotation matrices	Same structure with 3-D cross products

Progression of relative-motion topics through an engineering dynamics curriculum

The critical conceptual leap from this lesson to the next is recognizing that rotation of the frame introduces fictitious forces and additional kinematic terms (Coriolis, centripetal, and Euler accelerations) that have no counterpart in the translating-frame case. By mastering the simpler case first, you develop the physical intuition to interpret these additional terms when they appear. In professional practice—whether analyzing a spinning turbine blade or programming a robotic manipulator—the translating-frame decomposition remains the first step in any kinematic analysis; the rotation corrections are layered on top.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Car A drives north at 60 km/h and Car B drives south at 60 km/h on the same straight highway. Without calculating components, use the relative-velocity equation to explain why each driver perceives the other car approaching at 120 km/h. Why does reversing the subscripts (v_A/B vs. v_B/A) not change the speed, only the direction?

Show answer

Using v_B/A = v_B − v_A, with north positive: v_B/A = (−60) − (+60) = −120 km/h, meaning B appears to approach A from the south at 120 km/h. By the antisymmetry property v_A/B = −v_B/A = +120 km/h, meaning A appears to approach B from the north at the same speed. Reversing subscripts negates the vector, which flips the direction but preserves the magnitude. This is physically intuitive: both drivers see the same closing speed.

Question 2: BASIC CALCULATION

Particle A has velocity v_A = (3î + 4ĵ) m/s and particle B has velocity v_B = (−2î + 7ĵ) m/s, both measured in a fixed frame. Compute the relative velocity v_B/A and its magnitude.

Show answer

v_B/A = v_B − v_A = (−2 − 3)î + (7 − 4)ĵ = (−5î + 3ĵ) m/s. The magnitude is |v_B/A| = √(25 + 9) = √34 ≈ 5.83 m/s. The direction is θ = arctan(3/(−5)), which places the vector in the second quadrant at about 149° from the positive x-axis. This problem illustrates the straightforward component-by-component subtraction approach.

Question 3: INTERMEDIATE

A ship S moves at 12 knots due north. A patrol boat P moves at 16 knots at 30° east of north. At a certain instant the patrol boat is 2 nautical miles due east of the ship. (a) Find the velocity of P relative to S. (b) Determine whether P is getting closer to or farther from S at that instant by computing the rate of change of the distance between them.

Show answer

(a) Let x = east, y = north. v_S = (0, 12) kn; v_P = (16 sin 30°, 16 cos 30°) = (8, 13.86) kn. Thus v_P/S = (8, 1.86) kn, with magnitude √(64 + 3.46) ≈ 8.21 kn at θ = arctan(1.86/8) ≈ 13.1° north of east. (b) The relative position vector of P from S is r_P/S = (2, 0) nmi. The rate of change of distance is the dot product of the unit relative-position vector ê = (1, 0) with v_P/S, giving dr/dt = 8 kn > 0. Therefore P is moving away from S at that instant.

Question 4: APPLIED

An engineer on a moving truck (frame A) must throw a tool kit onto a loading dock that is stationary. The truck moves at v_A = 3 m/s due east. The dock (point B) is located 6 m north and 4 m east of the engineer at the moment of release. If the engineer throws the kit at 5 m/s relative to herself, at what angle (measured from east, in her reference frame) should she aim so that the kit travels directly toward the dock in the ground frame? Neglect gravity.

Show answer

In the ground frame, the kit must travel from the engineer's position toward the dock, which is at relative position (4, 6) m. The required ground-frame direction is arctan(6/4) = 56.3° north of east. The kit's ground-frame velocity must be along this direction; call its magnitude V. We need V(cos 56.3°, sin 56.3°) = v_A + v_kit/A, where |v_kit/A| = 5 m/s. So v_kit/A = (V cos 56.3° − 3, V sin 56.3°). Squaring and summing components gives (V cos 56.3° − 3)² + (V sin 56.3°)² = 25. Expanding: V² − 6V cos 56.3° + 9 = 25, or V² − 3.33V − 16 = 0, yielding V ≈ 5.73 m/s. Then v_kit/A = (5.73 × 0.555 − 3, 5.73 × 0.832) = (−0.18, 4.77) m/s. The angle in the truck frame is arctan(4.77/(−0.18)) ≈ 92.2° from east—nearly due north with a slight westward bias to compensate for the truck's eastward motion.

Question 5: CRITICAL THINKING

Prove that for any three points A, B, and C in 2-D, the relative-velocity equations satisfy the chain rule v_C/A = v_C/B + v_B/A. Then discuss: does this "chaining" property extend to accelerations in translating frames? Would it still hold if the intermediate frame were rotating? Justify your answer.

Show answer

From the fundamental equations: v_C = v_B + v_C/B and v_B = v_A + v_B/A. Substituting the second into the first: v_C = v_A + v_B/A + v_C/B. But also v_C = v_A + v_C/A. Comparing, v_C/A = v_B/A + v_C/B = v_C/B + v_B/A (QED). The same proof works for acceleration in translating frames because differentiation is linear and the frame unit vectors are constant. For rotating frames, the chain rule still holds for the full (absolute) relative quantities, but the decomposition into individual terms (Coriolis, centripetal) does not simply add—each intermediate frame contributes its own angular velocity cross-product terms. Thus naive term-by-term chaining of rotating-frame formulas fails; one must apply the transport theorem at each stage.

SUMMARY

Lesson Summary

Summary

This lesson introduced the relative motion framework for translating frames in 2-D. The fundamental relationship $r_B$ = $r_A$ + $r_{B/A}$ — and its time derivatives for velocity and acceleration — follow directly from the tip-to-tail vector addition rule. Because the translating frame's axes remain parallel to the fixed frame, no Coriolis or centripetal terms appear, and the equations reduce to simple component-by-component addition or subtraction.

The velocity triangle is the essential graphical tool: sketch it before computing to catch sign errors and build geometric intuition. The subscript notation (B/A means "of B as seen by A") and the antisymmetry property $v_{B/A}$ = − $v_{A/B}$ are conventions that will persist through rotating-frame and rigid-body kinematics. Mastering these basics now ensures a smooth transition to the rotating-frame transport theorem and its additional cross-product terms in subsequent lessons.