AP Physics 1: Relative Motion

Understanding motion is fundamental to physics, but motion itself is not absolute. How fast a plane is flying or where a boat will land depends entirely on who is measuring it. The study of relative motion provides the framework to analyze these situations by shifting perspectives between different observers. Mastering this concept is crucial for solving a wide array of practical problems, from navigation to predicting collisions, and forms a key skill tested on the AP Physics 1 exam.

Foundational Concepts: Frames of Reference and Relative Velocity

All descriptions of motion are made relative to a chosen frame of reference. This is a coordinate system from which an observer measures positions. The ground is a common frame of reference, but a moving car or a flowing river is equally valid. The central idea of relative motion is that the velocity of an object depends on the frame from which it is observed.

We describe this mathematically using subscript notation. The velocity of object A relative to object B is written as ${\vec{v}}_{AB}$. This is read as "the velocity of A as seen by an observer moving with B." The most critical equation in relative motion connects velocities measured from different frames: $${\vec{v}}_{AC}={\vec{v}}_{AB}+{\vec{v}}_{BC}$$ This states that the velocity of object A relative to frame C equals the velocity of A relative to frame B plus the velocity of frame B relative to frame C. Think of it as a "chain rule" for velocity vectors. The order of the subscripts matters: ${\vec{v}}_{AB}=-{\vec{v}}_{BA}$. If a train ($T$) moves east at 20 m/s relative to the ground ($G$), then ${\vec{v}}_{TG}=+20\text{ m/s }\hat{i}$ and ${\vec{v}}_{GT}=-20\text{ m/s }\hat{i}$ (the ground appears to move west relative to the train).

Vector Addition and Problem-Solving Strategy

Because velocities are vectors, solving relative motion problems is an exercise in careful vector addition. You cannot simply add or subtract magnitudes; you must account for direction. The first and most important step is to clearly define the frames and the vectors involved.

A reliable strategy is:

1. Define the frames: Identify all relevant observers or reference frames (e.g., Ground (G), Boat (B), Water (W), Air (A), Plane (P)).
2. Interpret the given information: Translate phrases like "the boat's engine propels it at 5 m/s north" into a relative velocity (${\vec{v}}_{BW}$ if the water is the reference).
3. Write the vector equation: Use the fundamental rule ${\vec{v}}_{AC}={\vec{v}}_{AB}+{\vec{v}}_{BC}$ to set up an equation for the desired unknown velocity.
4. Solve graphically or component-wise: Sketch the vectors tip-to-tail to visualize the problem, then break vectors into x- and y-components for precise calculation.
5. Interpret the result: State your final answer in the correct frame with both magnitude and direction.

Application: River-Crossing Problems

River-crossing problems are classic applications of 2D relative velocity. In these scenarios, a boat has a velocity relative to the water (${\vec{v}}_{BW}$), and the water has a velocity relative to the ground (${\vec{v}}_{WG}$). The velocity of the boat relative to the ground (${\vec{v}}_{BG}$) is the vector sum: $${\vec{v}}_{BG}={\vec{v}}_{BW}+{\vec{v}}_{WG}$$

For example, suppose a motorboat points directly across a river (north) with a speed of 4.0 m/s relative to the water (${\vec{v}}_{BW}=4.0\text{ m/s }\hat{j}$). The river current flows east at 3.0 m/s relative to the ground (${\vec{v}}_{WG}=3.0\text{ m/s }\hat{i}$). To find the boat's ground velocity:

• Vector Equation: ${\vec{v}}_{BG}={\vec{v}}_{BW}+{\vec{v}}_{WG}=(3.0\hat{i}+4.0\hat{j})$ m/s.
• Magnitude: $|{\vec{v}}_{BG}|=\sqrt{(3.0{)}^2+(4.0{)}^2}=5.0$ m/s.
• Direction: $\theta ={\tan }^{-1}(4.0/3.0)\approx 53.1^{\circ }$ north of east.

The boat does not land directly across from its starting point; it is carried downstream. To land directly across, the boat must aim upstream at an angle to counteract the current—a common exam twist that requires solving for the necessary heading of ${\vec{v}}_{BW}$.

Application: Airplane-Wind Problems

Airplane-wind problems are structurally identical to river-crossing problems but occur in the air. An airplane has an airspeed—its speed relative to the air (${\vec{v}}_{PA}$). The wind has a velocity relative to the ground (${\vec{v}}_{AG}$). The plane's velocity relative to the ground, or groundspeed (${\vec{v}}_{PG}$), is: $${\vec{v}}_{PG}={\vec{v}}_{PA}+{\vec{v}}_{AG}$$

A pilot must navigate by adjusting the plane's heading (the direction of ${\vec{v}}_{PA}$) to achieve a desired track over the ground (the direction of ${\vec{v}}_{PG}$). For instance, if a pilot wants to fly due north (${\vec{v}}_{PG}$ is north) but faces a strong west wind (${\vec{v}}_{AG}$ is west), she must point the plane's nose partly to the east. The vector triangle shows ${\vec{v}}_{PA}$ (plane relative to air) as the vector that, when added to the west wind, results in a purely northward ground velocity.

Relative Approach and Separation Velocities

A powerful application of relative velocity is finding how quickly two objects are moving toward or away from each other. This is essential for predicting collisions or rendezvous. The relative velocity of object A with respect to object B (${\vec{v}}_{AB}$) directly gives this rate.

For approach velocity, consider two cars on a highway. Car A moves east at 25 m/s (${\vec{v}}_{AG}$), and Car B moves east at 15 m/s (${\vec{v}}_{BG}$). The velocity of A relative to B is: $${\vec{v}}_{AB}={\vec{v}}_{AG}+{\vec{v}}_{GB}={\vec{v}}_{AG}-{\vec{v}}_{BG}=(25-15)\text{ m/s east}=10\text{ m/s east}.$$ From B's perspective, A is approaching at 10 m/s. If they were moving toward each other on the same line, you would add the magnitudes. The key is to use the relative velocity equation methodically, which automatically handles all directional cases for both one- and two-dimensional motion.

Common Pitfalls

1. Adding Magnitudes Instead of Vectors: The most frequent error is treating velocities as scalars. If a boat moves perpendicular to a current, its ground speed is not 4 m/s + 3 m/s = 7 m/s; it is the magnitude of the vector sum, which is 5 m/s. Always account for the angle between vectors.
2. Incorrect Vector Subtraction: Students often confuse ${\vec{v}}_{AB}$ with ${\vec{v}}_{BA}$. Remember, ${\vec{v}}_{AB}=-{\vec{v}}_{BA}$. To find the velocity of A relative to B, you subtract B's ground velocity from A's ground velocity: ${\vec{v}}_{AB}={\vec{v}}_A-{\vec{v}}_B$.
3. Misidentifying the Reference Frame: Confusing what a given velocity is relative to is a critical mistake. "The plane's speed is 200 m/s" is ambiguous. Is that airspeed (${\vec{v}}_{PA}$) or groundspeed (${\vec{v}}_{PG}$)? You must carefully label every velocity in your work (e.g., ${\vec{v}}_{BW}$, ${\vec{v}}_{AG}$) to avoid this.
4. Forgetting the Vector Nature in Headings: In navigation problems, a "northward" ground track requires accounting for wind or current. You cannot simply point the vehicle in the desired direction of travel over the ground. You must solve a vector equation to find the necessary heading.

Summary

• Motion is relative. The measured velocity of any object depends entirely on the chosen frame of reference.
• The fundamental equation is ${\vec{v}}_{AC}={\vec{v}}_{AB}+{\vec{v}}_{BC}$, which requires careful vector addition, not simple scalar arithmetic.
• River-crossing and airplane-wind problems are solved by summing the object's velocity relative to the medium and the medium's velocity relative to the ground.
• The relative velocity between two objects, ${\vec{v}}_{AB}$, directly tells you their rate of approach or separation.
• Success hinges on clear definitions: always label velocities with double subscripts (e.g., ${\vec{v}}_{BW}$) to specify "velocity of what, relative to what?" and then apply the vector addition rule systematically.