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AP Physics 1: Algebra Notes

1.4.3 Relative Velocity in One Dimension

AP Syllabus focus: ‘Observed velocity depends on both the object's velocity and the observer's reference frame, using vector addition or subtraction.’

Relative velocity in one dimension explains why different observers can report different velocities for the same object. The key is treating velocity as a signed vector along a chosen axis and relating measurements between observers.

Core Idea: Velocity Depends on the Observer

Reference frames in 1D

Choose a one-dimensional axis (often +x to the right). Any velocity measured along that line must include a sign that indicates direction relative to the axis.

A velocity statement is incomplete unless you know who measures it (the observer’s frame). For example, “the cart moves at +3 m/s+3\ \text{m/s}” implicitly means “+3 m/s+3\ \text{m/s} relative to the ground” (or another stated frame).

Relative velocity: the velocity of an object measured in the reference frame of an observer, found by combining the object’s velocity and the observer’s velocity using 1D vector addition/subtraction.

Relative velocity is still a velocity vector in 1D, so it can be positive, negative, or zero.

Using Vector Addition/Subtraction in One Dimension

Sign conventions do the directional work

In one dimension, opposite directions are represented by opposite signs.

Pasted image

Velocity vectors are drawn on a line to show how a negative velocity (opposite direction) combines with a positive velocity to produce an observed relative velocity. The diagram reinforces that 1D vector addition is just signed algebra once a positive direction is chosen. Source

This allows “vector addition” to be done with ordinary algebra, as long as you keep the sign convention consistent.

  • If two velocities point in the same direction, their magnitudes add (signs match).

  • If they point in opposite directions, one will be negative relative to the other, and the algebra naturally subtracts.

A useful mental model: relative velocity answers, “How fast does A approach or recede from B, as seen by B?”

Interpreting negative results

A negative relative velocity does not mean “wrong.” It means the actual direction is opposite the positive axis (or opposite the direction you expected). Always interpret the sign physically.

Standard Notation and Relationship

Let:

  • vA/Ev_{A/E} = velocity of object A relative to Earth/ground frame E

  • vB/Ev_{B/E} = velocity of observer/frame B relative to E

  • vA/Bv_{A/B} = velocity of A as measured by B

A single relationship connects them:

vA/B=vA/EvB/Ev_{A/B} = v_{A/E} - v_{B/E}

vA/Bv_{A/B} = velocity of A relative to B (m/s)

vA/Ev_{A/E} = velocity of A relative to Earth/ground E (m/s)

vB/Ev_{B/E} = velocity of B relative to Earth/ground E (m/s)

vA/E=vA/B+vB/Ev_{A/E} = v_{A/B} + v_{B/E}

vA/Ev_{A/E} = velocity of A relative to Earth/ground E (m/s)

vA/Bv_{A/B} = velocity of A relative to B (m/s)

vB/Ev_{B/E} = velocity of B relative to Earth/ground E (m/s)

This is the algebraic form of “observed velocity depends on both the object's velocity and the observer's reference frame,” and it is fundamentally vector subtraction in 1D.

Common Pitfalls (and How to Avoid Them)

Mixing up speed and velocity

Speed is a magnitude only, but relative motion requires velocity with direction. If a problem gives speeds, you must assign signs based on direction before combining.

Forgetting to define the positive direction

Before doing any subtraction, state a clear axis direction (e.g., “east is positive”). Then:

  • velocities to the east are positive

  • velocities to the west are negative

Confusing “A relative to B” with “B relative to A”

They are opposites:

  • vA/B=vB/Av_{A/B} = -v_{B/A}

Switching the order flips the sign, not the magnitude (in 1D).

Treating the observer as “at rest” when they are not

If the observer/frame is moving, you must include vB/Ev_{B/E} (or the relevant frame velocity). The “ground” frame is only special if the problem states it or implies Earth-based measurements.

FAQ

Because “relative to B” means you remove B’s motion from the ground-measured motion: $v_{A/B}=v_{A/G}-v_{B/G}$.

A and B have the same velocity in the ground frame, so A appears stationary to B even though both may be moving.

Yes. If two objects move in opposite directions, the relative speed magnitude can be the sum of magnitudes.

Pick a convenient direction (often “to the right” or “east”) and keep it consistent. Only the final sign depends on that choice.

Translate each motion into a signed velocity along one axis (one positive, the other negative), then apply $v_{A/B}=v_{A/G}-v_{B/G}$.

Practice Questions

(2 marks) A cyclist moves at +6,m/s+6,\text{m/s} relative to the ground. A runner moves at +2,m/s+2,\text{m/s} relative to the ground in the same direction. Find the cyclist’s velocity relative to the runner.

  • Uses vC/R=vC/GvR/Gv_{C/R}=v_{C/G}-v_{R/G} (1)

  • +6+2=+4,m/s+6-+2=+4,\text{m/s} (1)

(5 marks) Take east as positive. Train A moves at +20,m/s+20,\text{m/s} relative to the ground. Train B moves at 15,m/s-15,\text{m/s} relative to the ground.
(a) Determine vA/Bv_{A/B}. (3 marks)
(b) Determine vB/Av_{B/A} and comment on the relationship between your answers. (2 marks)

(a)

  • Correct relationship vA/B=vA/GvB/Gv_{A/B}=v_{A/G}-v_{B/G} (1)

  • Substitution with signs: vA/B=+20(15)v_{A/B}=+20-(-15) (1)

  • vA/B=+35,m/sv_{A/B}=+35,\text{m/s} east (1)

(b)

  • vB/A=vB/GvA/G=1520=35,m/sv_{B/A}=v_{B/G}-v_{A/G}=-15-20=-35,\text{m/s} (1)

  • States vB/A=vA/Bv_{B/A}=-v_{A/B} (equal magnitude, opposite sign) (1)

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