June 12, 2026

Reference Frames and Relative Motion

« Back to AP Physics Guide / Unit 1: Kinematics / Topic 1.4: Reference Frames and Relative Motion

Visualization of a person throwing a ball on a moving train, showing different velocity vectors for observers on the train versus on the ground.

Velocity is never absolute. It entirely depends on the reference frame of the observer.

Imagine you are sitting on a train moving at 50 \text{ m/s}. To the person sitting next to you, your velocity is zero. To a person standing on the ground outside, your velocity is 50 \text{ m/s}. Both observers are absolutely correct in their own Reference Frames.

In AP Physics C, we use Galilean relativity to translate measurements made by one observer into the reference frame of another observer. Because velocity is a vector, we do this using vector addition.

1. The Relative Velocity Equation

To find the velocity of an object A relative to an observer C, we can use an intermediate frame of reference B (like the ground, a river, or a moving train). The standard relative velocity formula is:

    \[\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}\]

Read as: “The velocity of A relative to C is equal to the velocity of A relative to B, PLUS the velocity of B relative to C.” Notice how the inner ‘B’ subscripts perfectly cancel out to leave A/C!

The Flipping Rule: If you know the velocity of the train relative to the ground (\vec{v}_{T/G} = 50 \text{ m/s}), then the velocity of the ground relative to the train is exactly equal and opposite: \vec{v}_{G/T} = -50 \text{ m/s}.

Mathematically: \vec{v}_{A/B} = -\vec{v}_{B/A}.

2. 2D Relative Motion (Rivers & Airplanes)

1D relative motion is simple addition or subtraction. But if an airplane flies North while a wind blows East, or a boat tries to cross a river while a current pushes it downstream, we must use 2D vector addition (breaking velocities into \hat{i} and \hat{j} components).

⚠️ The “Crossing Time” Trap: Notice in the simulator that the river current (x-axis) has zero effect on how long it takes to cross the river (y-axis)! Only the y-component of the boat’s engine speed (\vec{v}_{b/w} \cos\theta) determines crossing time (t = \Delta y / v_y). The current only dictates how far downstream the boat drifts!

3. Quick AP Practice

📚 Topic 1.4 Mastery Challenge

1. A flatbed truck is driving East at 20 \text{ m/s} relative to the ground. A student standing on the truck throws a ball East at 5 \text{ m/s} relative to the truck. What is the velocity of the ball relative to a stationary observer on the ground?

Check Answer Use the relative velocity formula: \vec{v}_{ball/ground} = \vec{v}_{ball/truck} + \vec{v}_{truck/ground}.
Since they are in the same direction (East is positive):
\vec{v}_{b/g} = 5 \text{ m/s} + 20 \text{ m/s} = \mathbf{25 \text{ m/s East}}.

2. An airplane’s compass indicates it is heading due North, and its airspeed indicator shows 100 \text{ m/s} (velocity relative to the air). However, there is a strong wind blowing due East at 50 \text{ m/s} relative to the ground. What is the magnitude of the airplane’s true velocity relative to the ground?

Check Answer \vec{v}_{plane/ground} = \vec{v}_{plane/air} + \vec{v}_{air/ground}.
In vector notation: \vec{v}_{p/g} = (100\hat{j}) + (50\hat{i}) = 50\hat{i} + 100\hat{j}.

To find the magnitude, use the Pythagorean theorem since the vectors are perpendicular:
|\vec{v}_{p/g}| = \sqrt{50^2 + 100^2} = \sqrt{2500 + 10000} = \sqrt{12500} \approx \mathbf{111.8 \text{ m/s}}.