Reference Frames and Relative Motion

Lesson ~10 min read 8 MCQs

In simple terms: In simple terms, relative motion is about how an object's movement looks different depending on whether you, the observer, are standing still or moving too.

Why this matters

Imagine you're at the airport in Atlanta, running late for a flight. You see one of those long moving walkways and jump on. As you walk on the moving surface, you feel like you're flying past the people standing still at the gates. But to your friend walking at the same pace on the regular floor next to you, you only seem to be moving a little bit faster. Who is right?

You both are. Your velocity depends entirely on who is measuring it. This is the core idea of reference frames. In this lesson, we'll break down how to describe and calculate motion from different points of view, whether you're on a moving walkway, a boat in a river, or a car on the highway.

Diagram

This diagram shows a boat crossing a river from a top-down perspective. The boat's intended path (north) and the river current (east) are shown as perpendicular vectors. The resulting actual path of the boat is a diagonal vector, which is the hypotenuse of the right triangle formed by the other two vectors. A separate box shows the vector addition triangle, explicitly labeling v_bs = v_bw + v_ws.

Concept map

flowchart TD
    A[Start: Problem with multiple moving objects] --> B{Is motion in 1D or 2D?};
    B -->|1D| C[Define positive direction and assign signs to velocities];
    B -->|2D| D[Separate motion into perpendicular components (x and y)];
    C --> E[Use v_ac = v_ab + v_bc];
    D --> F[Draw vectors tip-to-tail];
    F --> G[Use Pythagorean Theorem for magnitude];
    G --> H[Use tan_inv for direction];
    E --> I[Solve for the unknown velocity];
    H --> I;
    I --> J[Final Answer: Velocity relative to the desired frame];

Core explanation

What is a Reference Frame?

Let's start with a simple idea: to measure motion, you need a starting point and a set of directions (like a coordinate system). This "point of view" is what physicists call a reference frame.

Think about driving on a highway. If you're in a car moving at 60 mph, the trees on the side of the road seem to fly backward at 60 mph. Your reference frame is the car. But for someone standing on the side of the road, their reference frame is the Earth, and they see your car moving forward at 60 mph.

The key takeaway is that quantities like position and velocity depend on the reference frame you choose. There's no single "correct" reference frame—just different ones.

(This covers LO 1.4.A and EK 1.4.A.1)

Relative Velocity in One Dimension

Let's go back to the airport walkway.

Let's say the Walkway moves at v_wg = +2 m/s relative to the Ground. (We'll use + for forward).
You, Priya, are walking on the walkway at v_pw = +1 m/s relative to the Walkway.

How fast are you moving relative to the Ground? Your intuition probably tells you to just add them up: 2 m/s + 1 m/s = 3 m/s. And you're right!

We can write this with a clear formula using subscripts. The velocity of object A relative to observer B is written as v_ab.

Our rule is: v_ac = v_ab + v_bc

For Priya at the airport:

We want v_pg (Priya's velocity relative to the Ground).
We have v_pw (Priya's velocity relative to the Walkway) and v_wg (Walkway's velocity relative to the Ground).

So, v_pg = v_pw + v_wg v_pg = (+1 m/s) + (+2 m/s) = +3 m/s

What if your friend, Marcus, is standing still on the ground? His velocity relative to the ground is v_mg = 0. What does he see? He sees you moving at v_pg = +3 m/s.

What if you were walking against the motion of the walkway? Then v_pw would be -1 m/s. v_pg = v_pw + v_wg = (-1 m/s) + (+2 m/s) = +1 m/s. To someone on the ground, you would still be moving forward, but much more slowly.

(This covers EK 1.4.B.1 and part of EK 1.4.B.2.i)

Relative Motion in Two Dimensions: The Boat Problem

This is a classic AP Physics problem. Imagine Maya is trying to steer a boat across a river.

The river is 50 meters wide.
The boat's engine can push it at 4 m/s relative to the water. Maya points the boat straight across, toward the north bank.
The river has a current that flows at 3 m/s to the east.

Let's define our reference frames:

The boat (b)
The water (w)
The shore (s)

What do we know?

The velocity of the boat relative to the water is v_bw = 4 m/s, North. This is where the boat's engine is trying to make it go.
The velocity of the water relative to the shore is v_ws = 3 m/s, East. This is the river current.

What does someone on the shore see? We want to find the velocity of the boat relative to the shore, v_bs.

Using our subscript rule: v_bs = v_bw + v_ws

Let's sketch it out. The vectors form a right triangle:

The v_bw vector points up (North).
The v_ws vector points right (East).
The resultant vector, v_bs, is the hypotenuse of the triangle.

To find the magnitude of v_bs, we use the Pythagorean theorem: |v_bs|^2 = |v_bw|^2 + |v_ws|^2 |v_bs|^2 = (4 m/s)^2 + (3 m/s)^2 |v_bs|^2 = 16 + 9 = 25 |v_bs| = √25 = 5 m/s

So, a person on the shore sees the boat moving at 5 m/s.

To find the direction, we use trigonometry: tan(θ) = opposite / adjacent = |v_ws| / |v_bw| = 3 / 4 θ = tan⁻¹(3/4) ≈ 36.9°

The boat's velocity relative to the shore is 5 m/s at 36.9° East of North. The boat doesn't go straight across; it gets pushed downstream by the current.

(This covers LO 1.4.B, EK 1.4.B.2, and EK 1.4.B.2.i)

The Constant in All of This: Acceleration

So velocity is relative. What about acceleration?

Let's imagine two people: Carlos is standing on a train platform in Chicago. Aaliyah is on a train moving past him at a constant velocity (no speeding up, no slowing down). This is called an inertial reference frame—a frame that is not accelerating.

If Aaliyah drops her phone, what does she see? She sees it accelerate straight down at g = 9.8 m/s².

What does Carlos see? He sees the phone start with a horizontal velocity (the train's velocity) and then follow a parabolic path as it falls. But if he were to measure its vertical acceleration, he would also get g = 9.8 m/s².

The acceleration of an object is the same as measured from all inertial reference frames.

This is a huge, foundational idea in physics. If the train were accelerating (speeding up), it would no longer be an inertial frame. Aaliyah would see her dropped phone appear to fall backward as well as down, and the measured accelerations would not match. The AP exam will only ask you about non-accelerating, or inertial, reference frames.

(This covers EK 1.4.B.2.ii)

Worked examples

Example 1

Catching Up on the Highway

Two cars, a blue one and a red one, are driving east on a straight highway. The blue car is traveling at a constant 55 mph. The red car is traveling at 70 mph. At the moment the red car is right next to the blue car, what is the red car's velocity relative to the blue car?

1. Define Frames and Variables:

Let the ground be our stationary reference frame (g).
Let b be the blue car and r be the red car.
Let the eastward direction be positive.
v_bg = +55 mph (velocity of blue car relative to ground)
v_rg = +70 mph (velocity of red car relative to ground)

2. Identify the Goal:

We want to find the velocity of the red car relative to the blue car, which is v_rb.

3. Apply the Relative Velocity Formula:

The main formula is v_ac = v_ab + v_bc. We need to arrange it to solve for v_rb.
Let's set it up: v_rg = v_rb + v_bg. This follows our subscript rule.
Now, we just need to rearrange the equation to solve for v_rb: v_rb = v_rg - v_bg

4. Calculate the Solution:

v_rb = (+70 mph) - (+55 mph)
v_rb = +15 mph

Why this makes sense: From the perspective of someone in the blue car, the red car is pulling away from them in the forward (east) direction at 15 mph. This matches our everyday experience. The most common mistake here is getting the subtraction order wrong. If you calculated v_br (the blue car's velocity relative to the red), you would get 55 - 70 = -15 mph. This means from the red car's perspective, the blue car is moving backward at 15 mph. Both answers are correct, they just describe different points of view!

Example 2

The Boat Crossing the River (Calculation)

Let's use the numbers from our core explanation. A boat points straight north across a river with a velocity of 4 m/s relative to the water. The river current flows east at 3 m/s. If the river is 50 meters wide, how long does it take for the boat to cross, and how far downstream does it land?

1. Separate the Motion into Components:

This is a 2D motion problem. The key is to analyze the "across the river" motion (let's call it the y-direction) and the "down the river" motion (x-direction) completely separately.
Y-direction (North): The boat's speed dedicated to crossing the river is v_y = v_bw = 4 m/s. The width of the river is Δy = 50 m.
X-direction (East): The boat's speed downstream is determined by the current: v_x = v_ws = 3 m/s.

2. Calculate the Time to Cross:

The time it takes to cross the river depends only on the y-direction velocity and the river's width. The current doesn't make the boat cross faster or slower; it just pushes it sideways.
Using the basic kinematics equation distance = velocity × time: Δy = v_y * t t = Δy / v_y t = 50 m / 4 m/s = 12.5 s
It takes 12.5 seconds to cross the river.

3. Calculate the Downstream Distance:

During those 12.5 seconds, the current is pushing the boat downstream. The distance it travels downstream depends on the x-direction velocity and the time we just calculated.
Δx = v_x * t
Δx = (3 m/s) * (12.5 s) = 37.5 m
The boat lands 37.5 meters downstream from its starting point.

Try it yourself

An airplane is flying due north at an airspeed of 200 km/h. A strong wind is blowing from west to east at 75 km/h. What is the plane's actual speed and direction relative to an observer on the ground?
- Hint: Airspeed is the plane's speed relative to the air. The wind is the air's speed relative to the ground. This is a 2D vector addition problem, just like the boat in the river. Draw the triangle!
You are riding a city bus in Seattle that is moving forward at 8 m/s. You walk from the back of the bus to the front at a speed of 2 m/s relative to the bus. A person on the sidewalk sees you. At the same time, a person on a bicycle is riding toward the bus at 4 m/s. What is your velocity relative to the person on the bicycle?
- Hint: First, find your velocity relative to the ground. Then, remember that when objects move toward each other, their relative speed is the sum of their individual speeds. Be careful with your signs! Let "forward" for the bus be the positive direction.

TL;DR

In simple terms, relative motion is about how an object's movement looks different depending on whether you, the observer, are standing still or moving too.

Key terms

Relative motion Reference frames Relative velocity AP Physics 1 Kinematics Vector addition Boat in river problem Inertial reference frame Velocity vectors

You can now…

1.4.A: Describe the reference frame of a given observer.
1.4.B: Describe the motion of objects as measured by observers in different inertial reference frames.

Essential knowledge (exam-tested)

1.4.A.1: The choice of reference frame will determine the direction and magnitude of quantities measured by an observer in that reference frame.
1.4.B.1: Measurements from a given reference frame may be converted to measurements from another reference frame.
1.4.B.2: The observed velocity of an object results from the combination of the object's velocity and the velocity of the observer's reference frame.
1.4.B.2.i: Combining the motion of an object and the motion of an observer in a given reference frame involves the addition or subtraction of vectors.
1.4.B.2.ii: The acceleration of any object is the same as measured from all inertial reference frames.

Concept map

flowchart TD
    A[Start: Problem with multiple moving objects] --> B{Is motion in 1D or 2D?};
    B -->|1D| C[Define positive direction and assign signs to velocities];
    B -->|2D| D[Separate motion into perpendicular components (x and y)];
    C --> E[Use v_ac = v_ab + v_bc];
    D --> F[Draw vectors tip-to-tail];
    F --> G[Use Pythagorean Theorem for magnitude];
    G --> H[Use tan_inv for direction];
    E --> I[Solve for the unknown velocity];
    H --> I;
    I --> J[Final Answer: Velocity relative to the desired frame];

Read what Saavi narrates

Hi everyone, it's Saavi from Shrutam. Let's talk about something that feels tricky at first but is actually part of our everyday lives: relative motion.

Imagine you're at the airport in Atlanta, and you jump on one of those moving walkways. As you walk on it, you feel like you're zooming past everyone. But to your friend walking on the floor next to you, you're not moving *that* much faster. So, who's right about your speed? The surprising answer is... you both are.

Your measured speed, or velocity, depends on your point of view, which we call a reference frame. In physics, we need a way to translate motion from one reference frame to another. That's all this topic is about.

Let's work through a classic example. Imagine a boat trying to cross a river. The boat's engine pushes it north at 4 meters per second. But the river has a current flowing east at 3 meters per second. So, what does someone on the shore see?

We can't just add 4 and 3. Velocity has direction. We have to add them like vectors. The boat's velocity and the river's velocity form a right triangle. To find the boat's actual speed relative to the shore, we use the Pythagorean theorem. Four squared plus three squared is sixteen plus nine, which is twenty-five. The square root of that is five. So, the boat is moving at 5 meters per second relative to the shore. It's also moving diagonally, pushed downstream by the current.

A common mistake I see all the time is students just adding the numbers, like 4 plus 3 equals 7. But velocity is a vector! You have to account for direction. For a 2D problem like the boat, you need to draw your vectors and use the Pythagorean theorem.

The big idea is that what you observe depends on your own motion. By using simple vector math, we can understand and predict what different observers will see. Keep practicing this, and it will start to feel like second nature. You've got this.

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Settings & Accessibility

Reference Frames and Relative Motion

Why this matters

Diagram

Concept map

Core explanation

What is a Reference Frame?

Relative Velocity in One Dimension

Relative Motion in Two Dimensions: The Boat Problem

The Constant in All of This: Acceleration

Worked examples

Catching Up on the Highway

The Boat Crossing the River (Calculation)

Try it yourself

Practice — 8 questions