Relative Motion

You are sitting in a Rajdhani Express moving at 130 km/h. A Shatabdi Express on the next track races past in the opposite direction at 150 km/h. For that half-second of blurred windows, the Shatabdi seems to scream past at a terrifying speed — much faster than 150 km/h. But someone standing on the platform sees it pass at exactly 150 km/h, no more.

Who is right? Both of you. You measured a different velocity because you are watching from a different place — a moving train instead of a stationary platform. The Shatabdi's speed has not changed. What changed is your motion, and that changed what you observed. This is the heart of relative motion: every velocity measurement is made from some vantage point, and the answer depends on which vantage point you choose.

Physics calls that vantage point a reference frame. The platform is one frame. Your train seat is another. The Shatabdi's pantry car is a third. Each frame is equally valid, and each gives a different velocity for the same object. There is no "true" velocity — there is only velocity relative to a chosen frame.

Reference frames — choosing your vantage point

A reference frame is a coordinate system attached to an observer. When you say "the car is moving at 60 km/h," you mean 60 km/h relative to the ground. The ground is your reference frame. You do not normally say this out loud because the ground is the default — everyone around you is standing on it. But the moment you sit inside a moving train, the ground is no longer the obvious frame, and ambiguity creeps in.

Any object can serve as a reference frame: the ground, a train, a boat, a cyclist, the Sun. The only requirement is that you define a coordinate origin and a set of axes attached to that object. Once you do, every position, velocity, and acceleration you measure is "as seen from that frame."

For the problems in this article, all frames move at constant velocity relative to each other — no acceleration, no rotation. These are called inertial reference frames, and Newton's laws hold in every one of them. The physics is the same; only the numbers change.

Relative velocity — the subtraction rule

Suppose two cars, A and B, are driving along a straight highway. Car A moves at velocity \vec{v}_A and car B at velocity \vec{v}_B, both measured by a traffic camera fixed to the ground.

What is the velocity of A as seen by someone sitting in B?

Think of it this way. You are in car B. The ground is sliding backward under you at speed v_B — but you do not notice that, because your frame moves with B. Everything you see has B's motion subtracted out. A tree on the roadside, which is stationary to the traffic camera, appears to move backward at v_B from your point of view. And car A, which the camera sees at v_A, appears to you at:

Why: when you sit in B's frame, you subtract B's velocity from everything. The ground goes from 0 to -\vec{v}_B. Car A goes from \vec{v}_A to \vec{v}_A - \vec{v}_B. This subtraction is the frame transformation from the ground frame to B's frame.

This is the fundamental formula of relative motion. Read the subscript carefully: \vec{v}_{AB} means "velocity of A relative to B" — the velocity of A as measured by B. The first subscript is the object you are asking about; the second is the observer.

A useful symmetry: the velocity of B relative to A is \vec{v}_{BA} = \vec{v}_B - \vec{v}_A = -\vec{v}_{AB}. If you see B approaching you at 50 km/h, then B sees you approaching at 50 km/h in the opposite direction. The magnitudes are equal; the directions are opposite.

One-dimensional relative motion — same line, two objects

When two objects move along the same straight line, the vector subtraction simplifies to ordinary subtraction with signs. Set up a sign convention — say, rightward (or the direction of travel) is positive — and every velocity is just a signed number.

Same direction

Two autorickshaws on MG Road: rickshaw A at +30 km/h, rickshaw B at +20 km/h, both heading north.

Why: A is faster, so from B's viewpoint, A slowly pulls ahead at 10 km/h. The positive sign says A moves in the positive direction (northward) relative to B — which makes sense, since A is overtaking B.

From A's viewpoint, B falls behind:

The negative sign means B appears to drift southward (backward) at 10 km/h. Both observers agree on the magnitude (10 km/h) but disagree on the direction — as the symmetry rule guarantees.

Opposite directions

An Indian Railways express train heading north at +90 km/h passes a goods train heading south at -60 km/h (negative because southward).

Why: when two objects move toward each other, their relative speed is the sum of their individual speeds. This is why the Shatabdi in the opening seemed so fast — your Rajdhani's speed added to it.

This is the formula behind every "two trains approaching each other" problem. The gap between them closes at 150 km/h, even though neither train moves that fast on its own.

The crossing-time formula

When two objects approach each other along a line, the time to cross depends on the total length to be covered and the relative speed.

Two trains of lengths L_1 and L_2 approach each other on parallel tracks. From the instant the engines meet, the trains are "crossing" each other until the tail of one passes the engine of the other. The total distance covered during this crossing is L_1 + L_2 — because both trains must completely pass each other. The speed at which this gap closes is the relative speed |v_{AB}|.

Why: in B's frame, B is stationary and A approaches at |v_{AB}|. Train A must cover a distance equal to the sum of both train lengths to completely clear B. Time = distance / speed.

If two trains move in the same direction (overtaking), the crossing distance is still L_1 + L_2, but the relative speed is smaller (|v_A - v_B|), so the overtaking takes much longer than crossing from opposite directions.

Two-dimensional relative motion — vector subtraction

Real motion is not always along a line. A boat crosses a river while the current pushes it sideways. A plane flies northeast while the wind blows from the west. Two cars approach an intersection from perpendicular roads. In all these situations, the velocities point in different directions, and you must use vector subtraction.

The formula is the same:

But now \vec{v}_A and \vec{v}_B are vectors with components:

Subtract component by component:

Why: vector subtraction works independently on each component. The x-components subtract to give the x-component of the relative velocity, and the y-components do the same. This is why you resolved vectors into components in the first place — it turns a vector problem into two scalar problems.

The magnitude of the relative velocity is:

And the direction (angle with the x-axis):

The vector triangle

There is a neat geometric way to see the subtraction. Draw \vec{v}_A and \vec{v}_B from the same point. The vector from the tip of \vec{v}_B to the tip of \vec{v}_A is \vec{v}_{AB} = \vec{v}_A - \vec{v}_B.

This triangle is your best friend in 2D relative velocity problems. Instead of decomposing into components, you can often use the geometry of the triangle directly — applying the cosine rule when the angle between the velocities is known, or using Pythagoras when the velocities are perpendicular.

Minimum distance of approach

Two objects move in straight lines that do not intersect. At some instant they are far apart; they approach each other, reach a minimum separation, and then move apart again. Finding that minimum distance is a classic relative-velocity problem.

The trick is to work in the frame of one object. Say object B is your frame. In B's frame, B is stationary at a point, and A moves in a straight line at the relative velocity \vec{v}_{AB}. The minimum distance between A and the stationary B is simply the perpendicular distance from B to A's straight-line path.

How to compute d_{\min}:

Step 1. Find the relative velocity \vec{v}_{AB} = \vec{v}_A - \vec{v}_B.

Why: in B's frame, A moves at \vec{v}_{AB} in a straight line. B sits at the origin, unmoving.

Step 2. Find the initial relative position \vec{r}_{AB}(0) = \vec{r}_A(0) - \vec{r}_B(0). This is the vector from B to A at time t = 0.

Why: in B's frame, A starts at this position and then moves along the direction of \vec{v}_{AB}.

Step 3. The position of A relative to B at time t is:

The distance squared is:

Step 4. Minimise d^2 by differentiating with respect to t and setting the derivative to zero:

Why: this is the time at which A is closest to B. The dot product \vec{r}_{AB}(0) \cdot \vec{v}_{AB} captures how much of the initial separation is along the direction of approach. Dividing by |\vec{v}_{AB}|^2 converts that into a time.

Step 5. The minimum distance is:

Or, geometrically, d_{\min} is the perpendicular distance from B to the line of A's relative motion. You can compute this using the cross product:

Why: |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta gives the area of the parallelogram formed by \vec{a} and \vec{b}. Dividing by the base |\vec{b}| gives the height, which is the perpendicular distance.

This approach — switching to one object's frame — is far simpler than tracking both objects in the ground frame. In B's frame, the problem reduces to finding the distance from a point to a line, which is basic geometry.

Worked examples

Common confusions

• "Relative velocity is always the sum of speeds." Only when the objects move directly toward each other (opposite directions on the same line). If they move in the same direction, relative speed is the difference. In 2D, you must do a full vector subtraction — neither sum nor difference captures it.

• "There is one true velocity." There is not. Velocity is always relative to a frame. A passenger on a train is moving at 100 km/h relative to the ground, stationary relative to the seat, and moving at 200 km/h relative to a train coming the other way. All three are equally correct velocity measurements. Physics does not privilege one frame over another (as long as the frames are inertial).

• "Switching frames changes the physics." It does not. Newton's laws work in every inertial frame. The forces are the same, the accelerations are the same. Only the velocities and positions shift by a constant amount. If a cricket ball accelerates at 5 \text{ m/s}^2 in the ground frame, it accelerates at 5 \text{ m/s}^2 in the train frame too. Switching frames changes descriptions, not physics.

• "Relative velocity of A w.r.t. B is the same as relative velocity of B w.r.t. A." The magnitudes are the same, but the directions are opposite. \vec{v}_{AB} = -\vec{v}_{BA}. If you see B approaching from the east, B sees you approaching from the west.

• "The minimum-distance-of-approach formula only works for perpendicular motions." It works for any two objects moving in straight lines at constant velocity, regardless of the angle between their paths. The key step — switching to one object's frame so the problem becomes "distance from a point to a line" — works for every geometry.

Where this leads next

• River-Boat and Rain Problems — applying relative velocity to boats crossing rivers and pedestrians walking in rain. The velocity triangle from this article reappears in every one of those problems.
• Projectile Motion — Advanced Problems — problems that combine projectile motion with relative velocity, such as one projectile observing another.
• Uniform Circular Motion — when the reference frame itself rotates, you need to introduce fictitious forces. This article's inertial-frame assumption no longer holds.
• Velocity and Acceleration in Two Dimensions — the prerequisite on vector kinematics that underpins the 2D relative motion treatment here.
• Vector Addition and Subtraction — the mathematical machinery behind the velocity triangle and the component-by-component subtraction.