A train moves at 80 km/h. Relative to what? The ground, obviously — but not obviously to the passenger walking inside it. Every velocity is measured from somewhere. Here is what changes when you change that somewhere.

Relative Motion: Every Velocity Needs a Reference

You are sitting in a train moving at 80 km/h. A passenger walks toward the front of the train at 5 km/h. How fast is she moving?

If you answer 85 km/h, you are thinking from the ground's perspective. If you answer 5 km/h, you are thinking from your perspective — sitting in the same train. Both answers are correct. Neither is more correct than the other. They are measurements from different reference frames.

This is the core of relative motion: velocity is not an absolute property of an object. It is always a relationship between two things — the object and the observer measuring it. Change the observer, change the velocity.

For most of K01 and K02, this was hidden. We assumed a fixed ground, a fixed observer. That is fine for most problems. But the moment two objects are both moving, or the moment you need to find what one moving object looks like from another, you need the language of relative motion.

The notation

Let A and B be two objects. The velocity of A as seen from B — that is, relative to B — is written:

$\mathbf{v}_{AB}$

Read it as: "velocity of A with respect to B" or "velocity of A as seen by B."

The fundamental rule:

\mathbf{v}_{AB} = \mathbf{v}_A - \mathbf{v}_B

To find what A looks like from B: subtract B's velocity from A's velocity. Both measured from the same ground frame.

The reverse: what B looks like from A:

$\mathbf{v}_{BA} = \mathbf{v}_B - \mathbf{v}_A = -\mathbf{v}_{AB}$

Equal in magnitude, opposite in direction. If you see me moving right at 10 m/s, I see you moving left at 10 m/s.

1D first: two cars on a highway

Two cars, A and B, moving along the same straight road. Positive direction: east.

Case 1 — Same direction. Car A: +60 km/h. Car B: +80 km/h. Both moving east, B faster.

Velocity of A relative to B: v_AB = 60 − 80 = −20 km/h.

From B's perspective, A appears to move west at 20 km/h — slowly falling behind. This matches what you see when a faster car overtakes you: the slower car appears to drift backward.

Velocity of B relative to A: v_BA = 80 − 60 = +20 km/h. B appears to move east at 20 km/h as seen from A — pulling ahead.

Case 2 — Opposite directions. Car A: +60 km/h east. Car B: −70 km/h (moving west).

Velocity of A relative to B: v_AB = 60 − (−70) = +130 km/h.

From B's perspective, A is approaching at 130 km/h. This is why head-on collisions are so violent — the relative speed is the sum of the two speeds, not the difference.

Case 3 — One stationary. Car A: +60 km/h. Car B: stationary (v_B = 0).

v_AB = 60 − 0 = 60 km/h. When the observer is stationary, relative velocity equals absolute velocity. The ground frame is just a special case where the observer has zero velocity.

Relative acceleration

The same rule extends to acceleration:

$\mathbf{a}_{AB} = \mathbf{a}_A - \mathbf{a}_B$

If both objects have the same acceleration — both in free fall, for example — their relative acceleration is zero. From the perspective of one falling object, the other appears stationary. This is the principle behind weightlessness in orbit: the astronaut and the spacecraft are both falling at the same rate, so relative to each other, neither is falling at all.

2D: the river-boat problem

A boat can move at speed v_b relative to the water. The river flows at speed v_r relative to the ground (along the bank, say eastward). The boat aims to cross the river (northward). What actually happens?

The boat's velocity relative to the ground is the vector sum:

$\mathbf{v}_{boat, ground} = \mathbf{v}_{boat, water} + \mathbf{v}_{water, ground}$

If the boat points directly north at v_b and the river flows east at v_r, the resultant ground velocity points northeast — the boat is carried downstream while crossing. The crossing speed (how fast it covers the width) is still v_b. The drift downstream is determined by v_r and the crossing time.

Two standard questions arise from this setup:

Question 1 — Shortest time to cross. Aim the boat directly perpendicular to the bank. Crossing speed = v_b. Time = d/v_b where d is the width. The boat drifts, but crosses as fast as possible. Direction of boat: straight across.

Question 2 — Reach directly opposite (zero drift). The boat must aim upstream at some angle α such that the eastward component of v_b cancels v_r exactly:

$v_b \sin\alpha = v_r \implies \alpha = \sin^{-1}\left(\frac{v_r}{v_b}\right)$

The boat points upstream, the river pushes it downstream, and the two cancel — the boat travels straight across. This is only possible if v_b > v_r. If the river is faster than the boat, the boat cannot go straight across — it will always drift, no matter the angle. Best it can do is minimise the drift.

{/* DIAGRAM: River flowing east, boat aimed at angle α upstream, resultant velocity pointing straight north */}

The rain-man problem

Rain falls vertically at speed v_r (relative to the ground). A person walks horizontally at speed v_p (relative to the ground). What angle should the umbrella be tilted?

The velocity of rain relative to the person is:

$\mathbf{v}_{rain, person} = \mathbf{v}_{rain, ground} - \mathbf{v}_{person, ground}$

v_rain,ground is straight down: (0, −v_r). v_person,ground is horizontal: (v_p, 0).

$\mathbf{v}_{rain, person} = (0 - v_p,\ -v_r - 0) = (-v_p,\ -v_r)$

The rain appears to come from forward and above — diagonally. The umbrella should tilt forward at angle:

$\tan\alpha = \frac{v_p}{v_r}$

where α is measured from the vertical. The faster you walk, the more you tilt forward. When you stop, the rain is vertical again and the umbrella goes straight up.

{/* DIAGRAM: Person walking right, rain falling vertically, relative velocity vector showing diagonal rain from front */}

Note: 1. Establish a ground frame. Fix one reference — usually the ground or the earth. 2. Write all velocities relative to this ground frame with correct signs and directions. 3. Apply v_AB = v_A − v_B to find the velocity of A relative to B. 4. For 2D problems: treat x and y components separately. The vector subtraction applies to each component independently. 5. River-boat: decide the goal first — shortest time (aim perpendicular) or zero drift (aim upstream). The goal determines the boat's direction. 6. Rain-man: the umbrella angle equals the angle of v_rain,person, which is found by vector subtraction. 7. Check with physical intuition: if A is faster than B in the same direction, A should appear to move forward from B's frame. If the signs give you the opposite, recheck your convention.

Warning: Adding velocities instead of subtracting. v_AB = v_A − v_B, not v_A + v_B. The river-boat velocity is a sum because you are adding the boat's velocity relative to water to the water's velocity relative to ground — these are two different pairs. In the standard relative motion formula, it is always subtraction.

> > **Confusing v_AB and v_BA.** They are equal and opposite. v_AB = −v_BA. If A moves toward B at 10 m/s, B moves toward A at 10 m/s — same magnitude, opposite direction. > > **Ignoring the vector nature in 2D.** In the river-boat problem, the boat's speed relative to water and the river's speed relative to ground are perpendicular. You cannot add them as scalars. Draw the vector triangle. > > **Thinking zero drift requires pointing straight across.** It requires pointing upstream at an angle that exactly cancels the river's current. Pointing straight across gives shortest time but not zero drift. > > **Forgetting that drift occurs even if the boat reaches the other bank.** If the question asks "where does the boat land?", account for the downstream drift using crossing time × river speed.

A worked example

A river is 100 m wide and flows east at 3 m/s. A boat can travel at 5 m/s relative to the water. Find: (a) the time and drift if the boat aims straight north, (b) the angle to aim for zero drift and the resulting crossing time.

Part (a) — aim straight north.

Crossing speed = 5 m/s (northward component, unchanged by river). Time to cross: t = 100/5 = 20 s

Drift = river speed × time = 3 × 20 = 60 m east

The boat lands 60 m downstream from the directly opposite point.

Part (b) — zero drift.

Aim upstream (northwest) at angle α where: sinα = v_r/v_b = 3/5 → α = 37° west of north

Northward component of boat's velocity: v_b cosα = 5 × (4/5) = 4 m/s (The 3-4-5 triangle: sinα = 3/5, cosα = 4/5.)

Time to cross: t = 100/4 = 25 s

Zero drift — lands directly opposite. But takes 5 s longer than the straight-across approach.

Twist variant

Same river, same boat, but now the river flows at 6 m/s — faster than the boat's 5 m/s. Can the boat reach directly opposite?

No. sinα = 6/5 > 1 — no real solution. The river is too fast. The boat cannot cancel the drift no matter which direction it aims.

Best strategy: aim upstream at 90° (directly against the current). Even then, the boat is carried downstream. The minimum drift occurs at a different angle — found by minimising the drift as a function of the boat's heading — but zero drift is impossible when v_r > v_b.

What to remember

Every velocity is relative to a reference frame. There is no absolute velocity.

v_AB = v_A − v_B. Subtract the observer's velocity from the object's velocity, both measured in the ground frame.

In 1D: same direction means relative speed is the difference. Opposite directions means relative speed is the sum.

In 2D: use vector subtraction component by component. Draw the triangle.

River-boat: shortest time means aim perpendicular. Zero drift means aim upstream and requires v_b > v_r.

Rain-man: tilt the umbrella forward at angle tan⁻¹(v_person / v_rain).

K05 is the last piece of kinematics: circular motion. A particle moving in a circle at constant speed is accelerating — because the direction of velocity keeps changing. The acceleration points inward, toward the centre, and its magnitude is v²/r. This is the last new kinematic idea before Newton's laws take over in Chapter 3.