Module 3: Kinematics · Theory · Prerequisites: K12, Foundation 4
One subtraction, three costumes. Whether it is a river, a rainstorm or an aircraft in wind, the mathematics is identical.
Mahavakya
Relative motion is not a separate formula set. It is a change in what we call the zero of position — and everything follows from differentiating that change.
On this page
1. Deriving relative velocity
K7 introduced this in one dimension. The extension to two is one line.
Define the position of B relative to A as the difference of their position vectors:
rBA = rB − rA
Differentiate: vBA = vB − vA
Differentiate again:
aBA = aB − aA
Nothing was assumed beyond the definition of relative position, so these hold for any two particles moving in any way. The chain r → v → a applies to differences just as it does to individual quantities.
The most important consequence: if aA = aB, then aBA = 0, and the relative velocity is constant — regardless of how large or complex the individual accelerations are. Two projectiles, two objects in free fall, two charged particles in the same uniform field: their relative motion is uniform, and section 6 shows why that matters.
2. The subscript convention
Fixed in K7 and repeated here because it is the most common sign error in the topic.
vBA means the velocity of B as measured by A — equivalently, the velocity of B in a frame in which A is at rest.
Swap the subscripts and you negate: vAB = −vBA.
In the river problem: the water moves east at u relative to the ground. The boat moves at v in a chosen direction relative to the water. The velocity of the boat relative to the ground is:
vBG = vBW + vWG
Boat relative to ground = boat relative to water + water relative to ground. The intermediate subscript W cancels like a fraction. This is not a new formula — it is the triangle inequality in algebraic form, and it follows directly from the definition above.
3. The velocity triangle
Every relative-motion problem in two dimensions is a vector triangle problem. The three velocities — object relative to medium, medium relative to ground, and object relative to ground — form a triangle, and the question specifies some sides or angles and asks for others.
The three quantities
- vBW — what the boat’s engine produces in still water: magnitude fixed, direction chosen by the helmsman.
- vWG — the current: fixed by the river.
- vBG — the actual path over the ground: what the problem asks about.
The skill is deciding which of the three is unknown, and then applying the triangle equation. Draw before calculating — K2’s rule applies with extra force here, because placing the wrong vector on the wrong side of the triangle is the commonest error and it produces a plausible-looking answer.
4. River crossings
A river 120 m wide, current 3 m/s eastward, boat speed 5 m/s in still water. Module 1’s worked-examples post 6 solved the case where the current wins — here the boat is faster.
Same river, same boat, same current. The question asked decides the strategy.
Minimum time. Head straight across — maximise the component perpendicular to the current. Crossing time t = W/v = 120/5 = 24 s. Drift = 3 × 24 = 72 m downstream.
Zero drift. Aim upstream so the current exactly cancels the upstream component. Condition: v sin θ = u, so sin θ = 3/5 and θ = 36.9°. Effective crossing speed = √(25 − 9) = 4 m/s. Time = 120/4 = 30 s.
Zero drift is only possible when v > u. If the current is stronger than the boat, no heading angle cancels it. The boat can reduce drift but cannot eliminate it, and the problem changes character completely. Module 1’s post 6 covers that regime: the same 3 and 5 appear but with the boat slower, making zero drift impossible and the question one of minimising drift by calculus.
5. Rain and umbrella
Rain falls vertically at 4 m/s. A person walks east at 3 m/s. From the person’s frame, the rain appears to come from the north-east — the apparent velocity is the rain’s velocity relative to the person:
vrain, person = vrain, ground − vperson, ground = −4j − 3i
This vector points southward and westward — so the rain appears to come from the north-east, and the umbrella must tilt forward (north-east, toward the direction of apparent origin).
Direction of motion and direction of apparent origin are opposite. The person moves east; the rain appears to come from the east (from in front). The umbrella tilts toward the direction of travel, not away from it. Students tilt it the wrong way — and the algebra above is the reason, not a mnemonic.
The angle the umbrella makes with the vertical: tan α = 3/4, so α = 36.9°. Notice this is the angle of the apparent velocity vector, which is what the umbrella must align with.
The same relation governs a swimmer in a channel, an aircraft in crosswind, a drone in a gust. The medium changes; the triangle does not.
6. Equal accelerations and what they produce
This is the result worth pausing on.
Two particles both under gravity have aA = aB = −gj. The relative acceleration is:
aBA = aB − aA = 0
Zero relative acceleration means the relative velocity is constant. From A’s frame, B moves in a straight line at constant speed — regardless of how complicated either trajectory looks from the ground.
Worked example — interception
Particle A is projected from the origin at 20 m/s at 60°. Particle B is projected simultaneously from (60, 0) at 10 m/s at 30°, with g = 10. Do they collide?
Since both are under gravity, the relative velocity is constant:
vA = (10, 10√3) m/s vB = (5√3, 5) m/s
vBA = (5√3 − 10, 5 − 10√3) m/s (constant)
Initial relative position: rBA(0) = (60, 0) − (0, 0) = (60, 0).
Since the relative velocity is constant, the relative position is rBA(0) + vBAt. For collision, this must reach 0.
60 + (5√3 − 10)t = 0 → t = 60/(10 − 5√3) ≈ 17.1 s
Verify the y-component gives the same time — if it does, they collide; if not, they cross paths at different instants. This is the monkey-and-hunter result in general form: two objects under the same acceleration have constant relative velocity, and the collision question reduces to K7’s arithmetic.
7. The question that decides everything
Most errors in relative motion problems come not from wrong arithmetic but from not being clear about which question is being asked.
- Minimum time: what heading maximises the component perpendicular to the current?
- Zero drift: what heading makes the resultant velocity perpendicular to the banks?
- Minimum distance: what heading minimises the total path length?
- Target: what heading makes the resultant velocity point at the destination?
Each is a different optimisation, and each involves a different unknown in the same triangle. Identifying the question before drawing the triangle takes ten seconds. Discovering you solved the wrong question at the end takes considerably longer.
Contemplation
A boat crossing a river, a person walking in rain, an aircraft flying in crosswind, a swimmer in a tidal channel. Four physical situations, one mathematical structure. The medium flows; the object moves through it; the ground sees the sum.
That is the point of having a framework rather than a formula for each situation. Four separate formulas would cover these four cases, and would fail at the fifth. One subtraction covers all of them, because what they share is not the physics but the geometry — and it is the geometry that the mathematics describes.
This is also what Module 1 was building toward. The vectors and calculus were not preparation for projectiles; they were preparation for the moment when a physical situation and a mathematical structure turn out to be the same thing seen from two directions.
Common misconceptions
1. Relative velocity is a formula to memorise
It is the derivative of a difference of positions. Differentiate once to get relative velocity; differentiate again to get relative acceleration.
2. vAB = vBA
They differ by a sign. vBA = vB − vA; swap the subscripts and you negate the result.
3. Heading north means moving north
Heading is the direction the boat points. The actual path over the ground is the vector sum of the heading and the current. A north heading in an east current produces a north-east actual path.
4. Tilt the umbrella away from the direction of walking
Tilt it toward. The apparent rain comes from in front, because walking into it shifts the apparent source forward. The algebra in section 5 shows why.
5. Equal acceleration means equal velocity
Equal acceleration means constant relative velocity. The individual velocities can differ greatly; it is their difference that is constant.
Reflection
- In the river problem, what happens to the zero-drift solution as the current speed approaches the boat speed? What happens when it exceeds it?
- Two stones are thrown simultaneously from the same height — one horizontally, one at 45°. From the first stone’s frame, what does the second one do?
- The umbrella problem gives the angle to the vertical. At what walking speed would the umbrella be held at 45°, and what rain speed would that require?
Key takeaways
- vBA = vB − vA, derived by differentiating rBA = rB − rA.
- vBG = vBW + vWG: the intermediate subscript cancels.
- Three velocities form a triangle. Draw it before solving it.
- Minimum time: head perpendicular to the current. Zero drift: head upstream so sin θ = u/v. Zero drift requires v > u.
- Umbrella tilts toward the direction of motion, not away.
- Equal accelerations ⇒ zero relative acceleration ⇒ constant relative velocity. The relative motion is uniform.
- State what you are optimising before drawing the triangle.
What comes next
Relative motion asked how bodies move relative to each other when they are free. K14 asks what happens when they are not — when a string, a pulley or a contact surface links them so that the motion of one determines the motion of another.
Prerequisites: K12 and Foundation 4 · Reading: 16 min · Practice: 40 min · Difficulty: Intermediate to Advanced