Relative Velocity & Reference Frames

You are sitting in a train at a station when the train beside yours begins to slide smoothly backward. For two full seconds you are certain: they are pulling out. Then the other train clears the platform clock — which is moving too — and your stomach corrects the story: it was your train all along. Nothing in the gentle start gave it away, because there is nothing to give away: sealed inside a smoothly moving carriage, no experiment — dropped keys, poured water, a thrown ball — behaves differently than at rest. Motion is not a property an object carries around; it is a relationship between the object and whoever measures it. That single idea powers a whole class of exam problems. Two trains approaching at 60 km/h each close their gap at 120 km/h. A pilot aims the nose due north and arrives east of north because the air itself was moving. All of these are one calculation — the relative velocity $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ — applied with care about signs in one dimension and honest vector subtraction in two. CBSE, ICSE, JEE and NEET lean on this module hard, because it doubles as a problem-solving superpower: sit yourself on one moving object, and a messy two-body chase collapses into a one-body problem. You already own the tools — velocity as a vector, and vector addition. Here we point them at each other.

Key takeaways

Every velocity is measured by somebody: the same motion gets different, equally honest descriptions in different reference frames.
The velocity of A relative to B — what B sees A doing — is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ : subtracting the observer's velocity "zeroes out" the observer's own motion.
Subscript grammar does the bookkeeping: $\vec{v}_{AB} = -\vec{v}_{BA}$ , and frames chain as $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$ (inner subscripts touch and cancel).
Fix a positive direction first; the one subtraction then yields the difference of speeds for same-direction motion and the sum for head-on — via the double negative.
In 2D, subtract as vectors: reverse $\vec{v}_B$ and add tip-to-tail, or subtract components — never bare magnitudes.
A constant frame velocity drops out of any change in velocity, so acceleration is identical in all uniformly moving frames — two projectiles in flight have zero relative acceleration.
Key idea: to see the world as B sees it, subtract $\vec{v}_B$ from every velocity in the problem — the physics survives the move, and hard two-body problems become easy one-body problems.

The core insight

The velocity of A relative to B is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ — subtracting the observer's own velocity converts every question "what does B see?" into ordinary kinematics, because switching between frames that move uniformly changes every velocity by the same constant while leaving every acceleration untouched.

Where this lives today

Aviation runs on this module. The TCAS boxes in every airliner continuously compute closing speeds — relative velocities along the line joining two aircraft — and command climbs or descents when time-to-closest-approach falls below threshold: exactly the sit-on-B construction of the worked example above. Landing is a relative-velocity trade, too: aircraft land into the wind because touchdown needs a small velocity relative to the runway, while the wings fly on the velocity relative to the air. Orbital docking is the same subtraction with higher stakes: a spacecraft and the International Space Station both travel near 7.66 km/s, yet contact is safe because the crews null the relative velocity to centimetres per second — only $\vec{v}_{AB}$ matters. And every automotive radar measures relative velocity directly: the Doppler shift of the reflected wave encodes how fast the gap closes, so emergency-braking systems react to $v_{AB}$ , not to either speedometer.

The physical picture

Toss your phone straight up in a train cruising at 30 m/s. To you, it rises and falls along one vertical line, back into your hand — exactly as in your bedroom. To a farmer watching through the window, it traced a wide, flat parabola: it left your hand already moving forward at 30 m/s, kept that forward velocity throughout the flight (nothing horizontal touched it), and landed in your hand only because your hand kept pace the whole way.

Figure

One toss, two honest stories — a vertical line in the train frame, a parabola from the ground

Which story "really" happened? Both. Physics refuses to crown either observer, because the two descriptions differ only by the train's steady velocity, and no law of mechanics cares about a steady velocity — Galileo made exactly this argument aboard a ship in 1632. The lesson is liberating: you may choose your observer, and a wise choice flattens hard problems. The tool that moves you between observers is a subtraction: to know what B sees, take away B's own motion from everything. A car doing 100 km/h sees a car ahead doing 80 km/h creep forward at 20 km/h; it sees an oncoming 80 km/h car rush at it at 180 km/h. One rule generates both numbers — the signs do the thinking.

The rule: subtract the observer

The question "how does B see A move?" has a one-line answer. Take both velocities in any one common frame (usually the ground) and remove the observer's share:

\vec{v}_{AB} = \vec{v}_A - \vec{v}_B \qquad (1)

where: $\vec{v}_{AB}$ — the velocity of A relative to B, as measured by an observer riding on B; $\vec{v}_A, \vec{v}_B$ — the two velocities in the common frame. Read the subscripts as a sentence: "A, according to B". Subtracting $\vec{v}_B$ is what sitting on B feels like: your own motion becomes invisible (you are at rest in your own frame), and every other velocity in the world loses a $\vec{v}_B$ along with yours. Note the order — the observer's velocity is removed. Swap the roles and the answer flips: if B watches A drift east, A watches B drift west at the same rate:

\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = -\,\vec{v}_{AB} \qquad (2)

where: the two relative velocities are equal in magnitude and exactly opposite in direction. This is a free consistency check in exams: if you compute $\vec{v}_{AB} = +10$ m/s and the next part asks for $\vec{v}_{BA}$ , the answer is $-10$ m/s with no fresh work — and any option grid offering different magnitudes for the two is broken. Frames also chain. A passenger walks inside a train while the train moves over the ground; the passenger's ground velocity stacks the two:

\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC} \qquad (3)

where: A — the object, B — an intermediate frame (the train), C — the final frame (the ground). The grammar polices itself: the inner letters (B, B) touch and cancel, leaving "A relative to C". Walk at 1.5 m/s toward the front of a train doing 30 m/s and the ground assigns you $1.5 + 30 = 31.5$ m/s; walk rearward at 2 m/s and you still do $-2 + 30 = 28$ m/s forward. Chains of any length work — object/water, water/ground — while neighbouring subscripts match.

One dimension: let the signs do the work

In 1D, declare a positive direction before touching any numbers. With east positive, car A at $+30$ m/s and car B at $+20$ m/s (same direction) give $v_{AB} = 30 - 20 = +10$ m/s: A gains on B at the gentle overtaking rate you feel on highways. If instead B comes toward A at 20 m/s, its velocity is $-20$ m/s, and

v_{AB} = 30 - (-20) = 50 \ \text{m/s}

where: the double negative is the whole story — the rule never stops being "subtract", but subtracting a negative velocity adds the speeds. Head-on traffic closes fast not because the rule changed but because the observer's velocity carries the opposite sign. One formula, two outputs: same direction → the small difference; opposite directions → the sum.

Two dimensions: reverse, then add

When the two velocities are not along one line, equation (1) is a genuine vector subtraction, and the safe recipe is: reverse the observer's vector, then add tip-to-tail — $\vec{v}_{AB} = \vec{v}_A + (-\vec{v}_B)$ . Adding or subtracting bare magnitudes is meaningless unless the vectors are parallel. In components, subtract x from x and y from y:

\vec{v}_{AB} = (v_{Ax} - v_{Bx})\,\hat{i} + (v_{Ay} - v_{By})\,\hat{j} \qquad (4)

where: $\hat{i}, \hat{j}$ — unit vectors along your axes; magnitude from Pythagoras, direction from $\tan\theta$ . A car heading north at 20 m/s watched from a car heading east at 15 m/s: $\vec{v}_{AB} = -15\hat{i} + 20\hat{j}$ , magnitude $\sqrt{15^2 + 20^2} = 25$ m/s — the observed car slides north-west across your windscreen though neither car moves north-west. Relative velocity is real and measurable (it is what a radar on B reads), but it belongs to the pair, not to either car.

Figure

Vector subtraction the safe way — reverse the observer's velocity and add tip-to-tail

Acceleration doesn't care about the frame

The quiet punchline that sets up Newton's laws: if B moves at constant velocity, the change in $\vec{v}_{AB}$ between any two instants equals the change in $\vec{v}_A$ — a constant subtracts out of any difference. Divide by the time interval:

\vec{a}_{AB} = \vec{a}_A - \vec{a}_B \qquad (5)

where: $\vec{a}_{AB}$ — the acceleration of A as B measures it; a uniformly moving observer has $\vec{a}_B = 0$ , so every such observer measures the same acceleration for A. Velocity is a matter of opinion between frames; acceleration is a matter of fact — that is why the phone toss works on the cruising train. It also hands JEE a favourite one-liner: two projectiles in flight both have $\vec{a} = \vec{g}$ , so their relative acceleration is zero — each watches the other travel a straight line at constant velocity.

Which tool answers which question

the exam asks	reach for	watch for
"velocity of A with respect to B"	$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$	order: observer's velocity is subtracted
passenger / object inside a moving carrier	chain: $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$	inner subscripts must match
closing speed, time to meet or overtake	1D signs, then gap ÷ relative speed	opposite directions add via the double negative
angled velocities (wind, current, two roads)	components, equation (4)	reverse-then-add; never bare magnitudes
shortest distance between two movers	sit on one object; geometry of the straight relative path	perpendicular distance to the relative-velocity line

The exam superpower: sit on the moving object

Every chase problem is secretly a one-body problem. A police car at 30 m/s pursues a scooter at 20 m/s, currently 500 m ahead. Now sit on the scooter: it is parked in its own frame, and the police car approaches at $30 - 20 = 10$ m/s with 500 m to cover — time $500/10 = 50$ s, one line. The gap closes at the relative speed; nothing else matters. This collapse move is the highest-value habit this module teaches, and it powers the river-boat, rain-umbrella and train-crossing archetypes drilled in their own modules.

Derivation AWhy the relative velocity is the rate the separation changes

We want to justify equation (1) from raw positions — to show that $v_A - v_B$ is literally the speed at which the gap between A and B changes, which is why it answers every "when do they meet?" question. The strategy: write both positions as functions of time in one frame, build the separation, and watch its rate of change. The result must depend only on the difference of velocities, never on where the origin sits.

Step 1. Let both objects move uniformly along a line, with positions measured from one common origin: $x_A(t) = x_{A0} + v_A t$ and $x_B(t) = x_{B0} + v_B t$ , where $x_{A0}, x_{B0}$ are the starting positions.

Step 2. The separation — B's view of where A sits, since B always places itself at its own origin — is $s(t) = x_A(t) - x_B(t) = (x_{A0} - x_{B0}) + (v_A - v_B)\,t$ . Why this line matters: the origins and starting positions bundle into one constant, while time is multiplied by exactly one quantity, $v_A - v_B$ . The separation changes at that rate and no other.

Step 3. Read off the rate of change of the separation: in any interval $\Delta t$ , the gap changes by $\Delta s = (v_A - v_B)\,\Delta t$ , so the observed velocity is $\Delta s / \Delta t = v_A - v_B = v_{AB}$ . This is equation (1), and the same argument runs unchanged for each vector component in 2D.

Step 4. Extract the meeting condition: A and B meet when $s(t) = 0$ , i.e. at $t = -(x_{A0} - x_{B0})/(v_A - v_B)$ — initial gap over relative velocity, the sign telling you whether they ever meet at all. Verified: police car 30 m/s at $x_{A0} = 0$ , scooter 20 m/s at $x_{B0} = 500$ m: $s(t) = -500 + 10t$ , zero at $t = 50$ s — matching the sit-on-the-scooter shortcut, and at that instant both ground positions agree: $30 \times 50 = 1500$ m and $500 + 20 \times 50 = 1500$ m.

Go deeper · Derivation B · 1 minThe frame chain — why velocities stack across nested frames

Derivation BThe frame chain — why velocities stack across nested frames

We want equation (3), $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$ — the rule that combines a passenger's velocity relative to a train with the train's velocity over the ground. The strategy is bare algebra on definition (1), applied twice; watch how the intermediate frame's velocity enters and cancels. This chain is the licence behind every "boat relative to water, water relative to ground" setup later.

Step 1. Write the two given relative velocities using definition (1) in the common frame C: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ and $\vec{v}_{BC} = \vec{v}_B - \vec{v}_C$ . Since C is the frame we measure everything in, $\vec{v}_C = 0$ , so $\vec{v}_{BC} = \vec{v}_B$ .

Step 2. Add the two relative velocities: $\vec{v}_{AB} + \vec{v}_{BC} = (\vec{v}_A - \vec{v}_B) + \vec{v}_B = \vec{v}_A$ . Why the cancellation is the content: the intermediate frame B appears once with a minus (as an observer) and once with a plus (as an observed object), so it drops out — which is exactly what the touching inner subscripts record.

Step 3. Recognise the result: $\vec{v}_A$ measured in C is $\vec{v}_{AC}$ . So $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$ , and longer chains telescope the same way — $\vec{v}_{AD} = \vec{v}_{AB} + \vec{v}_{BC} + \vec{v}_{CD}$ — provided neighbouring subscripts match. Verified: passenger walking at $+1.5$ m/s relative to a train doing $+30$ m/s over the ground: $\vec{v}_{AC} = 1.5 + 30 = 31.5$ m/s; walking rearward at 2 m/s gives $-2 + 30 = 28$ m/s — both matching the direct ground-frame bookkeeping.

Go deeper · Derivation C · 1 minA crosswind flight, worked by components

Derivation CA crosswind flight, worked by components

We want the ground velocity of an aircraft whose nose points due north with airspeed 200 km/h while the wind blows due east at 60 km/h — the canonical 2D chain problem. The strategy: identify the chain (plane relative to ground = plane relative to air + air relative to ground), resolve into components, rebuild magnitude and direction. The two given vectors happen to be perpendicular, which is what licenses Pythagoras; other angles need components or the cosine rule.

Step 1. Set axes: $\hat{i}$ east, $\hat{j}$ north. The chain (3) reads $\vec{v}_{PG} = \vec{v}_{PA} + \vec{v}_{AG}$ with $\vec{v}_{PA} = 200\,\hat{j}$ (airspeed, along the nose) and $\vec{v}_{AG} = 60\,\hat{i}$ (the wind).

Step 2. Add components: $\vec{v}_{PG} = 60\,\hat{i} + 200\,\hat{j}$ km/h. The plane tracks east of north even though the nose never turns — the air it flies in is itself a moving carpet.

Step 3. Magnitude: $|\vec{v}_{PG}| = \sqrt{60^2 + 200^2} = \sqrt{3600 + 40000} = \sqrt{43600} \approx 209$ km/h.

Step 4. Direction: $\tan\theta = 60/200 = 0.3$ , so $\theta = \tan^{-1}(0.3) \approx 16.7^\circ$ east of north. A navigator states both: about 209 km/h, $16.7^\circ$ east of north. Verified: $209^2 = 43681 \approx 43600$ ✓ (rounding), and the 3-4-5-flavoured check $\sqrt{43600} = 208.8$ ; the drift angle is comfortably under $45^\circ$ as it must be, since the wind (60) is much smaller than the airspeed (200).

Limiting cases

case	result	meaning
observer at rest, $\vec{v}_B = 0$	$\vec{v}_{AB} = \vec{v}_A$	relative velocity generalises ordinary velocity, it doesn't replace it
equal velocities, $\vec{v}_A = \vec{v}_B$	$\vec{v}_{AB} = 0$	convoy driving: to each other, both cars are parked
head-on, equal speeds $v$	closing speed $2v$	the double negative at full strength
perpendicular velocities	$\|\vec{v}_{AB}\| = \sqrt{v_A^2 + v_B^2}$	Pythagoras — the 3-4-5 windscreen slide
swap the observers	$\vec{v}_{BA} = -\vec{v}_{AB}$	same magnitude, opposite arrow — free second answer
both in free flight	$\vec{a}_{AB} = 0$	each projectile sees the other travel a straight line

The last row rewards a second look: throw two balls at different angles from the same rooftop, and however curved their paths look from the ground, each ball sees the other recede along a perfectly straight line at constant speed — gravity accelerates both identically, so it vanishes from their relationship.

Common trap“They're moving toward each other, so I should add — but the formula says subtract!”

Both instincts are right, and confusing them is the top mark-loser in this chapter. The formula is always $v_{AB} = v_A - v_B$ — but velocities carry signs, and for head-on motion the observer's velocity is negative: $v_{AB} = 30 - (-20) = 50$ m/s. Subtracting a negative adds the speeds. Students who memorise "opposite means add" without the sign logic then also add for same-direction motion — where the honest subtraction gives the small difference $30 - 20 = 10$ m/s — and every option grid carries that wrong sum. Protect yourself mechanically: fix a positive direction, write each velocity with its sign, then apply the one formula. Never decide add-versus-subtract by feel; let the signs decide.

Worked examples

Board level — two trains, closing and overtaking (CBSE/ICSE pattern)

Train A moves east at 72 km/h and train B moves west at 90 km/h on parallel tracks, initially 4.5 km apart and approaching. Treat them as point objects. Find (a) the velocity of A relative to B, (b) the time until they pass each other. (c) If instead both move east — A at 90 km/h and B at 72 km/h with A starting 300 m behind B — how long does A take to catch B?

Step 1. Convert once, at the start: $72$ km/h $= 72 \times \tfrac{5}{18} = 20$ m/s; $90$ km/h $= 25$ m/s. Take east as positive.

Step 2. (a) Approaching case: $v_A = +20$ m/s, $v_B = -25$ m/s. By equation (1), $v_{AB} = 20 - (-25) = \boxed{45 \ \text{m/s, due east}}$ — B sees A rushing at it at 45 m/s.

Step 3. (b) Sit on B: A must cover the 4.5 km gap at 45 m/s. $t = \dfrac{4500}{45} = \boxed{100 \ \text{s}}$ .

Step 4. (c) Overtaking case: $v_A = +25$ m/s, $v_B = +20$ m/s, so $v_{AB} = 25 - 20 = 5$ m/s, and the 300 m gap closes in $t = 300/5 = \boxed{60 \ \text{s}}$ . Sanity check in the ground frame: in 60 s, A travels $1500$ m and B travels $1200$ m; A's extra 300 m is exactly the initial gap ✓.

Competitive level — closest approach of two cars (JEE-Main pattern)

Car B is at an intersection moving east at 15 m/s. At the same instant, car A is 200 m due south of the intersection moving north at 20 m/s. Find the minimum distance between the cars and the time at which it occurs.

Step 1. Collapse to one body: sit on B. The relative velocity is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = 20\,\hat{j} - 15\,\hat{i}$ , magnitude $\sqrt{15^2 + 20^2} = 25$ m/s. In B's frame, B is parked at the intersection and A travels along a straight line in the direction of $\vec{v}_{AB}$ (north-west-ish), starting 200 m due south.

Step 2. Geometry of the straight relative path: the line from A to B initially points due north. The angle $\phi$ between $\vec{v}_{AB}$ and this line satisfies $\cos\phi = \dfrac{20}{25} = 0.8$ , $\sin\phi = \dfrac{15}{25} = 0.6$ (components of the unit relative velocity along and across the A-to-B direction).

Step 3. Minimum distance = the perpendicular distance from the parked B to A's straight relative path: $d_{\min} = 200 \sin\phi = 200 \times 0.6 = \boxed{120 \ \text{m}}$ .

Step 4. Time to the foot of that perpendicular: A covers $200\cos\phi = 160$ m of relative path at 25 m/s, so $t = 160/25 = \boxed{6.4 \ \text{s}}$ . Component check at $t = 6.4$ s: separation $(-96, -72)$ , magnitude $\sqrt{9216 + 5184} = 120$ m ✓. Timing note: 90 seconds once the sit-on-B reflex exists — the entire problem is Step 1 plus one sine. Solving it in the ground frame with two position vectors and a minimised quadratic takes 4-5 minutes; that difference is the point of this module.

Exam traps

Deciding add-vs-subtract by feel. The formula is always subtract; the signs create the sum for head-on cases. Fix the positive direction first, carry every sign, and the arithmetic cannot betray you.

Reversing the wrong subscript. $v_{AB}$ means "A as seen by B" — the observer B is subtracted. Computing $v_B - v_A$ answers the other question, off by an overall minus — and on MCQs that sign is always a listed option.

Subtracting magnitudes at an angle. For non-parallel velocities you must reverse-then-add vectors or subtract components. $|\vec{v}_A - \vec{v}_B|$ is not $|v_A - v_B|$ ; for the 20/15 perpendicular pair it is 25, not 5.

Unit mixing. Trains in km/h, gaps in metres: convert with $\tfrac{5}{18}$ before dividing gap by relative speed. A 4.5 km gap at "45" km/h instead of 45 m/s inflates the answer by a factor of 3.6.

Forgetting that two projectiles have zero relative acceleration. Both accelerate at $\vec{g}$ , so $\vec{a}_{AB} = 0$ : the relative path is a straight line traced at constant velocity. JEE assertion-reason items bank on students assuming "curved paths → curved relative path".

Ignoring direction in closest-approach problems. The minimum separation is the perpendicular distance from the observer to the relative-velocity line — not the initial gap, and not zero unless the relative velocity points exactly along the line joining them.

Origins · 1 minErnst Mach and how this was discovered

Ernst Mach

1838–1916 · Vienna and Prague

Mach pushed "motion is relative" from a calculating trick to a philosophy of physics. In The Science of Mechanics (1883) he attacked Newton's absolute space — the God's-eye frame where "true" velocities supposedly live — arguing that only motion relative to other matter is observable: no instrument pointed at empty space reads a velocity. Every measurement in this module agrees: radars, speedometers and navigators all report relative velocities. His insistence that physics speak only of observables deeply influenced the young Einstein, who credited Mach's critique as a step toward relativity. And the Mach number of supersonic flight — speed relative to the surrounding air, in units of the local sound speed — keeps his name on exactly the right idea: even "the speed of a jet" is a relative velocity.

History

1632

— Galileo's Dialogue argues that no experiment below decks can reveal a ship's smooth motion — the principle of relativity.

1687

— Newton's Principia builds mechanics on absolute space, while its own laws work identically in every uniformly moving frame.

1883

— Mach's Science of Mechanics attacks absolute space: only relative motion is observable.

1905

— Einstein's special relativity replaces simple velocity subtraction near light speed; at everyday speeds the classical rule survives untouched.

1993

— TCAS collision-avoidance systems, built on computed closing speeds, become mandatory on large airliners in US airspace.

Going further

Texts. NCERT Physics XI Part 1, Ch. 3 (relative velocity) and Ch. 4 (vectors in a plane) — the CBSE/NEET canon · H.C. Verma, Concepts of Physics Vol. 1, Ch. 3 · Halliday, Resnick & Walker, Ch. 4 · Galileo, Dialogue, Second Day — the ship argument in its original voice. The full UG treatment is the deep dive Relative Motion & Galilean Transformations (aq-mech tier).

Provenance. Standard Galilean kinematics per NCERT Ch. 3-4. A verifier should recompute: $30 - (-20) = 50$ m/s head-on; passenger chain $1.5 + 30 = 31.5$ m/s; police chase $500/10 = 50$ s (ground check $30 \times 50 = 500 + 20 \times 50$ ); trains: closing $45$ m/s, $4500/45 = 100$ s, overtake $300/5 = 60$ s; aircraft $\sqrt{60^2 + 200^2} \approx 209$ km/h at $\tan^{-1}(0.3) = 16.7^\circ$ ; closest approach $d_{\min} = 200 \times 0.6 = 120$ m at $6.4$ s, component check $\sqrt{96^2 + 72^2} = 120$ m.

Prerequisites. Speed & Velocity — velocity as a signed/vector rate · Adding Vectors — the tip-to-tail machinery this module runs backwards. Leads to. River-Boat Crossing · Rain & the Moving Observer · Trains Crossing & Overtaking — the three drilled archetypes built on today's one formula.

If all motion is relative, is anything actually moving?

"Moving" is an incomplete sentence — like "to the left of". Every honest statement of motion names an observer: the book is at rest relative to the table, doing 30 m/s relative to the road, about 30 km/s relative to the Sun. None is the "real" one; each is exact in its own frame. What IS frame-independent here: the relative velocity between two named objects, time intervals, and accelerations. Physics doesn't lose its grip — it just refuses to answer "how fast?" until you finish the question with "…relative to what?"

Why is the closing speed a sum when the formula says subtract?

Because the formula subtracts signed velocities, not speeds. Head-on means the observer's velocity is negative in your convention: $v_{AB} = 30 - (-20) = 50$ m/s — the subtraction of a negative is where the addition comes from. The reliable habit: never ask "add or subtract?"; ask "what is each velocity with its sign?", then always subtract the observer. There is only one operation, so you cannot pick the wrong one.

Two balls are thrown at different angles — why does each see the other move in a straight line?

Both balls, once released, share the identical acceleration $\vec{g}$ , so $\vec{a}_{AB} = \vec{g} - \vec{g} = 0$ and the relative velocity fixed at the moment of release never changes in magnitude or direction. Constant velocity means a straight-line relative path, however dramatic both parabolas look from the ground. JEE turns this into one-liners: two projectiles separating from the same point are $|\vec{v}_{AB}|\,t$ apart at time $t$ .

My answer for v_AB came out negative — what does that mean?

In 1D the sign IS the direction: $v_{AB} = -10$ m/s says B sees A moving at 10 m/s in your chosen negative direction. Never "fix" the sign by dropping it — carry it forward, where it decides whether a gap opens or closes and whether a meeting time comes out positive (they meet) or negative (they never will ahead). Translate only in the final line: " $-10$ m/s" becomes "10 m/s westward".

In 2D, which vector do I reverse — the first or the second?

Reverse the observer's — the one after "relative to". $\vec{v}_{AB} = \vec{v}_A + (-\vec{v}_B)$ : draw $\vec{v}_A$ , from its tip draw $\vec{v}_B$ reversed; the arrow from start to final tip is $\vec{v}_{AB}$ . Reverse the wrong one and you get $\vec{v}_{BA}$ — right magnitude, opposite direction, the answer to the mirrored question. Components dodge the picture: subtract x from x and y from y, observer's from object's.

Does the subtraction rule still work if the objects are accelerating?

Yes — at every instant $\vec{v}_{AB}(t) = \vec{v}_A(t) - \vec{v}_B(t)$ , and the same subtraction gives $\vec{a}_{AB} = \vec{a}_A - \vec{a}_B$ . What you must NOT do is use an accelerating object's frame for dynamics: inside it, free objects seem to accelerate with no force (you feel this when a bus brakes). Kinematic bookkeeping is always safe; for Newton's laws stick to uniformly moving frames — pseudo-forces get their own module later.

When should I actually switch frames in an exam instead of working from the ground?

Switch when one object's motion is the question and another's is the complication: chases and overtakes (sit on the target — the gap closes at $v_{rel}$ ), closest-approach problems (sit on one car — the other travels a straight line, minimum distance is a perpendicular), anything phrased "as seen by". Stay on the ground when the question itself is ground-referenced (where does it land, how far from the station). The mechanical test: if two objects both move and the question involves only their separation — meeting, gap, closest distance — the relative frame wins every single time.