ARCHETYPE: Trains Crossing & Overtaking

Stand on a platform as an express thunders through and count seconds: the whole train takes time to pass you — nose first, tail last. In that gap between nose and tail lives a whole family of exam problems. A point particle "passes" a signal instantly; a 400-metre freight train does not. Once an object has length, every crossing question becomes: how far must the front travel? Its own length to clear a pole, its own length plus the platform to clear a platform, the sum of both lengths to clear another train. Examiners adore this archetype because it welds two ideas you already own — uniform motion from Speed & Velocity, and frame-hopping from Relative Velocity & Reference Frames — into questions with satisfyingly concrete answers. CBSE and ICSE pose the pole-and-platform versions for three marks; JEE-Main and NEET dress the same skeleton in opposite-direction crossings, overtakes that crawl at the difference of speeds, and the sly "how long to pass a passenger seated in the other train?" — the most reliable trap in the family. So treat the setup itself as the concept: one bookkeeping principle (track the front, add the lengths), one frame trick (sit on one train and freeze it), and the variants examiners recycle every season — poles, platforms, tunnels, crossings, overtakes, and the accelerating train whose coaches file past a signal in a square-root rhythm.

Key takeaways

A finite-length body passes an object when its front travels the body's own length plus the object's length: pole → $L$ , platform → $L + L_p$ .
Two trains cross or overtake at their relative speed: opposite directions $v_1 + v_2$ , same direction $|v_1 - v_2|$ , always over the distance $L_1 + L_2$ .
Passing a person seated in the other train needs only your own length over the relative speed — the classic trap.
Convert units first: $\mathrm{km/h} \times \tfrac{5}{18} = \mathrm{m/s}$ . Nearly every trains question is engineered to need it.
Accelerating trains obey the same length bookkeeping with suvat: equal coaches pass a pole at times proportional to $\sqrt{n}$ .
Key idea: replace each train by its front point, add every length that must be cleared, and divide by the (relative) speed.

The core insight

A train of length $L$ has "passed" something only when its front has travelled $L$ plus the obstacle's length, so every crossing or overtaking time is (total length to clear) ÷ (speed of approach — the relative speed if both bodies move).

Where this lives today

Railway signalling solves trains-crossing problems thousands of times an hour. Block signalling will not admit a train to a track segment until the previous one has fully cleared it — nose-plus-tail bookkeeping, automated. High-frequency metros (Delhi, London, Tokyo) run "moving block" systems where each train's protected zone is computed live from its length, speed, and braking curve — literally $L$ plus a safety margin over relative speed, recalculated every second. And every highway overtake is equation (4) wearing car costumes: "don't pull back in until you can see the whole vehicle" is your own length finishing its clearing. The next module in the gym, Chase, Catch & Meet, drops the lengths but frees the speeds — together the two archetypes cover most of relative motion's exam catalogue.

The setup

Geometry first, because in an archetype the geometry is the physics. A train is a rigid body of length $L$ moving at speed $v$ . Fix your eye on the front (any point works; the front is cleanest). The train has completely passed a roadside pole when its tail clears the pole — and the tail sits $L$ behind the front, so the front must travel exactly $L$ beyond the pole.

Figure

Passing a pole versus clearing a platform — what the front must travel

Now stretch the obstacle. To clear a platform of length $L_p$ , the front enters at the near end, and clearing means the tail leaves the far end — so the front travels $L_p$ and then the train's own $L$ : total $L + L_p$ . A bridge, a tunnel, a parked train: all are "platforms" of the appropriate length. Two conventions keep every trains problem honest:

Reduce each train to its front point. The full body is just the front point dragging a tail of known length $L$ . Every crossing statement — "enters", "passes", "clears" — becomes a statement about how far this single point travels, so uniform-motion machinery ( $d = vt$ ) applies immediately.

Convert km/h to m/s before anything else. Train speeds arrive in km/h, lengths in metres. The conversion $1\ \mathrm{km/h} = \tfrac{1000}{3600} = \tfrac{5}{18}\ \mathrm{m/s}$ is engineered into these questions: $54 \to 15$ , $72 \to 20$ , $36 \to 10\ \mathrm{m/s}$ . Examiners choose multiples of 18 so the numbers come out whole — a non-integer m/s value is a hint to re-check the conversion.

The master solution pattern

Every trains question is the same three moves: identify the total length that must be cleared, identify the speed of approach, divide. For a single train at speed $v$ passing a pole, the length to clear is just its own $L$ :

t_{\text{pole}} = \frac{L}{v} \qquad (1)

where: $L$ — the train's length; $v$ — its speed in m/s. A pole has no extent, so the only thing being "cleared" is the train itself. A 150 m train at 15 m/s needs 10 s — feel how long that is next time one passes you. For a platform (or bridge, or tunnel) of length $L_p$ , the front must cover both lengths:

t_{\text{platform}} = \frac{L + L_p}{v} \qquad (2)

where: $L_p$ — the platform's length. The train's own length never disappears: even a 1 km tunnel doesn't excuse the train from also clearing its own 150 m. Equation (2) with $L_p = 0$ collapses back to (1) — one formula, two costumes. When both bodies move, hop into the frame of one of them — the manoeuvre from Relative Velocity & Reference Frames. Sitting on train 2, train 1 approaches at the relative speed, and it must clear the combined length:

t_{\text{cross}} = \frac{L_1 + L_2}{v_1 + v_2} \quad \text{(opposite directions)} \qquad (3)

where: $v_1 + v_2$ — the closing speed for head-on approach; $L_1 + L_2$ — from noses meeting to tails separating. Opposite-direction crossings are quick: two ordinary trains at 20 and 10 m/s close at 30 m/s and swap past each other in seconds.

t_{\text{overtake}} = \frac{L_1 + L_2}{v_1 - v_2} \quad \text{(same direction, } v_1 > v_2\text{)} \qquad (4)

where: $v_1 - v_2$ — the creep speed of the faster train past the slower. Overtakes are slow: the same two trains take three times as long at a 10 m/s difference as they took to cross head-on. If $v_1 \le v_2$ , the overtake never completes — the denominator is telling you the physics.

situation	length to clear	speed of approach	time
pole, signal, standing person	$L$	$v$	$L/v$
platform, bridge, tunnel ( $L_p$ )	$L + L_p$	$v$	$(L+L_p)/v$
stationary train ( $L_2$ ) on next track	$L_1 + L_2$	$v_1$	$(L_1+L_2)/v_1$
crossing, opposite directions	$L_1 + L_2$	$v_1 + v_2$	$(L_1+L_2)/(v_1+v_2)$
overtaking, same direction	$L_1 + L_2$	$v_1 - v_2$	$(L_1+L_2)/(v_1-v_2)$
passing a passenger in the other train	$L_1$ only	$v_1 + v_2$ or $v_1 - v_2$	$L_1/(v_{\text{rel}})$

The last row is the trap row. A person seated at a window is a point, not a length: your train's nose reaches them, your tail leaves them, and only your own length has been cleared — at the relative speed, because they are moving too.

Derivation AThe passing principle — why lengths add

We derive the platform-crossing time, equation (2), by tracking a single coordinate: the position $x_f(t)$ of the train's front along the track. The strategy is to write down where the front is at the "start passing" and "finish passing" instants and subtract; the train's rigid length then enters the answer automatically. Watch how the definition of finished — the tail, not the front, clearing the far end — is where the extra $L$ comes from.

Step 1. Set up the axis. Let $x = 0$ be the near end of the platform and $x = L_p$ its far end. The train moves at constant speed $v$ , so its front obeys uniform motion: $x_f(t) = x_f(0) + vt$ .

Step 2. Define the start. Passing begins when the front arrives at the near end: $x_f = 0$ . Start the clock here, so $x_f(t) = vt$ . Why: any point of the train could be tracked, but starting the clock when the front arrives makes the initial position zero and saves an algebra step.

Step 3. Define the finish. Passing ends when the tail leaves the far end. The tail trails the front by the fixed length $L$ , so its position is $x_{\text{tail}}(t) = vt - L$ . The finish condition is $x_{\text{tail}} = L_p$ .

Step 4. Solve for the time:

vt - L = L_p \;\;\Rightarrow\;\; t = \frac{L + L_p}{v}

Why: the two lengths enter symmetrically — the front must sweep the platform's extent AND the train's own extent must drain past the far end. That symmetry is why a "150 m train, 300 m platform" question is unchanged if the numbers swap roles.

Step 5. Specialise. A pole is a platform of zero length: set $L_p = 0$ to recover $t = L/v$ , equation (1). Verified: 150 m train at 15 m/s, 300 m platform: $t = 450/15 = 30$ s, and the front alone would need only $20$ s — the extra 10 s is exactly $L/v$ , the pole-passing time. Limits: $L \to 0$ (a point object) gives $t = L_p/v$ ; $L_p \to 0$ gives $t = L/v$ . Both match the table.

Go deeper · Derivation B · 1 minCrossing and overtaking by freezing one train

Derivation BCrossing and overtaking by freezing one train

We derive equations (3) and (4) with the relative-velocity manoeuvre: transform into the reference frame of train 2, where train 2 stands still and all the motion belongs to train 1. The strategy is to reduce the two-body problem to the one-body problem we already solved — train 1 passing a stationary obstacle of length $L_2$ . The bookkeeping of which relative speed to use (sum or difference) falls out of the subtraction automatically.

Step 1. Assign velocities with signs on one axis. Opposite directions: $v_1 > 0$ and $-v_2 < 0$ . Same direction: $v_1$ and $v_2$ both positive, $v_1 > v_2$ .

Step 2. Transform to train 2's frame by subtracting $\vec{v}_2$ from every velocity. Train 2 becomes stationary. Train 1's velocity becomes $v_1 - (-v_2) = v_1 + v_2$ (opposite case) or $v_1 - v_2$ (same-direction case). Why: subtracting the observer's velocity is the whole content of relative velocity — "zero out my own motion and see what's left". Speeds add for head-on approach, subtract for a stern chase.

Step 3. State the reduced problem. In this frame, train 1 (length $L_1$ ) passes a stationary train of length $L_2$ : by the passing principle, the front of train 1 must travel $L_1 + L_2$ at the relative speed $v_{\text{rel}}$ .

Step 4. Read off the times:

t_{\text{cross}} = \frac{L_1+L_2}{v_1+v_2}, \qquad t_{\text{overtake}} = \frac{L_1+L_2}{v_1-v_2}

Step 5. The passenger variant. A person seated in train 2 is a point object in train 2's frame: train 1 passes them when only its own length $L_1$ has gone by. Hence $t = L_1/v_{\text{rel}}$ — no $L_2$ anywhere. Why: lengths belong to bodies, not to frames — a seated passenger has no 180-metre tail. Ask "whose nose-to-tail must drain past the reference point?" and only train 1's does. *Verified: 120 m and 180 m trains at 20 and 10 m/s. Opposite: $t = 300/30 = 10$ s. Same direction: $t = 300/10 = 30$ s. Passenger in the slower train (opposite case): $t = 120/30 = 4$ s. Time is Galilean-invariant, so the answers hold in the ground frame too.*

Go deeper · Derivation C · 1 minThe accelerating train — coaches file past in a square-root rhythm

Derivation CThe accelerating train — coaches file past in a square-root rhythm

We derive the times at which successive equal coaches of a train, starting from rest with uniform acceleration $a$ , pass a trackside signal — the standard competitive-level twist. The strategy: the passing principle still holds (the front must travel $n$ coach-lengths for $n$ coaches to have passed), but uniform motion is replaced by suvat from Kinematic Equations. Expect square roots, because distance grows as $t^2$ .

Step 1. Set up. The front of the train is at the signal at $t = 0$ , at rest. Each coach has length $\ell$ . The front's position obeys $x_f = \tfrac{1}{2}at^2$ .

Step 2. The $n$ -th coach has fully passed when $n$ coach-lengths have gone by the signal: $x_f = n\ell$ . Solve:

\tfrac{1}{2}at_n^2 = n\ell \;\;\Rightarrow\;\; t_n = \sqrt{\frac{2n\ell}{a}} = t_1\sqrt{n}

where $t_1 = \sqrt{2\ell/a}$ is the time for the first coach. Why: distance scales as the square of time under constant acceleration, so time scales as the square root of distance — the same $\sqrt{n}$ law as Galileo's odd-number rule for free fall, run sideways.

Step 3. The time for the $n$ -th coach alone is a difference:

\Delta t_n = t_n - t_{n-1} = t_1\left(\sqrt{n} - \sqrt{n-1}\right)

Why: each successive coach passes faster because the train is moving faster — $\Delta t_n$ shrinks like $1/(2\sqrt{n})$ for large $n$ , so a long accelerating train's coaches blur past at an ever-quickening tempo.

Step 4. Speed as the $n$ -th coach clears: $v_n = at_n = \sqrt{2n\ell a}$ — also obtainable from $v^2 = 2a(n\ell)$ with no clock at all. Verified: $t_1 = 4$ s means the second coach alone takes $4(\sqrt{2}-1) \approx 1.66$ s and both together take $4\sqrt{2} \approx 5.66$ s. Check by $v$ : with $\ell = 8$ m, $a = 1\ \mathrm{m/s^2}$ : $t_1 = 4$ s ✓, $v_2 = \sqrt{2 \cdot 2 \cdot 8 \cdot 1} \approx 5.66\ \mathrm{m/s} = a t_2$ ✓.

The variants examiners actually pose

Tunnel with a margin. "A 200 m train takes 30 s to cross a tunnel at 72 km/h — find the tunnel's length." Run equation (2) backwards: $L_p = vt - L = 20(30) - 200 = 400$ m. Any variant inverts — time given, length asked; lengths given, speed asked.

Two poles, one train. The train passes a pole in $t_a$ and a platform of known length in $t_b$ : two equations, two unknowns ( $L$ and $v$ ). Divide the equations and the algebra collapses — see Worked Example 2.

The walking passenger. A person walking along the platform at $u$ is a moving point: the train passes them over its own length $L$ only, at $L/(v-u)$ same-direction or $L/(v+u)$ opposite.

Distance during the pass. "How far does the faster train travel during the overtake?" — find $t$ from (4), then distance $= v_1 t$ in the ground frame. Frames tell you when; the ground frame tells you where.

Figure

Overtaking seen from the slower train — the two-body problem collapses

Limiting cases

case	result	meaning
$L \to 0$	$t_{\text{pole}} \to 0$	a point particle passes a point instantly — length is the whole story
$L_p \gg L$	$t \to L_p/v$	crossing a very long tunnel, the train is effectively a point
$v_1 = v_2$ (same direction)	$t_{\text{overtake}} \to \infty$	equal speeds never overtake: relative velocity zero, the trains hang frozen side by side
$v_2 = 0$	(3) and (4) both $\to (L_1+L_2)/v_1$	a stationary train is just a platform of length $L_2$
equal speeds, opposite	$t = (L_1+L_2)/2v$	head-on crossings halve the time — each train does half the clearing

Worked examples

1 · Board level — pole and platform (CBSE/ICSE pattern, 3 marks)

A 150 m long train runs at a uniform 54 km/h. Find the time it takes to pass (a) a signal post, (b) a 300 m long platform.

Step 1. Convert the speed: $54\ \mathrm{km/h} = 54 \times \tfrac{5}{18} = 15\ \mathrm{m/s}$ .

Step 2. Signal post — the train clears its own length only: $t = \dfrac{L}{v} = \dfrac{150}{15} = \boxed{10\ \mathrm{s}}$ .

Step 3. Platform — front travels $L + L_p = 150 + 300 = 450$ m: $t = \dfrac{450}{15} = \boxed{30\ \mathrm{s}}$ . Sanity: the platform answer must exceed the pole answer by $L_p/v = 20$ s ✓. State the conversion and the $L + L_p$ reasoning explicitly — both carry method marks.

2 · Board level — reverse-engineering length and speed (ISC pattern)

A train passes a standing man in 8 s and a 264 m long bridge in 30 s, moving uniformly. Find the train's length and speed.

Step 1. Two unknowns, two facts. Man (a point): $L = 8v$ . Bridge: $L + 264 = 30v$ .

Step 2. Substitute the first into the second: $8v + 264 = 30v \Rightarrow 22v = 264 \Rightarrow v = \boxed{12\ \mathrm{m/s}}$ ( $= 43.2$ km/h).

Step 3. Then $L = 8(12) = \boxed{96\ \mathrm{m}}$ .

Step 4. Cross-check in the bridge equation: $(96 + 264)/12 = 360/12 = 30$ s ✓. The "standing man" is the pole in disguise — human obstacles of zero length are the giveaway that only $L$ must be cleared.

3 · Competitive level — crossing, overtaking, and the passenger trap (JEE-Main/NEET pattern, ~3 minutes)

Train A (length 120 m) moves at 72 km/h and train B (length 180 m) at 36 km/h on parallel tracks. Find the time for A to pass B completely when they move (a) in opposite directions, (b) in the same direction. (c) In the opposite-direction case, how long does A take to pass a passenger seated at a window of B?

Step 1. Convert: $v_A = 20\ \mathrm{m/s}$ , $v_B = 10\ \mathrm{m/s}$ . Combined length $L_A + L_B = 300$ m.

Step 2. Opposite directions — closing speed $20 + 10 = 30\ \mathrm{m/s}$ : $t = \dfrac{300}{30} = \boxed{10\ \mathrm{s}}$ .

Step 3. Same direction — creep speed $20 - 10 = 10\ \mathrm{m/s}$ : $t = \dfrac{300}{10} = \boxed{30\ \mathrm{s}}$ .

Step 4. The passenger is a point moving with B. Only A's length clears them, at the closing speed: $t = \dfrac{120}{30} = \boxed{4\ \mathrm{s}}$ . Timing note: parts (a) and (b) are 30 seconds each once $\tfrac{5}{18}$ is automatic. Part (c) is where the marks hide — if you catch "passenger ⇒ own length only" instantly, the whole question is under 3 minutes; if you add $180$ m by reflex, you get 10 s, which sits among the options waiting.

4 · Competitive level — the accelerating train (JEE-Main stretch, ~4 minutes)

A train made of equal coaches starts from rest and accelerates uniformly. The first coach takes 4.0 s to pass a signal at the platform's end. Find (a) the time for the first two coaches to pass, (b) the time for the second coach alone, and (c) if each coach is 8.0 m long, the train's speed as the second coach clears the signal.

Step 1. By the $\sqrt{n}$ law (front travels $n\ell$ under $x = \tfrac12 at^2$ ): $t_n = t_1\sqrt{n}$ with $t_1 = 4.0$ s.

Step 2. Two coaches: $t_2 = 4\sqrt{2} = \boxed{5.66\ \mathrm{s}}$ .

Step 3. Second coach alone: $\Delta t_2 = t_2 - t_1 = 4(\sqrt{2} - 1) = \boxed{1.66\ \mathrm{s}}$ — much shorter than the first coach's 4 s, because the train is now moving.

Step 4. Acceleration from the first coach: $\ell = \tfrac12 a t_1^2 \Rightarrow 8 = \tfrac12 a(16) \Rightarrow a = 1.0\ \mathrm{m/s^2}$ . Speed at $t_2$ : $v = at_2 = 1.0 \times 5.66 = \boxed{5.7\ \mathrm{m/s}}$ . Cross-check with the no-clock route: $v^2 = 2a(2\ell) = 2(1)(16) = 32 \Rightarrow v = \sqrt{32} = 5.66\ \mathrm{m/s}$ ✓. Timing note: recognise "equal coaches + from rest + uniform acceleration" and jump straight to $t_n = t_1\sqrt{n}$ — deriving it from scratch in the exam costs two minutes you don't have. The $\sqrt{n}$ family (here, Galileo's free-fall odd-number rule, drops from a leaking tap) is one memorised pattern with many faces.

Exam traps

Forgetting the train's own length. The single biggest mark-killer: using $L_p/v$ for a platform crossing. The train's length appears in every crossing, no matter how long the obstacle. Five-second check: platform time must exceed pole time for the same train.

The passenger/man-in-train trap. Passing a train needs $L_1 + L_2$ ; passing a person in that train needs only $L_1$ . Lengths belong to bodies. Underline what is being passed before writing anything.

Wrong relative speed. Opposite directions → add speeds; same direction → subtract. Writing $v_1 - v_2$ for a head-on crossing gives a nonsense answer three times too large in Example 3. If your overtake time comes out smaller than your crossing time for the same trains, the signs are backwards.

Botched unit conversion. $\mathrm{km/h} \to \mathrm{m/s}$ is $\times \tfrac{5}{18}$ , not $\times \tfrac{18}{5}$ . Anchor: $36\ \mathrm{km/h} = 10\ \mathrm{m/s}$ (a city bus), so 72 km/h must be 20 m/s. A train doing " $194$ m/s" should scream at you.

Mixing frames mid-solution. Time from the relative frame, distance from the ground frame — fine, but only if you're deliberate. "How far did train A travel during the overtake?" is $v_1 t$ (ground frame), not $(v_1 - v_2)t$ (that's how far it travelled past B).

Accelerating trains: which point reaches the signal? In coach problems the clock starts when the front is at the signal with the train at rest there. If the train already has speed $u$ at the signal, $\sqrt{n}$ scaling breaks — use full suvat, $n\ell = ut_n + \tfrac12 at_n^2$ . And keep $4\sqrt{2}$ symbolic until the last line: rounding $t_2$ early shifts $\Delta t_2$ enough to miss a two-significant-figure answer box.

Origins · 1 minAlbert Einstein and how this was discovered

Albert Einstein

1879–1955 · Bern patent office → Berlin → Princeton

Einstein made trains the laboratory of the imagination. Working in the Bern patent office in 1905 — a city whose clocks were synchronised to railway timetables — he asked what a light flash looks like to a passenger on a moving train versus an observer on the platform, and found that simultaneity itself depends on the frame. His popular 1916 book teaches relativity almost entirely through trains passing embankments: the same nose-tail, frame-hopping bookkeeping you use here, pushed to speeds where Galilean rules finally break. At train speeds the corrections are one part in $10^{14}$ — your $v_1 + v_2$ arithmetic is safe on every exam board on Earth.

History

1632

— Galileo's ship argument establishes that uniform motion is undetectable from inside: the birth of reference frames.

1830

— The Liverpool–Manchester railway opens; timetabling finite-length trains on shared tracks makes crossing-and-passing arithmetic an engineering necessity.

1884

— Railway scheduling chaos drives the adoption of standard time zones — clocks, trains, and frames become one problem.

1905

— Einstein's special relativity paper reimagines the train-and-platform as the definitive thought experiment about frames.

1916

— Einstein's popular book teaches relativity through trains and embankments, fixing the imagery for every physics classroom since.

Going further

Texts. NCERT Physics Class 11, Ch. 3 (Motion in a Straight Line — relative velocity section) · HC Verma, Concepts of Physics Vol. 1, Ch. 3 · SL Arora, New Simplified Physics, kinematics problem sets (the trains bank).

Provenance. All results are standard uniform-motion and relative-velocity kinematics. A verifier should recompute: Example 1's 10 s and 30 s (150 m, 15 m/s, 300 m platform); Example 2's $v = 12$ m/s, $L = 96$ m from $22v = 264$ ; Example 3's 10 s / 30 s / 4 s (120+180 m at 30 and 10 m/s relative; 120 m at 30 m/s); Example 4's $4\sqrt{2} = 5.66$ s, $4(\sqrt{2}-1) = 1.66$ s, $v = \sqrt{32} = 5.66$ m/s from $\ell = 8$ m, $a = 1\ \mathrm{m/s^2}$ ; the tunnel inversion $L_p = 20(30) - 200 = 400$ m.

Prerequisites. Relative Velocity & Reference Frames · Speed & Velocity. Leads to. Chase, Catch & Meet · Graph Reading at Speed.

Why do I add the platform's length AND the train's length? It feels like double counting.

Follow the front. When it reaches the near end, nothing is cleared. It then sweeps the whole platform ( $L_p$ ) — but at that moment the train's body still lies across the platform, and the tail exits only after the front travels one more train-length $L$ . Two different extents drain past the far end, so the lengths genuinely add. The derivation makes it one subtraction: finish ( $vt - L = L_p$ ) minus start.

How do I decide between adding speeds and subtracting them?

Ask what the gap between the trains is doing. Head-on: both motions eat the gap, so the closing speed is the sum. Same direction: the faster train's motion eats the gap while the slower one's motion rebuilds it, so only the difference survives. Formally it is one rule, not two: $\vec{v}_{\text{rel}} = \vec{v}_1 - \vec{v}_2$ with signed velocities — opposite directions make $v_2$ negative and the subtraction becomes addition on its own.

For the passenger question, why doesn't the other train's length count?

Because a seated passenger is a point object — they have no 180-metre tail that must drain past your window. Your train has passed them when your nose arrives and your tail leaves: exactly your own length, covered at the relative speed (they are moving too, so the approach rate is still $v_1 \pm v_2$ ). The other train's length only matters when the whole other train is the thing being passed.

Does it matter which train I "sit on" when I freeze a frame?

No — time intervals are the same in every uniformly-moving frame (Galilean relativity, good to one part in $10^{14}$ at train speeds). Sitting on B, A covers $L_1 + L_2$ at $v_{\text{rel}}$ ; sitting on A, B covers the same distance at the same rate. Pick whichever frame makes the picture clearest — usually the train whose passenger the question mentions.

What if the trains are on the same track rather than parallel tracks?

Then "crossing" for opposite directions is a collision, not a passage. One-track questions ask when and where the trains meet (nose touches nose) — that is a Chase, Catch & Meet question about positions, and lengths drop out. Lengths matter only for complete passage on parallel tracks. Read the track layout before choosing the archetype.

My answer for an overtake came out negative. What did I do wrong?

You subtracted the speeds in the wrong order — or the overtake is impossible. $t = (L_1+L_2)/(v_1-v_2)$ assumes train 1 is the faster one. If $v_1 < v_2$ , the "overtaking" train never catches up; the negative time is the algebra telling you the tails separate in the past, not the future. State clearly which train overtakes, put the larger speed first, and sanity-check that the result is positive.

Is there a fast mental check for the 5/18 conversion?

Memorise three anchors: $18 \to 5$ , $36 \to 10$ , $72 \to 20$ (km/h to m/s). Exam speeds are almost always these or simple sums ( $54 = 36+18 \to 15$ ; $90 \to 25$ ; $108 \to 30$ ). If your converted speed is not a whole number, suspect the conversion before the physics — setters build these questions to come out clean.