Two media, one idea: relative speed. A train has length — it must clear itself. A boat has a river — it is pushed or slowed. Add lengths for distance; add or subtract speeds for the medium.
Unlike a point object, a train must travel its own length to clear an object completely:
Distance to cover = Length of Train + Length of Object
— the train's tail must pass the point, not just its head.
| Scenario | Total Distance | Speed Used |
|---|---|---|
| Train crossing Pole / Man / Tree (point) | LT |
ST (train speed) |
| Train crossing Platform / Bridge / Tunnel | LT + LP |
ST |
| Train crossing Another Moving Train | LT₁ + LT₂ |
Relative Speed Srel |
| Direction | Formula | Intuition |
|---|---|---|
| Opposite (towards each other) | Srel = S₁ + S₂ |
Closing speed is sum — they rush toward each other |
| Same (same direction) | Srel = Sfast − Sslow |
Faster must gain the gap — subtract |
300 / (60×5/18) = 300 / 16.67 =18s. Two trains 100m & 150m at 50 & 70
km/h opposite → distance 250m, Srel=120 km/h=33.33 m/s → time 7.5s.
Water itself moves — helping or hindering the boat.
B (or u): speed of boat in still water (engine speed)C (or v): speed of current (stream/river)D = B + CU = B − C (if C > B, you go backward)
Given downstream D and upstream U:
B = (D + U)/2 — average of D and U (current cancels)
C = (D − U)/2 — half the difference (boat cancels)
Boat goes distance d downstream and returns upstream:
2dd/D + d/U2d / (d/D + d/U) = 2DU/(D+U) = 2(B²−C²)/(2B) = (B²−C²)/B
This is the harmonic mean of D and U — same as Module 12. Not
(D+U)/2.
Two trains start simultaneously and meet, then take times
t₁ and t₂ to reach destinations after meeting:
S₁ / S₂ = √(t₂ / t₁) — note inverse: t₂ on top for S₁.
Intuition: Faster train covers its remaining distance quicker, so its
t is smaller. The ratio is inverse-square-root of times.
Boat tries to cross width W perpendicularly:
√(B² − C²) (vector triangle). Requires B > C; else you always
drift.
W / B, drift = Time × C = W·C/B.
√(B²−C²).
| Conversion | Multiply by |
|---|---|
km/h → m/s |
5/18 — divide by 3.6 |
m/s → km/h |
18/5 — multiply by 3.6 |
Memory: 18 km/h = 5 m/s, 36→10, 72→20, 90→25. Check units before applying LT+LP (meters) to ST (km/h) — convert ST to m/s first.
| Symbol | Meaning |
|---|---|
LT, LP |
Length of train, platform |
ST |
Speed of train |
Srel |
Relative speed |
B |
Boat in still water |
C |
Current |
D = B+C |
Downstream |
U = B−C |
Upstream |
Train crossing a pole — distance?
Train crossing a platform — distance?
Two trains opposite direction — Srel?
Downstream speed?
Given D and U, boat speed B = ?
Meeting theorem: S1/S2 = ?
km/h → m/s multiply by?
Length matters ⇒ distance = LT + object (point→LT only)
Pole/man ⇒ LT ; Platform/bridge ⇒ LT+LP ; Two trains ⇒ LT₁+LT₂
Srel ⇒ opposite + (S₁+S₂) ; same − (Sfast−Sslow)
B ⇒ still water ; C ⇒ current ; D=B+C (down) ; U=B−C (up)
B=(D+U)/2 ; C=(D−U)/2
Round trip avg ⇒ 2DU/(D+U) ; Meeting S₁/S₂=√(t₂/t₁) (inverse)
Crossing: shortest path → aim upstream, speed √(B²−C²) ; shortest time → aim ⊥, time W/B, drift W·C/B
Units ⇒ km/h→m/s ×5/18
Primary source: none pinned yet — drop your preferred video/book resource into RESOURCES.md and it will be linked here. Ask me anything that's unclear.
Questions? Ask your agent — you can follow up on any concept, quiz answer, or get extra practice problems tuned to this module.