5.2 Train Problems: Passing Poles, Platforms, Moving Objects & Relative Speed in Opposite/Same Directions

Fundamental Crossing Postulates & Effective Length

A train cannot be treated as a point mass when its physical length is significant relative to the obstacle being cleared. The fundamental kinematic equation $T = D / S$ governs all cases, but the effective distance $D$ depends strictly on the physical nature of the object.

CASE 1: TRAIN PASSING A POINT OBJECT (Pole / Person)
   [ Train: Length L_t, Speed S_t ]  ──►  │ (Signal Pole)
   Effective Distance = L_t
   Time = L_t / S_t

CASE 2: TRAIN PASSING AN EXTENDED OBJECT (Platform / Bridge)
   [ Train: Length L_t, Speed S_t ]  ──►  [ Platform: Length L_p ]
   Effective Distance = L_t + L_p
   Time = (L_t + L_p) / S_t

1. Crossing Stationary Point Objects

Point objects have negligible dimension along the track axis:

• Telegraph poles, Overhead Equipment (OHE) masts, signal posts, kilometer stones
• A standing passenger, trackman (gangman), or station master on the platform

$$\text{Effective Distance } (D) = L_{\text{train}}$$ $$\text{Crossing Time } (t) = \frac{L_{\text{train}}}{S_{\text{train}}}$$

2. Crossing Stationary Extended Structures

Extended structures have significant physical dimension along the track axis:

• Station platforms, passenger shelters
• Girder bridges, viaducts, rail culverts
• Tunnels, cutting walls
• A stationary freight consist or stabled rake on an adjacent loop line

$$\text{Effective Distance } (D) = L_{\text{train}} + L_{\text{structure}}$$ $$\text{Crossing Time } (t) = \frac{L_{\text{train}} + L_{\text{structure}}}{S_{\text{train}}}$$

3. Dual-Condition Traversal Problems

A frequent examination question presents a train passing two different extended objects (or a point object and an extended object) in differing times:

• Condition 1: Train passes platform of length $P_1$ in time $t_1$:

$$L_{\text{train}} + P_1 = S_{\text{train}} \times t_1$$

• Condition 2: Train passes platform of length $P_2$ in time $t_2$:

$$L_{\text{train}} + P_2 = S_{\text{train}} \times t_2$$

Subtracting Condition 1 from Condition 2 eliminates $L_{\text{train}}$:

$$P_2 - P_1 = S_{\text{train}} (t_2 - t_1) \implies S_{\text{train}} = \frac{P_2 - P_1}{t_2 - t_1}$$

Once $S_{\text{train}}$ is found, substitute back into either equation to determine $L_{\text{train}}$.

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Relative Speed Mechanics

Relative speed is the rate at which the distance between two moving bodies changes. The sign of the relative velocity vector depends on their relative heading.

Two Moving Trains Crossing Each Other

When two trains of lengths $L_1$ and $L_2$ with speeds $S_1$ and $S_2$ traverse each other:

• Total physical distance required to clear each other from the leading ends coming abreast until the trailing ends clear is always:

$$D_{\text{total}} = L_1 + L_2$$

1. Approaching on Parallel Tracks (Opposite Directions):

$$t_{\text{crossing}} = \frac{L_1 + L_2}{S_1 + S_2}$$

1. Overtaking on Parallel Tracks (Same Direction, where $S_1 > S_2$):

$$t_{\text{overtake}} = \frac{L_1 + L_2}{S_1 - S_2}$$

Train Crossing a Moving Person or Vehicle

When a train of length $L_t$ and speed $S_t$ passes a pedestrian or trackman moving at speed $S_p$:

• The person is a point object: $D = L_t$.
• If the person moves in the same direction as the train:

$$t = \frac{L_t}{S_t - S_p}$$

• If the person moves in the opposite direction to the train:

$$t = \frac{L_t}{S_t + S_p}$$

Observer Seated Inside a Moving Train

$$t = \frac{L_{\text{passing}}}{S_1 \pm S_2}$$

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Simultaneous Departure, Crossing & Post-Meeting Transit Formula

The following constant-speed model concerns two trains starting simultaneously from opposite terminals towards each other.

Station A ───► Train 1 (Speed S_1)         Train 2 (Speed S_2) ◄─── Station B
                       Meeting Point P
                          │
  ◄─── Train 2 takes T_2 ──┼── Train 1 takes T_1 ───►

The Theorem Formulation

Two trains $T_1$ and $T_2$ depart at the same instant from stations A and B towards each other. They meet at point P after an elapsed time $t$. After crossing each other at P, Train 1 takes $T_1$ hours to reach Station B, and Train 2 takes $T_2$ hours to reach Station A.

Complete Mathematical Proof

1. Before Meeting (Time $t$ elapsed for both trains):
   • Train 1 covers distance $AP$: $AP = S_1 \cdot t$
   • Train 2 covers distance $BP$: $BP = S_2 \cdot t$
2. After Meeting:
   • Train 1 covers remaining distance $PB$ in time $T_1$: $PB = S_1 \cdot T_1$
   • Train 2 covers remaining distance $PA$ in time $T_2$: $PA = S_2 \cdot T_2$
3. Equating Segment Lengths:
   • Distance $AP$: $S_1 \cdot t = S_2 \cdot T_2 \implies t = \frac{S_2 T_2}{S_1}$
   • Distance $BP$: $S_2 \cdot t = S_1 \cdot T_1 \implies t = \frac{S_1 T_1}{S_2}$
4. Equating the two expressions for meeting time $t$:

$$\frac{S_2 T_2}{S_1} = \frac{S_1 T_1}{S_2} \implies \frac{S_1^2}{S_2^2} = \frac{T_2}{T_1}$$

Taking the square root on both sides yields the canonical formula:

$$\frac{S_1}{S_2} = \sqrt{\frac{T_2}{T_1}}$$

Corollaries: Meeting Time and Total Track Distance

• Elapsed Time until Meeting ($t$): Multiplying the two expressions for $t$:

$$t^2 = \left(\frac{S_2 T_2}{S_1}\right) \left(\frac{S_1 T_1}{S_2}\right) = T_1 T_2 \implies t = \sqrt{T_1 T_2}$$

The time elapsed until the trains meet is the geometric mean of their post-crossing transit times!

• Total Distance Between Terminals ($D$):

$$D = AP + PB = S_1 t + S_1 T_1 = S_1 (t + T_1) = S_1 (\sqrt{T_1 T_2} + T_1)$$

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Worked Examples

Worked Example 1: Traversal of Point and Extended Structures

Problem: An express train traveling at a constant speed passes an OHE electrification mast in $15 \text{ seconds}$ and completely crosses a $360\text{-meter}$ long girder bridge in $33 \text{ seconds}$. Determine:

1. The operational speed of the train in kilometers per hour ($\text{km/h}$).
2. The physical length of the express train in meters.

• Step 1: Set up the kinematic traversal equations: Let train length be $L$ (in meters) and speed be $S$ (in $\text{m/s}$).
  • Crossing the mast (point object): $L = S \times 15$
  • Crossing the bridge (extended object): $L + 360 = S \times 33$
• Step 2: Solve for speed $S$: Substitute $L = 15S$ into the bridge equation:

$$15S + 360 = 33S \implies 18S = 360 \implies S = \frac{360}{18} = 20 \text{ m/s}$$

• Step 3: Convert speed to $\text{km/h}$:

$$S = 20 \times \frac{18}{5} = 4 \times 18 = 72 \text{ km/h}$$

• Step 4: Calculate train length $L$:

$$L = 15 \times S = 15 \times 20 = 300 \text{ meters}$$

• Conclusion: The speed of the train is 72 km/h, and its length is 300 meters.

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Worked Example 2: Parallel Track Overtaking and Opposite Crossing

Problem: Two passenger trains of lengths $280 \text{ meters}$ and $220 \text{ meters}$ operate on parallel broad-gauge tracks. When moving in the same direction, the faster train completely overtakes the slower train in $60 \text{ seconds}$. When moving in opposite directions, they completely pass each other in $12 \text{ seconds}$. Calculate the individual speeds of both trains in $\text{km/h}$.

• Step 1: Determine total distance for clearance:

$$D_{\text{total}} = L_1 + L_2 = 280 + 220 = 500 \text{ meters}$$

• Step 2: Formulate relative speed equations: Let the speeds of the trains be $S_1$ and $S_2$ in $\text{m/s}$ (with $S_1 > S_2$).
  • In the same direction (overtaking):

$$S_1 - S_2 = \frac{D_{\text{total}}}{t_{\text{overtake}}} = \frac{500 \text{ m}}{60 \text{ s}} = \frac{25}{3} \text{ m/s} \quad \text{--- (Equation 1)}$$

• In opposite directions (crossing):

$$S_1 + S_2 = \frac{D_{\text{total}}}{t_{\text{crossing}}} = \frac{500 \text{ m}}{12 \text{ s}} = \frac{125}{3} \text{ m/s} \quad \text{--- (Equation 2)}$$

• Step 3: Solve the simultaneous linear system: Add Equation 1 and Equation 2:

$$2S_1 = \frac{25}{3} + \frac{125}{3} = \frac{150}{3} = 50 \implies S_1 = 25 \text{ m/s}$$

Substitute $S_1 = 25$ into Equation 2:

$$S_2 = \frac{125}{3} - 25 = \frac{125 - 75}{3} = \frac{50}{3} \text{ m/s}$$

• Step 4: Convert both speeds to $\text{km/h}$:

$$S_1 = 25 \times \frac{18}{5} = 5 \times 18 = 90 \text{ km/h}$$ $$S_2 = \frac{50}{3} \times \frac{18}{5} = 10 \times 6 = 60 \text{ km/h}$$

• Conclusion: The speed of the faster train is 90 km/h, and the slower train is 60 km/h.

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Worked Example 3: Post-Meeting Transit Formula Application

Problem: Two superfast express trains, Train A and Train B, depart simultaneously from fictional stations X and Y towards each other. After crossing each other at an en-route junction station, they take $4 \text{ hours } 48 \text{ minutes}$ and $3 \text{ hours } 20 \text{ minutes}$ respectively to reach their destination terminals. If Train A is traveling at a booked speed of $75 \text{ km/h}$, determine:

1. The speed of Train B.
2. The elapsed time from departure until the two trains crossed.
3. The total rail distance between fictional stations X and Y.

• Step 1: Convert post-meeting times to improper fractions of hours:
  • $T_1$ (Train A) $= 4 \text{ hours } 48 \text{ minutes} = 4 + \frac{48}{60} = 4 + \frac{4}{5} = \frac{24}{5} \text{ hours}$
  • $T_2$ (Train B) $= 3 \text{ hours } 20 \text{ minutes} = 3 + \frac{20}{60} = 3 + \frac{1}{3} = \frac{10}{3} \text{ hours}$
• Step 2: Apply the post-meeting speed ratio formula:

$$\frac{S_1}{S_2} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{10/3}{24/5}} = \sqrt{\frac{10}{3} \times \frac{5}{24}} = \sqrt{\frac{50}{72}} = \sqrt{\frac{25}{36}} = \frac{5}{6}$$

• Step 3: Calculate the speed of Train B ($S_2$):

$$\frac{75}{S_2} = \frac{5}{6} \implies S_2 = 75 \times \frac{6}{5} = 15 \times 6 = 90 \text{ km/h}$$

• Step 4: Calculate the elapsed time until meeting ($t$):

$$t = \sqrt{T_1 T_2} = \sqrt{\frac{24}{5} \times \frac{10}{3}} = \sqrt{8 \times 2} = \sqrt{16} = 4 \text{ hours}$$

• Step 5: Calculate total rail distance ($D$):

$$D = S_1(t + T_1) = 75 \left(4 + \frac{24}{5}\right) = 75 \times \frac{44}{5} = 15 \times 44 = 660 \text{ km}$$

(Verification using $S_2$: $D = S_2(t + T_2) = 90 \left(4 + \frac{10}{3}\right) = 90 \times \frac{22}{3} = 30 \times 22 = 660 \text{ km}$. Exactly matches!)

• Conclusion: Train B speed is 90 km/h, elapsed meeting time is 4 hours, and total distance is 660 km.

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Exam Traps & Common Pitfalls