11.2 Problems on Trains, Platforms, and Boats & Streams

11.2 Problems on Trains, Platforms, and Boats & Streams

Problems on Trains and Boats & Streams extend foundational kinematic relationships into multi-dimensional scenarios involving physical lengths and fluid currents. In banking logistics, currency consignments can move between regions by rail under security escort, and boat-based banking outlets have been used to reach communities in riverine and lake regions. In the SBI Clerk Examination, train crossings and river kinetics feature regularly, testing candidate speed and spatial reasoning.

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1. Train Kinematics: Point vs Extended Objects

A train cannot be treated as a dimensionless point mass because its physical length ($L_T$) is significant relative to the obstacles it traverses. The critical factor in formulating train equations is identifying whether the obstacle has measurable length.

Scenario 1: Train Crossing a Point Object (Pole / Person)
[==== Train (Length L_T) ====>]  | (Pole)
Distance Covered to Clear = L_T

Scenario 2: Train Crossing an Extended Object (Platform / Bridge)
[==== Train (Length L_T) ====>]  [======== Platform (Length L_P) ========]
Distance Covered to Clear = L_T + L_P

Category A: Objects of Negligible Length (Point Objects)

When a train passes a stationary entity with zero physical length along the track (telegraph pole, signal post, kilometer stone, or standing person), the train completes the crossing the moment its final coach clears the object:

$\text{Distance Traversed } (D) = L_T$ $\text{Time to Cross } (T) = \frac{L_T}{S_T}$ Where $L_T$ is in meters and $S_T$ is in meters per second ($\text{m/s}$).

Category B: Objects of Extended Length (Dimensional Objects)

When a train traverses an obstacle possessing substantial physical length (station platform, railway bridge, tunnel, or another stationary train parked on a siding), the front of the locomotive enters the structure, and the entire train must pass through until its tail clears the opposite end:

$\text{Distance Traversed } (D) = L_T + L_{\text{object}}$ $\text{Time to Cross } (T) = \frac{L_T + L_{\text{object}}}{S_T}$

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2. Master Comparison Matrix of Train Crossing Scenarios

The following reference matrix outlines the governing kinematic formulas across all standard train problem variations appearing in banking exams:

[!WARNING] The Classic Passenger-in-a-Train Trap:
When an exam question asks for the time taken by Train A to cross a passenger seated inside Train B, examinees often reflexively add both train lengths ($L_A + L_B$). This is incorrect. Because the target being crossed is the passenger (a dimensionless point entity traveling at Train B's velocity), the distance traversed is strictly the length of the passing train ($L_A$)! The length of Train B ($L_B$) has zero mathematical relevance.

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3. Boats & Streams: Kinematic Vector Foundations

Unlike terrestrial locomotion where the medium of travel is stationary, nautical motion takes place within a flowing fluid medium. The velocity of the water stream directly alters the net ground velocity of the vessel.

Variable Conventions

• Let $u = \text{Speed of the boat / swimmer in still water (in km/h)}$
• Let $v = \text{Speed of the water stream / river current (in km/h)}$ (Physical requirement for navigable upstream motion: $u > v$)

Vector Kinematics

1. Downstream Motion (With the Current): The river current pushes the boat forward, reinforcing engine thrust: $\text{Downstream Speed } (D) = u + v$

2. Upstream Motion (Against the Current): The river current pushes backward against the boat, retarding forward progress: $\text{Upstream Speed } (U) = u - v$

The Inverse Resolution Identities

Given downstream speed ($D$) and upstream speed ($U$), the still-water speed and current speed can be solved simultaneously by addition and subtraction:

$D + U = (u + v) + (u - v) = 2u \implies u = \frac{D + U}{2}$

$D - U = (u + v) - (u - v) = 2v \implies v = \frac{D - U}{2}$

• Boat speed in still water ($u$): Arithmetic mean of downstream and upstream speeds.
• Stream speed ($v$): Half the difference between downstream and upstream speeds.

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4. Round Trip Kinematics & Advanced Boat Shortcuts

In a standard banking exam scenario, a watercraft journeys from dock A to dock B and returns to dock A over a fixed one-way distance $d$.

Total Round Trip Travel Time ($T_{\text{total}}$)

$T_{\text{total}} = T_{\text{down}} + T_{\text{up}} = \frac{d}{u + v} + \frac{d}{u - v}$

$T_{\text{total}} = d \left[\frac{(u - v) + (u + v)}{(u + v)(u - v)}\right] = \frac{2du}{u^2 - v^2}$

From this formulation, the one-way distance $d$ between docks can be computed directly without solving quadratic forms:

$d = \frac{T_{\text{total}} \left(u^2 - v^2\right)}{2u}$

Travel Time Differential ($\Delta T$)

Similarly, the additional time required to navigate upstream compared to downstream over distance $d$ is:

$\Delta T = T_{\text{up}} - T_{\text{down}} = \frac{d}{u - v} - \frac{d}{u + v} = \frac{2dv}{u^2 - v^2}$

The Multiplier Ratio Shortcut ($T_{\text{up}} = n \times T_{\text{down}}$)

When a boat takes $n$ times as long to row upstream as it does to row downstream over the same distance:

$\frac{T_{\text{up}}}{T_{\text{down}}} = \frac{d / (u - v)}{d / (u + v)} = \frac{u + v}{u - v} = n$

Applying Componendo & Dividendo to this ratio:

$\frac{(u + v) + (u - v)}{(u + v) - (u - v)} = \frac{n + 1}{n - 1} \implies \frac{2u}{2v} = \frac{n + 1}{n - 1} \implies \frac{u}{v} = \frac{n + 1}{n - 1}$

[!TIP] Speed Ratio Key: If an exam question states that a boat takes "thrice as long upstream as downstream" ($n = 3$), the ratio of boat speed to current speed is immediately $\frac{u}{v} = \frac{3 + 1}{3 - 1} = \frac{4}{2} = 2 : 1$. If $u = 12 \text{ km/h}$, then $v = 6 \text{ km/h}$.

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5. Step-by-Step Worked Transit Problems

Problem 1: Dual-Platform Train System

Problem: A passenger train running at uniform velocity passes an 80-meter-long railway platform in 15 seconds and completely clears a second platform 180 meters long in 20 seconds. Determine the physical length of the train and its operational speed in $\text{km/h}$.

1. Formulate Speed Equivalence:
   Let $L_T$ be train length in meters. Speed remains constant across both platforms: $S_T = \frac{L_T + 80}{15} = \frac{L_T + 180}{20}$
2. Eliminate Denominators via Cross-Multiplication:
   $20(L_T + 80) = 15(L_T + 180) \implies 4(L_T + 80) = 3(L_T + 180)$ $4L_T + 320 = 3L_T + 540 \implies L_T = 540 - 320 = \mathbf{220 \text{ meters}}$
3. Compute Operational Velocity:
   $S_T = \frac{220 + 80}{15} = \frac{300}{15} = 20 \text{ m/s}$
4. Convert to $\text{km/h}$:
   $S_T = 20 \times \frac{18}{5} = 4 \times 18 = \mathbf{72 \text{ km/h}}$

Problem 2: Floating Branch River Logistics

Problem: A watercraft operating a mobile banking facility on a river navigates 36 km downstream and returns to the base station, completing the entire round trip in 5 hours. If the speed of the river current is 3 km/h, find the cruising speed of the craft in still water.

1. Set Up the Total Time Equation:
   $T_{\text{down}} + T_{\text{up}} = \frac{36}{u + 3} + \frac{36}{u - 3} = 5$
2. Apply the Combined Closed Formula:
   $\frac{2du}{u^2 - v^2} = \frac{2(36)u}{u^2 - 3^2} = \frac{72u}{u^2 - 9} = 5$
3. Formulate the Quadratic Equation:
   $5(u^2 - 9) = 72u \implies 5u^2 - 72u - 45 = 0$
4. Factor the Quadratic:
   Product $= 5 \times (-45) = -225$. Factors summing to $-72$ are $-75$ and $+3$: $5u^2 - 75u + 3u - 45 = 0 \implies 5u(u - 15) + 3(u - 15) = 0$ $(5u + 3)(u - 15) = 0$ Since physical velocity must be positive, $u = \mathbf{15 \text{ km/h}}$.
   (Verification: $D = 15 + 3 = 18 \text{ km/h} \implies T_d = 36/18 = 2 \text{ h}$; $U = 15 - 3 = 12 \text{ km/h} \implies T_u = 36/12 = 3 \text{ h}$. Total time $= 2 + 3 = 5 \text{ hours}$.)