3.6 Speed, Distance, Time, Trains & Boats/Streams
Key Takeaways
- Converting speed between km/h and m/s requires multiplying by 5/18 or 18/5 respectively, which is critical for train length calculations.
- Average speed over equal distances equals the harmonic mean 2s1s2 / (s1 + s2), not the simple arithmetic mean of speeds.
- Relative speed equals the sum of speeds (s1 + s2) for objects moving in opposite directions and the difference (s1 - s2) for same-direction movement.
- When a train passes a platform or bridge, total distance covered equals the sum of the train's length and the platform's length.
- Boat speeds combine vectorially: Downstream speed D = u + v and Upstream speed U = u - v, yielding still water speed u = (D + U)/2 and stream speed v = (D - U)/2.
Speed, Distance, Time, Trains & Boats/Streams
Kinematics word problems are central to SBI PO Quantitative Aptitude. Whether analyzing two trains crossing on parallel tracks or calculating river current velocity against a boat's rowing speed, precision in unit conversions and relative motion principles is required.
1. Kinematics Fundamentals & Unit Conversions
Core Equation
Unit Conversion Formulas
- To convert $\text{km/h}$ to $\text{m/s}$, multiply by $\frac{5}{18}$:
- To convert $\text{m/s}$ to $\text{km/h}$, multiply by $\frac{18}{5}$.
Average Speed Formulas
- Equal Distances: If a body covers distance $d$ at speed $s_1$ and another equal distance $d$ at speed $s_2$, average speed is the Harmonic Mean:
- Equal Time Intervals: If a body travels for time $t$ at speed $s_1$ and for time $t$ at speed $s_2$, average speed is the Arithmetic Mean:
2. Relative Speed & Problems on Trains
Relative Speed Principles
- Opposite Directions: When two bodies move toward each other at speeds $s_1$ and $s_2$:
- Same Direction: When two bodies move in the same direction at speeds $s_1 > s_2$:
Train Distance Scenarios
| Object Passed by Train (Length $L_1$, Speed $S_1$) | Effective Distance Covered ($D$) | Relative Speed ($S_{\text{rel}}$) |
|---|---|---|
| Stationary Pole / Tree / Standing Person | $L_1$ | $S_1$ |
| Platform / Bridge / Tunnel (Length $L_2$) | $L_1 + L_2$ | $S_1$ |
| Moving Person (Speed $S_2$, Same Direction) | $L_1$ | $S_1 - S_2$ |
| Moving Person (Speed $S_2$, Opposite Direction) | $L_1$ | $S_1 + S_2$ |
| Another Train (Length $L_2$, Speed $S_2$, Opposite) | $L_1 + L_2$ | $S_1 + S_2$ |
| Another Train (Length $L_2$, Speed $S_2$, Same) | $L_1 + L_2$ | $S_1 - S_2$ |
3. Boats and Streams Vector Mechanics
Let speed of boat in still water $= u \text{ km/h}$, and speed of stream current $= v \text{ km/h}$.
- Downstream Speed ($D$): Moving with the stream current:
- Upstream Speed ($U$): Moving against the stream current:
- Speed of Boat in Still Water ($u$):
- Speed of Stream Current ($v$):
4. Step-by-Step Worked Math Examples
Worked Example 1: Boat Downstream & Upstream Round Trip
Problem: A boat travels 24 km downstream in 2 hours and covers the same distance upstream in 4 hours. Find the speed of the boat in still water and the speed of the stream current.
Step-by-Step Solution:
- Calculate Downstream Speed $D = \frac{\text{Distance}}{\text{Time}} = \frac{24}{2} = 12 \text{ km/h}$.
- Calculate Upstream Speed $U = \frac{\text{Distance}}{\text{Time}} = \frac{24}{4} = 6 \text{ km/h}$.
- Speed of boat in still water $u = \frac{D + U}{2} = \frac{12 + 6}{2} = \frac{18}{2} = 9 \text{ km/h}$.
- Speed of stream current $v = \frac{D - U}{2} = \frac{12 - 6}{2} = \frac{6}{2} = 3 \text{ km/h}$.
Worked Example 2: Two Trains Crossing Each Other
Problem: Two trains of lengths 140 meters and 160 meters are running on parallel tracks in opposite directions at speeds of 60 km/h and 48 km/h respectively. In how many seconds will they cross each other completely?
Step-by-Step Solution:
- Total distance to be covered $D = L_1 + L_2 = 140 + 160 = 300 \text{ meters}$.
- Since moving in opposite directions, relative speed $S_{\text{rel}} = S_1 + S_2 = 60 + 48 = 108 \text{ km/h}$.
- Convert relative speed to m/s: $108 \times \frac{5}{18} = 6 \times 5 = 30 \text{ m/s}$.
- Time required to cross $= \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{300}{30} = 10 \text{ seconds}$.
Worked Example 3: Train Crossing Platform & Moving Man
Problem: A train running at 72 km/h crosses a 200-meter long platform in 22 seconds. How long will it take to cross a man running at 18 km/h in the same direction?
Step-by-Step Solution:
- Convert train speed: $72 \times \frac{5}{18} = 20 \text{ m/s}$.
- Let length of train be $L$.
- When crossing platform, distance $= L + 200$, time $= 22 \text{ s}$.
- $L + 200 = \text{Speed} \times \text{Time} = 20 \times 22 = 440 \text{ m} \implies L = 440 - 200 = 240 \text{ meters}$.
- Convert man's speed: $18 \times \frac{5}{18} = 5 \text{ m/s}$.
- Relative speed when moving in same direction $= 20 - 5 = 15 \text{ m/s}$.
- Distance to cross man $= L = 240 \text{ meters}$.
- Time taken to cross man $= \frac{240}{15} = 16 \text{ seconds}$.
5. Exam Pitfalls & Calculation Traps
- Forgetting Unit Conversion: Never mix km/h with meters or seconds. Convert all quantities to m/s and meters before calculating time.
- Ignoring Platform Length: When a train crosses a bridge or tunnel, distance is $L_{\text{train}} + L_{\text{platform}}$, not just $L_{\text{platform}}$.
A boat can travel 36 km downstream in 3 hours and 24 km upstream in 4 hours. What is the speed of the stream current?
Two trains 150 m and 100 m long are moving on parallel tracks in opposite directions at 45 km/h and 45 km/h respectively. How long will they take to clear each other?
A man covers a certain distance at 20 km/h and returns to the starting point at 30 km/h. What is his average speed for the entire journey?
A train 180 meters long takes 18 seconds to cross a platform. If the speed of the train is 54 km/h, what is the length of the platform?