Time, Speed and Distance
Overview
Time, Speed and Distance (TSD) is a high-frequency arithmetic topic in SBI Clerk Prelims, appearing both as standalone questions and within Data Interpretation sets. The core relationship is deceptively simple—Distance = Speed × Time—but examiners test your ability to apply this formula across varied scenarios: relative motion, unit conversions, and special cases like trains and boats.
Mastery here offers quick marks since these problems follow predictable patterns. The key is building speed through formula recall and recognizing problem types instantly.
Focus on unit conversion (km/hr to m/s and vice versa), relative speed concepts, and the standard boat-stream formulas. These three areas cover nearly every TSD question at Clerk level.
Key Concepts
- Fundamental relationship: Distance = Speed × Time. Rearrange as needed: Speed = Distance/Time, Time = Distance/Speed.
- Unit conversion is non-negotiable: Most train problems give speed in km/hr but lengths in meters. Converting incorrectly is the #1 source of errors.
- Relative speed determines time in motion problems: When two objects move, what matters is how fast the gap between them changes, not their individual speeds.
- Same direction = subtract speeds; Opposite direction = add speeds: This applies to trains crossing each other and any two-body motion problem.
- Train crosses an object by covering (Train length + Object length): The "distance" in train problems is always the total length the train must travel for complete crossing.
- Still water speed vs. current speed: A boat has an inherent speed in still water. The stream either aids (downstream) or opposes (upstream) this speed.
- Average speed ≠ arithmetic mean of speeds: When covering equal distances at different speeds, use the harmonic mean formula, not simple averaging.
Formulas / Key Facts
| Formula | Context |
|---|---|
| Speed = Distance / Time | Core formula—memorize all three rearrangements |
| km/hr to m/s: multiply by 5/18 | 72 km/hr = 72 × 5/18 = 20 m/s |
| m/s to km/hr: multiply by 18/5 | 25 m/s = 25 × 18/5 = 90 km/hr |
| Relative speed (same direction) = S₁ − S₂ | Faster object catching up with slower |
| Relative speed (opposite direction) = S₁ + S₂ | Objects approaching each other |
| Time for train to cross pole = Train length / Speed | Pole/person has negligible length |
| Time for train to cross platform = (Train length + Platform length) / Speed | Both lengths matter |
| Time for two trains to cross = (L₁ + L₂) / Relative speed | Use sum or difference based on direction |
| Downstream speed = Boat speed + Stream speed | Current aids the boat |
| Upstream speed = Boat speed − Stream speed | Current opposes the boat |
| Speed in still water = (Downstream + Upstream) / 2 | Derived from adding the two equations |
| Speed of stream = (Downstream − Upstream) / 2 | Derived from subtracting the two equations |
| Average speed for equal distances = 2S₁S₂ / (S₁ + S₂) | Harmonic mean of two speeds |
Worked Examples
Example 1: Basic Train Crossing a Platform
Problem: A train 150 m long crosses a 250 m platform in 20 seconds. Find its speed in km/hr.
Solution:
- Total distance covered = Train length + Platform length = 150 + 250 = 400 m
- Time = 20 seconds
- Speed = 400/20 = 20 m/s
- Convert to km/hr = 20 × 18/5 = 72 km/hr
Example 2: Two Trains Crossing Each Other
Problem: Two trains of lengths 120 m and 180 m are running in opposite directions at 54 km/hr and 36 km/hr. How long will they take to cross each other completely?
Solution:
- Total distance = 120 + 180 = 300 m
- Relative speed (opposite direction) = 54 + 36 = 90 km/hr
- Convert to m/s = 90 × 5/18 = 25 m/s
- Time = 300/25 = 12 seconds
Example 3: Boats and Streams
Problem: A boat covers 24 km upstream in 4 hours and 36 km downstream in 3 hours. Find the speed of the boat in still water and the speed of the stream.
Solution:
- Upstream speed = 24/4 = 6 km/hr
- Downstream speed = 36/3 = 12 km/hr
- Speed in still water = (12 + 6)/2 = 9 km/hr
- Speed of stream = (12 − 6)/2 = 3 km/hr
Example 4: Average Speed
Problem: A person travels from A to B at 40 km/hr and returns at 60 km/hr. Find the average speed for the entire journey.
Solution:
- Since distances are equal, use harmonic mean
- Average speed = 2 × 40 × 60 / (40 + 60) = 4800/100 = 48 km/hr
- Note: Simple average would give 50 km/hr—this is wrong!
Common Mistakes
- Forgetting to add train length in platform problems → The train hasn't "crossed" until its tail clears the platform. Always add both lengths.
- Using km/hr directly with meter-based lengths → You'll get nonsense answers. Convert speed to m/s first when lengths are in meters.
- Adding speeds when trains move in the same direction → Same direction means the faster train slowly overtakes. Subtract speeds to get relative speed.
- Confusing upstream and downstream formulas → Remember: downstream = going with the flow = faster = add. Upstream = against = slower = subtract.
- Calculating average speed as (S₁ + S₂)/2 → This only works when time spent at each speed is equal, not when distance is equal. For equal distances, always use 2S₁S₂/(S₁ + S₂).
- Ignoring the direction of current in round-trip problems → A boat going upstream and returning covers the same distance but at different speeds—you need both calculations.
Quick Reference
- D = S × T — the foundation of every problem
- km/hr → m/s: × 5/18 | m/s → km/hr: × 18/5
- Opposite direction: ADD speeds | Same direction: SUBTRACT speeds
- Train crossing anything = (Train length + Object length) / Speed
- Still water speed = (Down + Up) / 2 | Stream speed = (Down − Up) / 2
- Average speed for equal distances = 2ab/(a+b), never (a+b)/2