Time, Speed and Distance
Overview
Time, Speed and Distance (TSD) is a cornerstone topic in the TN TET Mathematics section, appearing consistently in both Paper I and Paper II. This topic tests your ability to apply a single fundamental relationship across varied real-world contexts—moving trains, boats in streams, and journeys with varying speeds.
Mastery here requires comfort with unit conversions, the ability to set up equations from word problems, and recognising which formula variant applies to a given scenario. The good news: once you internalise the core formula and its rearrangements, every problem becomes a structured exercise rather than a puzzle. Expect 2–4 questions from this area, often combining TSD with ratio or algebra concepts.
Key Concepts
- **The fundamental relationship**: Distance = Speed × Time. All TSD problems derive from rearranging this single equation.
- **Unit consistency is critical**: Speed in km/hr with time in hours gives distance in km. Mixing units (minutes with km/hr) is the most common source of errors.
- **Conversion factors**: To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.
- **Relative speed for same-direction motion**: When two objects move in the same direction, relative speed = difference of their speeds.
- **Relative speed for opposite-direction motion**: When objects move towards each other, relative speed = sum of their speeds.
- **Boats and streams**: Downstream speed = boat speed + stream speed. Upstream speed = boat speed − stream speed.
- **Average speed for equal distances**: When the same distance is covered at two different speeds, average speed = 2ab/(a+b), not the arithmetic mean.
- **Train problems are length problems**: A train crossing a platform or another train must cover the combined lengths of both objects.
Formulas / Key Facts
| Formula | Context | |---------|---------| | Distance = Speed × Time | Core relationship | | Speed = Distance / Time | Finding rate of travel | | Time = Distance / Speed | Finding duration | | km/hr to m/s: multiply by 5/18 | Unit conversion | | m/s to km/hr: multiply by 18/5 | Unit conversion | | Downstream speed = B + S | Boat speed B, stream speed S | | Upstream speed = B − S | Boat speed B, stream speed S | | Boat speed in still water = (Downstream + Upstream)/2 | Deriving boat's own speed | | Stream speed = (Downstream − Upstream)/2 | Deriving current speed | | Average speed (equal distances) = 2ab/(a+b) | Speeds a and b for same distance | | Time to cross = (Length of train + Length of object) / Speed | Train crossing platform/bridge/another train | | Time to cross a person/pole = Length of train / Speed | Object of negligible length |
Worked Examples
### Example 1: Basic Train Problem *A train 150 m long passes a pole in 10 seconds. Find its speed in km/hr.*
**Solution:**
- Distance covered = length of train = 150 m
- Time = 10 s
- Speed = 150/10 = 15 m/s
- Converting to km/hr: 15 × 18/5 = 54 km/hr
**Answer: 54 km/hr**
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### Example 2: Boat and Stream *A boat travels 24 km downstream in 2 hours and returns upstream in 3 hours. Find the speed of the boat in still water and the speed of the stream.*
**Solution:**
- Downstream speed = 24/2 = 12 km/hr
- Upstream speed = 24/3 = 8 km/hr
- Boat speed in still water = (12 + 8)/2 = 10 km/hr
- Stream speed = (12 − 8)/2 = 2 km/hr
**Answer: Boat speed = 10 km/hr, Stream speed = 2 km/hr**
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### Example 3: Average Speed *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 distance is same both ways, use the formula: Average speed = 2ab/(a+b)
- Average speed = 2 × 40 × 60 / (40 + 60)
- = 4800/100 = 48 km/hr
**Note:** The arithmetic mean would give 50 km/hr—this is wrong because more time is spent at the slower speed.
**Answer: 48 km/hr**
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### Example 4: Two Trains Crossing Each Other *Two trains of lengths 120 m and 80 m are running towards each other at 50 km/hr and 40 km/hr. In how many seconds will they cross each other?*
**Solution:**
- Total distance to cover = 120 + 80 = 200 m
- Relative speed (opposite directions) = 50 + 40 = 90 km/hr
- Convert to m/s: 90 × 5/18 = 25 m/s
- Time = 200/25 = 8 seconds
**Answer: 8 seconds**
Common Mistakes
- **Forgetting to add lengths in train problems** → When a train crosses a platform or another train, you must add both lengths. Only when crossing a pole or person (negligible length) do you use train length alone.
- **Using arithmetic mean for average speed** → Students often calculate (speed₁ + speed₂)/2. This is correct only when equal time is spent at each speed. For equal distances, always use 2ab/(a+b).
- **Mixing units without conversion** → A speed of 72 km/hr used directly with time in seconds will give wrong answers. Convert 72 km/hr to 20 m/s first (72 × 5/18 = 20).
- **Confusing relative speed direction** → Same direction: subtract speeds. Opposite direction: add speeds. Visualise the movement to avoid sign errors.
- **Ignoring stream direction in boat problems** → Downstream means water helps (add stream speed). Upstream means water opposes (subtract stream speed). Reversing these destroys the calculation.
Quick Reference
- **D = S × T** — the one formula that solves everything.
- **km/hr to m/s: × 5/18 | m/s to km/hr: × 18/5**
- **Opposite direction: add speeds | Same direction: subtract speeds**
- **Boat still water speed = (downstream + upstream)/2**
- **Average speed for equal distances = 2ab/(a+b), never (a+b)/2**
- **Train crossing anything = (train length + object length) / speed**