Time, Speed and Distance
Overview
Time, Speed and Distance (TSD) is a core quantitative aptitude topic that appears consistently in TNPSC Group II/IIA prelims. Questions typically involve trains crossing platforms or other trains, boats moving in streams, and problems requiring calculation of average speed over multiple journeys. Mastering TSD is essential because it forms the foundation for related topics like races, circular motion, and clock problems.
This topic tests your ability to apply basic formulas while handling unit conversions and relative motion concepts. TNPSC questions usually involve moderate calculations, so speed and accuracy in setting up equations matter more than complex arithmetic. Students who build strong conceptual clarity here find it easier to tackle Data Interpretation problems involving speed-time-distance scenarios.
Key Concepts
- Basic Relationship: Speed = Distance/Time. This can be rearranged to Distance = Speed × Time and Time = Distance/Speed. All TSD problems reduce to manipulating this relationship.
- Unit Conversion: To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5. This conversion appears in almost every train problem.
- Relative Speed (Same Direction): When two objects move in the same direction, relative speed = difference of their speeds. The faster object "gains" on the slower one.
- Relative Speed (Opposite Direction): When two objects move towards each other or away from each other, relative speed = sum of their speeds.
- Average Speed: For equal distances at different speeds, average speed = 2ab/(a+b) where a and b are the two speeds. This is the harmonic mean, not the arithmetic mean.
- Train Problems: A train covers its own length plus the length of the object it crosses. For a pole/person, distance = train length. For a platform/bridge, distance = train length + platform length.
- Boats and Streams: Downstream speed = boat speed + stream speed. Upstream speed = boat speed − stream speed. The boat's speed in still water is the average of downstream and upstream speeds.
- Meeting Point: When two objects start simultaneously towards each other, time to meet = total distance/(sum of speeds).
Formulas / Key Facts
| Formula | Context |
|---|---|
| Speed = Distance/Time | Fundamental relationship |
| km/hr × (5/18) = m/s | Convert km/hr to m/s |
| m/s × (18/5) = km/hr | Convert m/s to km/hr |
| Time to cross pole = L/S | L = train length, S = train speed |
| Time to cross platform = (L + P)/S | P = platform length |
| Time for two trains to cross = (L₁ + L₂)/Relative Speed | Use sum or difference based on direction |
| Downstream speed = B + S | B = boat speed, S = stream speed |
| Upstream speed = B − S | Boat moves against current |
| Boat speed in still water = (Downstream + Upstream)/2 | Finding boat's own speed |
| Stream speed = (Downstream − Upstream)/2 | Finding current speed |
| Average speed (equal distances) = 2ab/(a + b) | Not simple average |
Worked Examples
Example 1: Train Crossing a Platform
A train 150 m long crosses a platform 250 m long in 20 seconds. Find the speed of the train in km/hr.
Solution:
- Total distance covered = Train length + Platform length = 150 + 250 = 400 m
- Time = 20 seconds
- Speed = 400/20 = 20 m/s
- Converting to km/hr = 20 × (18/5) = 72 km/hr
Example 2: Two Trains Moving in Opposite Directions
Two trains of lengths 120 m and 80 m are running in opposite directions at 50 km/hr and 40 km/hr. In what time will they completely cross each other?
Solution:
- Total distance to cross = 120 + 80 = 200 m
- Relative speed (opposite direction) = 50 + 40 = 90 km/hr
- Converting to m/s = 90 × (5/18) = 25 m/s
- Time = 200/25 = 8 seconds
Example 3: Boat and Stream
A boat travels 24 km downstream in 3 hours and returns upstream in 4 hours. Find the speed of the boat in still water and the speed of the stream.
Solution:
- Downstream speed = 24/3 = 8 km/hr
- Upstream speed = 24/4 = 6 km/hr
- Boat speed in still water = (8 + 6)/2 = 7 km/hr
- Stream speed = (8 − 6)/2 = 1 km/hr
Example 4: 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 distances are equal, average speed = 2ab/(a + b)
- Average speed = (2 × 40 × 60)/(40 + 60) = 4800/100 = 48 km/hr
- Note: Simple average would wrongly give 50 km/hr
Common Mistakes
- Using arithmetic mean for average speed → When distances are equal, always use the harmonic mean formula 2ab/(a+b). Arithmetic mean applies only when time spent at each speed is equal.
- Forgetting train length in crossing problems → When a train crosses any object, it must travel its own length completely past the object. Always add train length to stationary object length.
- Wrong relative speed direction → Students often add speeds when trains move in the same direction. Remember: same direction = subtract, opposite direction = add.
- Mixing units without conversion → Train lengths come in metres, speeds often in km/hr. Always convert to consistent units (typically m/s for train problems) before calculating.
- Confusing boat speed with downstream/upstream speed → Boat speed in still water is the boat's own capacity. Downstream and upstream speeds are effective speeds after accounting for current.
- Assuming upstream time equals downstream time → Upstream always takes longer for the same distance. If a question gives total time for round trip, set up equations for both legs separately.
Quick Reference
- km/hr to m/s: multiply by 5/18
- m/s to km/hr: multiply by 18/5
- Train crosses pole: Time = (Train length)/Speed
- Two trains crossing: Distance = Sum of both lengths
- Boat in still water: (Downstream speed + Upstream speed)/2
- Average speed for equal distances: 2ab/(a+b), never (a+b)/2