Time, Speed & Distance

Time, Speed & Distance

🟢 Lite

Key Rule / Formula

Speed = Distance/Time. Average speed (same distance at speeds a and b) = 2ab/(a+b). For trains: Time to cross a pole = Length of train / Speed. Time to cross a platform = (Length of train + Length of platform) / Speed.

Memory Trick

“Meeting trains” = add lengths, relative speed = sum of speeds. For trains moving in opposite directions, relative speed = S₁ + S₂; same direction = |S₁ − S₂|.

1-Sentence Summary

All TSD problems stem from Speed = Distance/Time; convert between them fluently and remember that average speed is harmonic mean when the same distance is covered at two different speeds.

Quick Example

Q: A train 150m long crosses a pole in 10 seconds. Find its speed in km/h. A: 54 km/h — 150/10 = 15 m/s × (18/5) = 54 km/h.

🟡 Standard

Concept

Speed, time, and distance are related by the fundamental equation: Distance = Speed × Time. From this single equation, you can derive everything. The trick in SSC is knowing which form to use and when. Average speed deserves special attention — many students wrongly take arithmetic mean of two speeds, but this is only correct when equal time is spent at each speed, not equal distance.

Train Problems: The most distinctive TSD question type in SSC. When a train crosses a stationary object (pole, person, tree), the distance covered equals the train’s own length. When it crosses a platform or another train, distance = sum of both lengths. Always convert speed to m/s when mixing with lengths in metres.

Conversions: 1 km/h = (1000m)/(3600s) = 5/18 m/s. Conversely, 1 m/s = 18/5 = 3.6 km/h.

Boats and Streams: Upstream speed = Boat speed in still water − Stream speed. Downstream speed = Boat speed + Stream speed. The current (stream) subtracts upstream and adds downstream.

Key Points

• When same distance d is covered at speeds a and b: Average speed = 2ab/(a+b).
• When same time t is spent at speeds a and b: Average speed = (a+b)/2.
• For boats: Downstream speed − Upstream speed = 2 × Stream speed. Downstream speed + Upstream speed = 2 × Boat speed.
• If speed changes in arithmetic progression over equal time intervals, average speed = (first + last)/2.
• For two trains moving in opposite directions: relative speed = S₁ + S₂. Same direction: |S₁ − S₂|.

Worked Example

Q: A boat covers a fixed distance of 48 km downstream in 4 hours and the same 48 km upstream in 6 hours. Find the speed of the stream and the speed of the boat in still water.

Approach: When a problem fixes one distance and gives both times, compute the two speeds directly and then split them into the boat and stream components.

1. Downstream speed = distance ÷ time = 48 ÷ 4 = 12 km/h.
2. Upstream speed = distance ÷ time = 48 ÷ 6 = 8 km/h.
3. Downstream speed = b + s and upstream speed = b − s, where b is the boat’s speed in still water and s is the stream speed.
4. Adding the two: (b + s) + (b − s) = 12 + 8, so 2b = 20 and b = 10 km/h.
5. Subtracting the two: (b + s) − (b − s) = 12 − 8, so 2s = 4 and s = 2 km/h.

Answer: Stream speed = 2 km/h, boat speed in still water = 10 km/h. This is the standard template for every boats-and-streams question: convert each leg to a speed, then use the half-sum for the boat and the half-difference for the stream.

SSC Pattern / Tips

• Always draw a diagram for train problems — label train length, platform length, and the gap between trains.
• For upstream/downstream, the key is that the current’s effect is additive in one direction and subtractive in the other.
• When a train overtakes another, relative speed = difference of speeds (same direction).
• Average speed formula 2ab/(a+b) only applies for equal distances at speeds a and b.

🔴 Extended

Full Concept

Advanced Train Problems: Two trains of lengths L₁ and L₂ running at speeds S₁ and S₂: time to cross each other completely (from front meeting to rear separation) = (L₁ + L₂) / (S₁ + S₂) for opposite directions, or (L₁ + L₂) / |S₁ − S₂| for same direction. The key insight: when trains cross each other in opposite directions, they effectively “add” their lengths relative to each other. When one overtakes another, the relative speed is the difference.

Meeting Point Problems: If two objects start from points A and B towards each other, their meeting point divides the distance AB in the ratio of their speeds. If they start at the same time: Distance covered by A from start to meeting / Distance covered by B from start to meeting = Speed of A / Speed of B. If one starts earlier, adjust for time difference first.

Circular Tracks: When two runners run on a circular track of length L at speeds u and v: time to meet = L / (u+v) (opposite directions), or L / |u−v| (same direction). Number of meetings in time t = (u+v)t/L (opposite) or |u−v|t/L (same).

Acceleration-Based Problems: Some Tier 2 questions involve speed changing at constant acceleration. Use: v = u + at, s = ut + ½at², v² = u² + 2as. These come from physics but are solved algebraically.

Speed Ratio and Time Ratio: If A is x times as fast as B, then A covers the same distance in 1/x of B’s time. This inverse relationship between speed and time is fundamental.

SSC CGL Deep Analysis

• Frequency: 1–2 questions per paper. Train problems with platform crossing and boats/stream appear every year.
• Difficulty: Medium. Train problems with multiple stages (stop, change speed) and circular track questions are the hardest.
• Recent trend: Questions involving trains starting from two stations toward each other, meeting time calculation, and then one train stopping at a station.
• Newer patterns: “A train starts from A at 60 km/h. After 30 minutes, another train starts from A at 80 km/h. After how much time does the second train catch up?” This delayed start pattern.
• Total weight in Tier 2: Roughly 2–3% of the quant paper.

High-Scoring Strategy

1. For any train problem, identify: is it crossing a stationary object (pole) or a moving object (another train)? This determines whether to add lengths.
2. For delayed start problems: first train’s head start distance = speed × time difference. Catch-up time = head start distance / relative speed.
3. In circular track problems, two runners meet again at the starting point when (u+v)t/L is a whole number for both runners; the number of distinct meeting points on the track is governed by the ratio of their speeds.
4. For boats/stream: always state b+s and b-s clearly before solving. Write “boat speed in still water = b, stream = s.”
5. Average speed for unequal distances: use 2ab/(a+b). For unequal times: use (a+b)/2. Know which applies.
6. When converting m/s to km/h, multiply by 18/5. When converting km/h to m/s, multiply by 5/18.

SSC-Level Practice

Q1: Two trains 200m and 300m long run at 45 km/h and 60 km/h respectively in opposite directions. How long do they take to cross each other? Answer: ≈ 17.14 seconds — Working: Total length to clear = 200 + 300 = 500 m. Since the trains move in opposite directions, relative speed = sum of speeds = 45 + 60 = 105 km/h. Convert: 105 × 5/18 = 525/18 = 175/6 m/s. Time = total length ÷ relative speed = 500 ÷ (175/6) = 500 × 6/175 = 3000/175 = 17.14 seconds. (Cross-check via hours: 0.5 km ÷ 105 km/h = 1/210 h = 3600/210 = 17.14 s.)

Q2: A man rows upstream at 8 km/h and downstream at 14 km/h. Find his speed in still water and speed of stream. Answer: Boat speed = 11 km/h, Stream speed = 3 km/h — Working: Boat + stream = 14, Boat − stream = 8. Adding: 2 Boat = 22 → Boat = 11. Subtracting: 2 Stream = 6 → Stream = 3.

Common Traps

• Trap 1: Confusing when to add speeds vs subtract speeds. Same direction = relative speed is difference; opposite direction = sum.
• Trap 2: Using arithmetic mean for average speed when distances are unequal. Average speed = total distance / total time, not (S₁+S₂)/2 unless times are equal.
• Trap 3: Forgetting to convert km/h to m/s when the distance is given in metres and time in seconds. Mixing units gives wrong answers.

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