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Time & Distance

Chapter Outline & Learning Goals

By the end of this chapter, you will be able to:

1. Convert speeds between km/h and m/s in under 3 seconds.
2. Solve problems on trains, boats & streams, and races without writing long equations.
3. Handle relative speed, average speed, and circular track questions confidently.
4. Spot the traps examiners set in SSC CGL Tier-II, IBPS PO Mains, CAT, and CSAT papers.

The Hook: A Train, A Pole, and 9 Seconds

A 180-metre-long train crosses a pole in 9 seconds. That's it — that's the whole question. And from this one line, the examiner expects you to find the train's speed in km/h.

Most students freeze. Not because the maths is hard, but because nobody told them the secret: when a train crosses a pole, it travels a distance equal to its own length. That's all. 180 metres in 9 seconds = 20 m/s = 72 km/h. Done in 10 seconds flat.

Time & Distance is not a chapter of formulas. It is a chapter of *one* formula wearing a hundred disguises. Once you see through the disguises, this becomes the highest-scoring topic in your quant paper.

Why This Chapter Feels Hard (And Why It Isn't)

Let's be honest. You've probably memorised "Speed = Distance/Time" a dozen times, and still blanked out in the exam hall when the question said *"a man covers two-thirds of the journey at 4 km/h and the rest at 5 km/h..."*

The struggle is real, and here's why: school taught you the formula, but exams test the relationships — what happens to time when speed doubles, what "relative" really means when two objects move, why average speed is *not* the average of speeds. This chapter fixes exactly that.

Chunk 1: The One Formula and Its Three Faces

Everything in this chapter comes from:

Speed = Distance ÷ Time

Rearranged: Distance = Speed × Time and Time = Distance ÷ Speed

But the real power lies in the proportions:

• Speed constant → Distance ∝ Time (double the time, double the distance)
• Time constant → Distance ∝ Speed
• Distance constant → Speed ∝ 1/Time (double the speed, *half* the time)

That last one is the examiner's favourite. If a man increases his speed in the ratio 5:6, his time ratio becomes 6:5. No calculation needed — just flip the ratio.

🧠 Active Recall: A car covers a fixed distance. If its speed becomes 3/4 of the original, what happens to the time taken? *(Answer: time becomes 4/3 of the original — flip the ratio.)*

Chunk 2: Unit Conversion — The 5/18 Reflex

Examiners love mixing km/h and m/s. Make this conversion a reflex:

• km/h → m/s: multiply by 5/18
• m/s → km/h: multiply by 18/5

Why 5/18? Because 1 km = 1000 m and 1 hour = 3600 s, so 1 km/h = 1000/3600 = 5/18 m/s.

Quick anchors to memorise: 36 km/h = 10 m/s, 54 km/h = 15 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s. Most train problems use exactly these numbers.

🧠 Active Recall: Convert 45 km/h to m/s without a pen. *(45 × 5/18 = 12.5 m/s.)*

Chunk 3: Average Speed — The Trap Everyone Falls Into

Average speed is NOT the average of the two speeds. Repeat that aloud.

If a man travels equal distances at speeds *a* and *b*:

Average Speed = 2ab / (a + b) (the harmonic mean)

Example: 60 km at 30 km/h, then 60 km at 60 km/h. Average = (2 × 30 × 60)/(30 + 60) = 3600/90 = 40 km/h — not 45!

If he travels for equal *times* instead, then average speed is simply (a + b)/2. The examiner's trap: mixing up "equal distance" with "equal time." Read the question twice.

For three equal distances at speeds a, b, c: Average Speed = 3abc / (ab + bc + ca)

🧠 Active Recall: A cyclist goes to market at 10 km/h and returns the same way at 15 km/h. Average speed? *(2×10×15/25 = 12 km/h.)*

Chunk 4: Trains — Length Is Everything

Train problems have one golden rule: identify what distance the train actually covers.

Relative speed: opposite directions → add speeds; same direction → subtract.

Example: Two trains of 120 m and 180 m move towards each other at 40 km/h and 50 km/h. Time to cross each other = (120 + 180) / [(40 + 50) × 5/18] = 300 / 25 = 12 seconds.

🧠 Active Recall: A 250 m train crosses a 150 m platform in 20 s. Speed in km/h? *(Distance = 400 m; speed = 20 m/s = 72 km/h.)*

Chunk 5: Boats & Streams — The River Helps or Fights

• Downstream speed (with the current) = boat speed + stream speed = B + S
• Upstream speed (against the current) = B − S

From these two, everything follows:

• B = (Downstream + Upstream) / 2
• S = (Downstream − Upstream) / 2

Example: A boat goes 30 km downstream in 2 h and the same distance upstream in 3 h. Downstream = 15 km/h, upstream = 10 km/h. So B = 12.5 km/h, S = 2.5 km/h.

🧠 Active Recall: If downstream speed is 18 km/h and the stream is 3 km/h, what is the upstream speed? *(B = 15, so upstream = 12 km/h.)*

Chunk 6: Races & Circular Tracks

Race language decoded:

• "A beats B by 20 m in a 100 m race" → when A runs 100 m, B runs 80 m. Their speed ratio is 100:80 = 5:4.
• "A gives B a start of 10 m" → B starts 10 m ahead, so B runs only 90 m.

Circular tracks: Two runners on a circular track of length L:

• Same direction: they meet every L / (difference of speeds)
• Opposite directions: they meet every L / (sum of speeds)
• Time to meet at the starting point = LCM of their individual lap times

🧠 Active Recall: In a 500 m race, A beats B by 50 m. If A's speed is 10 m/s, what is B's speed? *(Ratio 500:450 = 10:9, so B = 9 m/s.)*

🎯 Mnemonic: "DST — Don't Stop Thinking"

Remember the formula triangle D-S-T: cover the one you want.

• Cover D → D = S × T
• Cover S → S = D / T
• Cover T → T = D / S

And for unit conversion, remember "5/18 is small, 18/5 is tall" — going from big units (km/h) to small (m/s), multiply by the small fraction 5/18; going the other way, multiply by the tall one, 18/5.

Summary Table

📌 Review this chapter after 2 days. Redo the Active Recall prompts without looking at the answers, and re-derive the average speed formula once from scratch. Spaced repetition is what moves this chapter from "I read it" to "I own it."