Time, Speed and Distance: Formulas, Shortcuts and Worked Problems

Time, speed and distance is one of the highest-frequency topics across campus placement aptitude tests, from TCS NQT to AMCAT to company-specific rounds.

The core relationship is a single equation with three rearrangements. Get that right with clean unit handling, and most TSD problems reduce to two or three arithmetic steps. TSD shares the same ratio-and-proportionality logic with time and work problems, so practice time here multiplies across both topics.

The core formula and unit conversions

Three rearrangements of one equation cover every TSD problem:

• Speed = Distance / Time
• Distance = Speed x Time
• Time = Distance / Speed

Handling units

Exam problems mix km/h and m/s constantly. Two multipliers handle every conversion:

Convert from	To	Multiply by	Example
km/h	m/s	5/18	72 km/h = 20 m/s
m/s	km/h	18/5	10 m/s = 36 km/h

Quick check: km/h values are always numerically larger than their m/s equivalents. If your conversion gives a smaller km/h result, you applied the multiplier backwards.

Proportional relationships

Three patterns follow directly from the core formula:

• Same speed: distance is proportional to time.
• Same time: distance is proportional to speed.
• Same distance: time is inversely proportional to speed.

The third catches students most often. Double the speed, halve the time. Triple the speed, cut time to a third. They move in opposite directions.

Five formula shortcuts

1. Average speed for equal distances

When you travel from A to B at speed u and return from B to A at speed v, the average speed over the full trip is:

Average speed = 2uv / (u + v)

This is the harmonic mean. The arithmetic mean (u + v) / 2 overestimates because you spend more time at the slower speed. A worked example:

• Example: A to B at 40 km/h, return at 60 km/h.
• Harmonic mean (correct): 2 x 40 x 60 / (40 + 60) = 4800 / 100 = 48 km/h.
• Arithmetic mean (wrong): (40 + 60) / 2 = 50 km/h.
• The 2 km/h gap matters when verifying your answer under time pressure.

2. Relative speed

Direction of motion	Relative speed
Same direction	Absolute difference: faster minus slower speed
Opposite directions	Sum: u + v

Two objects approaching each other close the gap at the sum of their speeds. Two objects moving in the same direction open or close the gap at only the difference.

3. Ratio proportionality

Same travel time: distances are in the same ratio as speeds. Two points to remember:

• Train A at 60 km/h, train B at 80 km/h, same duration: distance ratio = 3:4 (matching the speed ratio).
• Same distance: travel times are in the inverse ratio. If speed ratio is 3:4, time ratio is 4:3.

4. Train problems

Treat the train as a physical object with length. Distance covered during a crossing is the total gap the front must travel before the tail clears the obstacle:

• Passing a pole or person: distance = train length
• Passing a platform: distance = train length + platform length
• Two trains crossing each other: distance = sum of both train lengths

5. Boats and streams

Let b = boat speed in still water, s = stream speed:

• Downstream (with current): b + s
• Upstream (against current): b - s

From measured upstream speed U and downstream speed D:

• Still water speed: b = (D + U) / 2
• Stream speed: s = (D - U) / 2

These follow from adding and subtracting the two equations above. Khan Academy’s distance-rate-time module covers the underlying algebra for those who want the full derivation.

Worked problems

All four problems are verified from first principles.

Problem 1: Speed conversion

• Q: A person crosses a 400 m street in 6 minutes. Find their speed in km/h.
• Step 1: Convert time to seconds: 6 x 60 = 360 s.
• Step 2: Speed = 400 / 360 = 10/9 m/s.
• Step 3: Convert to km/h: (10/9) x (18/5) = 180/45 = 4 km/h.
• Answer: 4 km/h.

Problem 2: Same distance, halved time

• Q: A train covers 200 km in 4 hours. At what speed must it run to cover the same 200 km in 2 hours?
• Step 1: Distance is fixed at 200 km.
• Step 2: Required speed = 200 / 2 = 100 km/h.
• Answer: 100 km/h. (Halving the time doubles the required speed, consistent with the constant-distance inverse proportionality rule.)

Problem 3: Speed ratio

• Q: Two trains run at speeds in ratio 6:8. The second train covers 300 km in 3 hours. Find the speed of the first train.
• Step 1: Speed of second train = 300 / 3 = 100 km/h.
• Step 2: Speed of first train = 100 x (6/8) = 75 km/h.
• Answer: 75 km/h.

Problem 4: Train crossing a platform

• Q: A 200 m train running at 72 km/h crosses a 400 m platform. How long does the crossing take?
• Step 1: Total distance = 200 + 400 = 600 m.
• Step 2: Speed in m/s = 72 x (5/18) = 20 m/s.
• Step 3: Time = 600 / 20 = 30 seconds.
• Answer: 30 seconds.

Two traps that cost marks

Trap 1: Arithmetic mean instead of harmonic mean

The average speed formula 2uv / (u + v) applies when the two legs cover equal distances. For equal time (not distance), the arithmetic mean (u + v) / 2 is correct. Round trips and “same route at different speeds” problems almost always involve equal distances.

Diagnostic: if the problem specifies the same route in both directions, use the harmonic mean. If it specifies the same duration for both legs, use the arithmetic mean.

• Example confirming the gap: round trip at 60 km/h one way and 40 km/h the other.
• Harmonic mean (correct): 2 x 60 x 40 / (60 + 40) = 4800 / 100 = 48 km/h.
• Arithmetic mean (wrong): (60 + 40) / 2 = 50 km/h.

Trap 2: Relative speed and direction

When two objects approach each other, relative speed is u + v. A candidate who applies the difference formula instead will compute double the actual meeting time. Before applying any relative speed formula, confirm the direction of motion explicitly.

“Moving toward each other” and “moving away from each other” both involve the closing or opening speed, which is the sum. “Moving in the same direction” uses the difference.

Drilling both traps before a campus placement aptitude test addresses the two most common formula-selection errors in TSD. Quant-heavy hirers like Mu Sigma use TSD variants that specifically target these two confusions.

---

Aptitude clears the gate into interviews. The proportional reasoning in TSD (halving travel time doubles required speed, doubling relative speed halves meeting time) is structurally identical to the throughput constraints in AI system design. If AI application roles are on your roadmap, TinkerLLM at ₹299 covers the LLM foundations that now appear in fresher AI interviews, including rate-limit reasoning and resource-constraint optimization, as a practical next step from this kind of quantitative prep.