Time, Speed and Distance for UPSC CDS — Trains, Relative Speed & Classic Traps

Concept

Time, Speed and Distance (TSD) is the backbone of roughly 4–6 questions in every CDS Mathematics paper. The core relationship is deceptively simple:

$\text{Distance} = \text{Speed} \times \text{Time}$

But the exam never tests this formula directly. It tests your ability to manipulate it under constraints — stoppages, relative motion, partial journeys, fractions of usual speed. That is where aspirants lose marks.

Think of it this way: distance is a fixed corridor. If you walk slower, it takes longer. If you sprint, you finish early. Speed and time are always pulling against each other while distance stays constant. This inverse relationship between speed and time (when distance is fixed) is the single most exploited idea in CDS TSD questions.

A useful real-world anchor: imagine two soldiers starting from opposite ends of a 800 km supply route. The moment they move toward each other, they are consuming the gap at the sum of their speeds — that is relative speed in the "toward" direction. If they move in the same direction, the faster one is only gaining on the slower one at the difference of their speeds.

Unit conversions are where you bleed time in the exam hall. Burn these into muscle memory:

$1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \qquad 1 \text{ m/s} = \frac{18}{5} \text{ km/hr}$

The fraction $\frac\{5\}\{18\}$ converts km/hr to m/s. Its reciprocal $\frac\{18\}\{5\}$ does the reverse. You will use these in virtually every train problem.

There are four question archetypes in CDS TSD:

1. Speed-fraction problems — speed becomes $\frac{p}{q}$ of usual; find time gained/lost
2. Train crossing problems — crossing a pole, platform, or another train
3. Meeting/Overtaking problems — two objects starting at different times or places
4. Stoppages — effective speed vs. actual running speed

Master these four and you have covered the entire chapter.

Deep Dive

The Inverse Speed-Time Relationship

When distance is constant, speed and time are inversely proportional:

$S_1 \cdot T_1 = S_2 \cdot T_2$

If speed becomes $\frac\{3\}\{5\}$ of usual, time becomes $\frac\{5\}\{3\}$ of usual. The extra time compared to normal is:

$\Delta T = \frac{5}{3}T - T = \frac{2}{3}T$

This is the direct route to "speed-fraction" problems. If you're given the extra time ($\Delta T$), solve for $T$ immediately:

$T = \frac{3}{2} \cdot \Delta T$

Do not set up a distance equation with variables for speed and distance. That approach takes 90 seconds. The ratio approach takes 15.

Train Problems — The Length Factor

Every train problem hides an invisible variable: the length of the train (or trains). The train "passes" an object only when its entire body has cleared that object.

• Passes a pole / person standing still: Distance = Length of train
• Passes a platform / bridge: Distance = Length of train + Length of platform
• Two trains crossing (opposite directions): Distance = $L_1 + L_2$, Relative speed = $S_1 + S_2$
• Two trains crossing (same direction): Distance = $L_1 + L_2$, Relative speed = $|S_1 - S_2|$
• Train overtakes a walking person: Distance = Length of train, Relative speed = Train speed − Person speed

The formula that ties overtaking to walking persons:

$L = (v_{\text{train}} - v_{\text{person}}) \times t \times \frac{5}{18}$

where $t$ is in seconds and speeds are in km/hr.

Relative Speed — The Unifying Concept

Relative speed is just the speed of one object as seen from the reference frame of the other.

• Same direction: Relative speed = $|S_A - S_B|$
• Opposite directions: Relative speed = $S_A + S_B$

Meeting/overtaking problems collapse into simple TSD problems once you frame them in terms of relative speed and the "gap" to be covered.

Gap to be covered: If A starts $t_0$ hours before B, the head start distance is $S_A \times t_0$. B then covers this gap at a relative speed of $S_B - S_A$ (assuming $S_B > S_A$, same direction).

Time for B to catch A:

$t_{\text{catch}} = \frac{S_A \times t_0}{S_B - S_A}$

Meeting point distance from B's starting station:

$d = S_B \times t_{\text{catch}}$

Stoppages — Effective vs. Running Speed

A train runs at speed $v$ but stops for $m$ minutes every hour. Its effective speed is:

$v_{\text{eff}} = v \times \frac{60 - m}{60}$

Given $v_{\text{eff}}$, find $m$:

$m = 60 \times \frac{v - v_{\text{eff}}}{v}$

Or equivalently: stoppage time per hour = $\frac{v - v_{\text{eff}}}{v} \times 60$ minutes.

For CDS, the shortcut is even cleaner — the loss in speed directly gives you the loss in distance per hour, and at the running speed, that translates to stoppage time. No algebra needed.

The Two-Equation Setup for Train-Overtaking a Person

When a train overtakes two persons walking in the same direction, you get two equations for two unknowns (train speed and train length). Write both length equations, set them equal, solve for speed, then back-calculate length. This is one of CDS's favourite question types — the structure repeats across years.

$L = (v - u_1) \cdot t_1 \cdot \frac{5}{18} = (v - u_2) \cdot t_2 \cdot \frac{5}{18}$

The $\frac\{5\}\{18\}$ cancels from both sides, simplifying the algebra considerably.

Memory Tricks & Shortcuts

Fast-Solving Framework

Read the question and classify it in the first 10 seconds:

Is speed given as a fraction of usual? → Use the Speed-Fraction Flip. Time is the reciprocal fraction of usual. Delay or saving = fractional difference × usual time. Solve directly.

Is a train crossing something? → Identify what it is crossing. Set: Distance = train length + (platform/other train length if applicable). Relative speed = sum (opposite) or difference (same direction).

Two trains starting at different times? → Compute head-start distance. Divide by relative speed to get catch time. Multiply by the chasing train's speed for meeting distance.

Stoppages? → Loss in speed per hour is directly proportional to stoppage time. Stoppage minutes = (speed loss / running speed) × 60.

Train overtakes two walking persons? → Write two length equations, set equal, solve for train speed, then find length. Cancel the 5/18 factor before solving.

If none of the above applies clearly, fall back to: identify knowns, identify the one unknown, write D = S × T, solve. Don't reach for quadratics or simultaneous equations unless you have two genuine unknowns.

Solved PYQs

Why this question: The CDS 2023 train-overtaking question is a perfect two-equation TSD setup. It tests unit conversion awareness and the cancellation shortcut simultaneously.

Solving path: Let train speed = $v$ km/hr. Write two length expressions using relative speed × time × 5/18. Note that 5/18 appears in both — cancel it before solving. You get: $9(v-3) = \frac\{75\}\{8\}(v-4)$. Cross-multiply: $72(v-3) = 75(v-4)$ → $72v - 216 = 75v - 300$ → $3v = 84$ → $v = 28$ km/hr. Then $L = 9 \times (28-3) \times \frac\{5\}\{18\} = 9 \times 25 \times \frac\{5\}\{18\} = \frac\{1125\}\{18\} = 62.5$ m.

Why this question: The stoppages question (CDS 2026) is a one-liner if you know the shortcut — but aspirants who set up distance equations waste nearly a minute here.

Solving path: Speed without stoppages = 60 km/hr. Speed with stoppages = 48 km/hr. The train "loses" 12 km every hour due to stops. At 60 km/hr running speed, 12 km takes $\frac\{12\}\{60\} \times 60 = 12$ minutes. That is the stoppage duration per hour.

Why this question: The speed-fraction problem (CDS 2025) is the canonical test of the inverse relationship. If you don't know the ratio trick, you will set up two distance equations and burn 60+ seconds.

Solving path: Speed = 3/5 usual → time = 5/3 usual. Extra time = $\frac\{5\}\{3\}T - T = \frac\{2\}\{3\}T = 20$ min. Therefore $T = 30$ minutes. Two lines of working. Done.

Why this question: The Delhi-Hyderabad overtaking problem (CDS 2024) appears twice in the spec with minor wording differences — CDS regularly recycles this structure. It tests head-start gap computation and meeting-point distance.

Solving path: Train A leaves at 7:00 a.m., Train B at 7:50 a.m. — a 50-minute = 5/6 hour gap. Head-start distance = $80 \times \frac\{5\}\{6\} = \frac\{200\}\{3\}$ km. Relative speed (same direction) = $100 - 80 = 20$ km/hr. Time for B to catch A = $\frac{200/3}\{20\} = \frac\{10\}\{3\}$ hr. Meeting distance from Delhi = $100 \times \frac\{10\}\{3\} = \frac\{1000\}\{3\}$ km.

Why this question: The P-Q opposite-direction problem (CDS 2024) pairs a meeting-time question with a meeting-distance question — both solvable from one setup. Learn the setup once, answer two questions.

Solving path: Train A starts at 7:00 p.m., Train B at 4:00 a.m. — a 9-hour gap. A covers $60 \times 9 = 540$ km before B starts. Remaining gap = $800 - 540 = 260$ km. Closing speed = $60 + 90 = 150$ km/hr. Time after B starts = $\frac\{260\}\{150\} = \frac\{26\}\{15\}$ hr = $\frac\{26\}\{15\} \times 60 = 104$ minutes = 1 hr 44 min. Meeting time = 4:00 a.m. + 1:44 = 5:44 a.m.

Common Mistakes

• Forgetting the train's own length. When a train "passes" a platform or person, the total distance is the train length plus the platform length — or just the train length for a stationary point object. Aspirants routinely use only the platform length or only the train length. Write "D = L_train + L_obstacle" before plugging in any numbers.

• Using the wrong relative speed direction. Same direction = difference of speeds. Opposite direction = sum of speeds. A common error is adding speeds when trains move in the same direction. Quick check: if they move together, the relative speed should be small; if they head toward each other, it should be large. Sanity-check your answer.

• Applying the 5/18 conversion at the wrong stage. Either convert all speeds to m/s at the beginning, or keep them in km/hr and convert only the final length at the end. Mixing km/hr for one train and m/s for another inside the same equation produces wrong answers every time.

• Confusing "distance from Q" with "distance from P." In opposite-direction meeting problems, always state clearly which station you're computing distance from. In the P-Q problem, B travels 156 km from Q — not 644 km, not 504 km. Label your answer before computing it.

• Speed-fraction problems with percentage framing. "20% faster" means speed = 1.2S, so new time = T/1.2 = 5T/6. Time saving = T/6. Many aspirants treat 20% faster as 4/5 of usual time — that is wrong. The fraction you invert is the ratio of new speed to original, not the percentage directly.

• Ignoring head-start in stoppage problems. Stoppages questions occasionally embed a head-start (e.g., the train already covered some distance before stoppages began). Read the problem for "without stoppages" vs. "with stoppages" phrasing carefully — the question is about per-hour stoppage, not total journey stoppage.

Related Topics

• ratio and proportion
• percentage
• simple equations
• boats and streams
• work and time
• averages

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