Speed, distance, and time form a triangle of relationships that appear in nearly every quantitative section of government exams. The core idea is simple: if you move at a certain rate for a certain duration, you cover a fixed distance.
The fundamental relationship is:
Distance = Speed × Time
Flip it either way: Speed = Distance / Time and Time = Distance / Speed.
Think of it like a water tap. Speed is how fast water flows (litres per minute), Time is how long you keep the tap open, and Distance is the total water collected. If you know any two, you can find the third.
Where candidates lose marks is not the formula — everyone knows it. The problem is unit mismatch, relative speed setups for trains, and the boats-and-streams variant. UP Police Constable papers consistently test these exact scenarios, so getting the mechanics right is non-negotiable.
One more thing to internalize before you go deeper: speed and time are inversely proportional when distance is fixed. If a man doubles his speed, he takes half the time. This inverse relationship is the backbone of ratio-based shortcuts that save 30-40 seconds per question.
The unit conversion that trips up most candidates:
1 km/h = 5/18 m/s
1 m/s = 18/5 km/h = 3.6 km/h
Burn these into memory. Every train-tunnel and train-platform question starts by converting km/h to m/s, and every fumbled conversion costs you the question.
D = S × T
When two of the three are given, you solve for the third. This is direct and most candidates handle it fine. The exam tests your speed, so practice until this takes under 5 seconds.
Always match units before applying the formula. If speed is in km/h and distance is in metres, convert before calculating.
54 km/h = 54 × (5/18) = 15 m/s
A reliable pattern: divide km/h by 3.6 to get m/s. Or multiply the km/h value by 5, divide by 18.
Here's where many candidates go wrong. Average speed is not the arithmetic mean of two speeds.
Average Speed = Total Distance / Total Time
If a man travels distance d₁ at speed s₁ and d₂ at speed s₂:
Average Speed = (d₁ + d₂) / (d₁/s₁ + d₂/s₂)
Special case: same distance at two different speeds s₁ and s₂.
Average Speed = 2s₁s₂ / (s₁ + s₂)
This is the harmonic mean. Use it only when both distances are equal. If the problem says "he travels 60% of the journey at one speed and 40% at another," you cannot use the harmonic mean formula — go back to total distance divided by total time.
When two objects move:
|S₁ - S₂|S₁ + S₂For train problems, what "passes" means matters:
This last point is critical — PYQ 6a3ab4744bc500be68467db7 directly tests it.
Two types appear in exams:
Type 1 — Find stoppage time per hour:
Speed without stops = s₁, speed with stops = s₂.
Distance lost per hour due to stops = s₁ - s₂.
Stoppage time per hour = (s₁ - s₂)/s₁ hours.
Type 2 — Find total distance given stop durations:
Total time with stops = Total time without stops + stop time.
D/s₂ - D/s₁ = total stop time
Solve for D.
If speeds are in ratio a:b, times are in ratio b:a (for equal distances). This is the backbone of PYQ 6a3b88f77812e6eabf35574c:
Speed ratio A:B = 2:3 → Time ratio A:B = 3:2.
If the difference in time = 15 min and the ratio difference = 3-2 = 1 unit, then 1 unit = 15 min. A takes 3 units = 45 minutes. At double the speed, A's time halves: 45/2 = 22.5 minutes.
Let boat speed in still water = b, stream speed = s.
b + sb - sTo find still water speed: b = (downstream + upstream) / 2
To find stream speed: s = (downstream - upstream) / 2
Always compute upstream and downstream speeds first from the given distance and time, then apply these formulas.
When a question gives speed in km/h but distance in metres, multiply km/h by 5 and divide by 18. Better pattern to memorize: km/h ÷ 3.6 = m/s. For common exam values: 54 km/h = 15 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s, 36 km/h = 10 m/s. Burn these four pairs. If the exam gives one of these values, you skip the calculation entirely. Standard method: multiply by 5, divide by 18 (3 steps, ~15 seconds). Using memorized pairs: 0 steps, ~2 seconds.
Stoppage per hour question: you do not need algebra. The formula is: minutes stopped per hour = (speed without stops - speed with stops) / speed without stops × 60. For PYQ with 66 km/h and 55 km/h: (66-55)/66 × 60 = 11/66 × 60 = 10 minutes. One calculation, no variable setup. Standard method (setting up equations with variables): ~45 seconds. This pattern: ~10 seconds.
When speeds are in ratio a:b, immediately write times as b:a. Then find one unit from the difference. Speed ratio 2:3 → time ratio 3:2 → difference = 1 unit = 15 min → total time for slower = 3 units = 45 min. At double speed, time halves. This eliminates equation setup entirely. Standard algebraic approach: ~60 seconds. Ratio flip: ~15 seconds.
For 'find distance given two speeds and stop time' problems: use D/slower_speed - D/faster_speed = stop_time. Factor D out: D × (1/slower - 1/faster) = stop_time. The fraction (1/50 - 1/60) = (6-5)/300 = 1/300. So D × 1/300 = 1 hour → D = 300 km. Write it as a single-line substitution. Standard method of writing full time expressions separately: ~50 seconds. Direct substitution: ~20 seconds.
Before solving any train-passing problem, write down exactly what distance is being covered. Eliminate wrong setups first. Passing a pole = 1 length. Passing a platform = 2 lengths (train + platform). Two trains crossing each other completely = 2 lengths (both trains). Slower train passing the driver of faster train = 1 length (slower train only). Forgetting this last rule adds an extra 800 m to the numerator and gives a wrong answer. Identify the scenario before touching numbers: ~5 seconds saved, and the question cannot trick you.
When you see a speed-distance-time question in the exam hall, run through this checklist in under 10 seconds:
If the question has a two-step structure (find distance first, then time), do not try to combine them mentally — write the two steps clearly and solve sequentially.
Why this question: This is the most direct train-tunnel format you will see. It tests unit conversion and the "two lengths" rule simultaneously.
Solving path: Convert 54 km/h to m/s: 54 × 5/18 = 15 m/s. Total distance = 600 m (train) + 1200 m (tunnel) = 1800 m. Time = 1800 ÷ 15 = 120 seconds. The only trap is forgetting to add both lengths, or failing to convert 1.2 km to 1200 m.
Why this question: This is the most commonly confused train-crossing variant. "Pass the driver" does not mean full crossing — only one train's length is covered.
Solving path: Relative speed = 80 + 60 = 140 km/h (opposite directions). Convert: 140 × 5/18 = 38.89 m/s. Distance covered = 800 m (only the slower train's length, because we are timing how long until the slower train's front clears the driver's position). Time = 800 / 38.89 ≈ 20.57 seconds.
Why this question: Average speed with given time (not speed) for each segment — the trap is averaging 60% and 40% speeds rather than using total distance over total time.
Solving path: Total distance = 80 km. Total time = 45 + 55 = 100 minutes = 100/60 hours = 5/3 hours. Average speed = 80 ÷ (5/3) = 80 × 3/5 = 48 km/h. There are no individual speeds given, so there is no temptation to use harmonic mean — if you go straight to total/total, the question dissolves in 20 seconds.
Why this question: Stop-time to distance is a very common pattern. The equation setup with D/50 - D/60 is the key move.
Solving path: Total stop time = 4 × 15 = 60 minutes = 1 hour. Let distance = D. D/50 - D/60 = 1. Find LCM of 50 and 60 = 300. D(6 - 5)/300 = 1 → D/300 = 1 → D = 300 km.
Why this question: The stoppage-minutes-per-hour formula. Pure pattern recognition once you have seen it.
Solving path: In one hour without stops, bus covers 66 km. With stops, it covers 55 km. The 11 km difference was lost due to stoppages. Time spent stopped = 11/66 hours = 1/6 hour = 10 minutes.
Why this question: The speed-ratio → time-ratio flip, followed by a "double the speed" halving.
Solving path: Speed A:B = 2:3, so time A:B = 3:2. Let times be 3k and 2k. Difference: 3k - 2k = k = 15 min. A's time = 3k = 45 min. At double speed, A's time = 45/2 = 22.5 minutes.
Why this question: Boats and streams — find upstream and downstream speeds, then use the averaging formula.
Solving path: Upstream speed = 80/16 = 5 km/h. Downstream speed = 72/12 = 6 km/h. Still water speed = (5 + 6)/2 = 5.5 km/h. Note: the answer key for this question shows a discrepancy — verify using the formula and your own calculation, as 5.5 km/h is the mathematically correct result from the given data.
Averaging speeds instead of using total distance / total time. When the question gives different speeds for different parts of a journey, you cannot add the speeds and divide by 2 unless both distances are explicitly equal. Always go back to the definition: total distance over total time.
Not converting km/h to m/s before applying the formula in train problems. Every train-length question uses metres and seconds. If your speed is still in km/h when you divide, your answer will be off by a factor of 3.6 and will not match any option — which wastes time as you recheck.
Adding both train lengths when the question asks only about passing the driver. "Pass the driver of the faster train" means the slower train's front travels the length of the slower train only, not the sum of both lengths.
Using harmonic mean (2ab/(a+b)) when distances are unequal. The harmonic mean formula for average speed is valid only when both segments cover the same distance. If the problem says "60% of distance at one speed," use total distance / total time.
Forgetting to convert stop time to hours before subtracting. In the Kochi-Coimbatore type problem, the stop time is given in minutes. You must convert to hours before writing the equation D/s₁ - D/s₂ = stop_time.
In boats and streams, confusing which is upstream and which is downstream. Upstream means against the current — harder, slower. Downstream means with the current — easier, faster. If you mix these up, your still-water formula gives you the stream speed and vice versa. Write "upstream = going against river" at the top of the workspace before starting.