Time, Speed, and Distance (TSD) is built on one relationship: Distance = Speed × Time. Every question in this chapter — no matter how it's dressed up with trains, boats, aircraft, or cyclists — is a manipulation of this single formula.
Here is the analogy that locks it in: think of a highway. The distance is the road itself — fixed, unchanging. Speed is how fast you're moving. Time is how long you've been driving. If you know any two of these three, you can always find the third. The AFCAT paper never breaks this rule. It only changes the costume.
What makes TSD questions feel harder than they are is the layering — two vehicles moving simultaneously, or a boat fighting a current, or a person covering different segments at different speeds. But once you strip that context away, you're always solving for one of three things: distance, speed, or time.
The unit discipline habit. Before you write a single equation, confirm that speed is in km/h and time is in hours — or speed in m/s and time in seconds. Mixing units is the single most common source of wrong answers. If speed is given in km/h and time in minutes, convert time to hours (÷ 60) before multiplying. The formula does not forgive unit mismatches.
The conversion you must have on reflex:
1 km/h = 5/18 m/s1 m/s = 18/5 km/h = 3.6 km/hFor AFCAT, most questions stay in km/h, but train questions sometimes drop into m/s without warning.
The real power of TSD comes from the concept of relative speed — when two objects are moving, you can treat the system as if one object is stationary and the other is moving at the combined or differential speed. This single idea handles the majority of "two vehicles" questions in under 30 seconds.
D = S × T rearranges to:
S = D / TT = D / SNothing exotic. The skill is knowing which form to reach for based on what the question asks.
When two objects move toward each other, their relative speed is the sum of their individual speeds. When they move in the same direction, relative speed is the difference.
Think of it this way: if you're running toward someone, the gap closes at both your speeds combined. If you're chasing someone who's faster, the gap widens at the difference of your speeds — and if you're faster, you close at the difference.
Formula summary:
Relative Speed = S₁ + S₂Relative Speed = |S₁ - S₂|Once you have relative speed, the problem reduces to: Time = Distance / Relative Speed. That's it.
This is where most candidates drop marks. If you travel from A to B at u km/h and return at v km/h, the average speed for the whole journey is not (u + v) / 2.
The correct formula:
Average Speed = 2uv / (u + v)
This is the harmonic mean of the two speeds. Use the arithmetic mean only when the time spent at each speed is equal. Use the harmonic mean when the distance covered at each speed is equal (which is what "going and returning" means).
For the general case — different distances at different speeds — you must go back to first principles:
Average Speed = Total Distance / Total Time
Compute total distance and total time separately, then divide.
When a train crosses a stationary object (a pole, a person), it travels a distance equal to its own length. When it crosses another train or a platform, it travels a distance equal to the sum of its length and the object's length.
Time = Length of train / SpeedTime = (Length of train + Length of platform) / SpeedTime = (L₁ + L₂) / Relative SpeedThe relative speed follows the same rule — add for opposite directions, subtract for same direction.
This is the same physics as trains and relative speed, just renamed. Let:
b = speed of boat in still waterw = speed of stream (current)Then:
b + wb - wRecovering b and w:
b = (Downstream + Upstream) / 2w = (Downstream − Upstream) / 2The same formulas apply to an aircraft flying with or against wind — this directly appears in AFCAT given the military aviation context.
A very common AFCAT pattern: "A covers a distance at speed u. If speed were v, time would differ by Δt. Find distance."
Set up: d/u - d/v = Δt (assuming u < v, so faster means less time).
Solve: d(1/u - 1/v) = Δt → d × (v - u)/(uv) = Δt → d = Δt × uv / (v - u).
You can derive this in 20 seconds once you know the pattern. Don't memorize the end formula — memorize the structure.
When two objects start from the same point and travel in perpendicular directions (north and east, for example), their separation at time t is found using the Pythagorean theorem:
Distance = √[(S₁t)² + (S₂t)²] = t × √(S₁² + S₂²)
Recognize the Pythagorean triple immediately — if speeds are 60 and 80, that's a 3-4-5 triple scaled by 20, giving a resultant of 100. AFCAT loves clean triples.
The moment you see two objects moving toward or away from each other, write: "RS = S₁ + S₂ (opposite) or S₁ − S₂ (same)." Then reduce the problem to a single-body problem with that combined speed. For the PYQ with cars at 45 and 55 km/h moving toward each other: RS = 100 km/h, need to cover 200 km → 2 hours. Standard approach (setting up equations for each car separately): ~60 seconds. Relative speed approach: under 15 seconds. Apply it every time without exception.
Whenever a question gives two speeds for equal distances (go and return), use Average Speed = 2uv/(u+v). For speeds 60 and 40: 2 × 60 × 40 / (60 + 40) = 4800/100 = 48 km/h. If you mistakenly average: (60+40)/2 = 50 km/h — wrong answer, wrong option selected. The arithmetic mean trap is intentionally placed in AFCAT options. The moment you see "same route, different speeds," reach for the harmonic mean. Standard equation method: ~45 seconds. Direct formula: ~10 seconds.
Given downstream speed D and upstream speed U: still-water speed = (D+U)/2, current speed = (D−U)/2. No algebra required. For the boat PYQ: downstream = 10 km/h, upstream = 5 km/h → still water = (10+5)/2 = 7.5 km/h. Instant read-off, zero equation-solving. Standard method (setting up simultaneous equations): ~50 seconds. Direct formula: ~8 seconds.
When two perpendicular speeds are given, check if they form a 3-4-5, 5-12-13, or 8-15-17 ratio. Speeds of 60 and 80 → ratio 3:4 → resultant = 5 × 20 = 100. Speeds of 30 and 40 → resultant = 50. This eliminates the square root computation entirely. Standard computation √(60² + 80²) = √(3600 + 6400) = √10000: ~30 seconds under exam pressure. Triple recognition: ~5 seconds.
For "same distance, time differs by T hours" problems, use d = T × (product of speeds) / (difference of speeds). Speeds 12 and 15, time difference 1 hour: d = 1 × (12 × 15) / (15 − 12) = 180/3 = 60 km. Compare: setting up d/12 − d/15 = 1, finding LCM, solving — that is 4 steps. Direct formula: 1 step. Saves 2-3 steps on every question of this type.
Read the question and immediately classify it:
Step 1 — Identify the setup:
Step 2 — Write the one equation: Do not write two or three equations if one suffices. Relative speed collapses two-body problems into one body.
Step 3 — Check units before computing: All speeds in km/h, all times in hours? Proceed. Mismatch? Convert first.
Step 4 — Backcheck with options: AFCAT options are clean numbers. If your answer is not a clean number, you have a unit error or a sign error. Re-examine before moving on — don't submit a fractional answer when all options are integers.
Why this question: Tests pure relative speed with an added layer — "100 km apart" rather than "meet," which trips up candidates who assume the question ends at collision.
Solving path: Relative speed = 45 + 55 = 100 km/h (moving toward each other). For them to be 100 km apart after starting 300 km apart, they need to close 200 km of the gap. Time = 200 / 100 = 2 hours. The trap: some candidates solve for the time they meet (when gap = 0), which gives 3 hours — that's option C, placed deliberately.
Why this question: Classic "same distance, different speeds, time difference" pattern. Tests whether you can build the correct fractional equation.
Solving path: Using the direct formula: d = 1 × (12 × 15) / (15 − 12) = 180 / 3 = 60 km. Or set up d/12 − d/15 = 1, LCM of 12 and 15 is 60: (5d − 4d)/60 = 1 → d = 60 km.
Why this question: Return-journey average speed. Tests whether you use the harmonic mean correctly rather than the arithmetic mean.
Solving path: Let distance = d. Time A→B = d/60, time B→A = d/40. Total: d/60 + d/40 = 5. LCM of 60 and 40 is 120: (2d + 3d)/120 = 5 → 5d = 600 → d = 120 km. Cross-check: time = 120/60 + 120/40 = 2 + 3 = 5 hours. Confirmed.
Why this question: Two-segment journey with different speeds per segment. Tests the total-distance / total-time approach, not average of speeds.
Solving path: Let each half = d km. d/4 + d/6 = 5. LCM = 12: (3d + 2d)/12 = 5 → 5d = 60 → d = 12. Total distance = 2 × 12 = 24 km. The arithmetic-mean trap: (4 + 6)/2 = 5 km/h, then 5 × 5 = 25 km — option C is placed for this exact mistake.
Why this question: Directly mirrors aircraft-and-wind problems that AFCAT favors given the aviation context. Tests the boats formula transposed to aviation.
Solving path: Speed with wind = 1200/3 = 400 km/h. Speed against wind = 1200/4 = 300 km/h. Still-air speed = (400 + 300)/2 = 350 km/h. Wind speed = (400 − 300)/2 = 50 km/h. Apply "add and halve / subtract and halve" — the answer is immediate once you have the two effective speeds.
Averaging speeds arithmetically for equal-distance journeys. (u + v)/2 is only valid when time spent at each speed is equal. For equal-distance segments (go and return), always use 2uv/(u+v). AFCAT option setters place the arithmetic mean answer as a distractor every time.
Forgetting to add lengths in train problems. When a train crosses a platform or another train, the distance covered is the sum of both lengths, not just the train's length. Crossing a pole or a standing person — then it's only the train's length.
Solving for "when they meet" instead of "when they are X km apart." Two objects moving toward each other and 300 km initially apart — if asked when they are 100 km apart, the answer is NOT when gap = 0. Close only 200 km of the gap. Read the question condition precisely.
Mixing km/h and minutes. Speed in km/h, time given in minutes — multiply directly and you get a nonsense answer. Convert minutes to hours (divide by 60) or convert km/h to km/min (divide by 60) before computing. Pick one and be consistent.
Using downstream speed directly as the boat speed. Downstream speed = boat speed + current. The boat's speed in still water is (downstream + upstream) / 2. Downstream speed alone is not the boat's "real" speed.
Missing the Pythagorean triple in perpendicular-path problems. Computing √(60² + 80²) the hard way when 60 and 80 are obviously 3-4-5 scaled by 20 costs 20-25 seconds. Always check for the triple first. If the ratio is not a recognizable triple, then compute — but check first.