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Speed, Time & Distance — Trains, Boats & Streams for SSC GD Constable

intermediate 20 min read

Concept

Speed, Time, and Distance is the single most recycled topic in SSC GD Maths. The core relationship is:

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

Everything else — trains, boats, chasing thieves, runners — is just this one formula dressed in different clothes.

Think of it this way: distance is fixed ground (a road, a river, a track). Speed is how fast you eat that ground. Time is how long you're eating it. Change any one, the others shift. That's the entire topic.

Why this trips people up: the formula itself is trivial. The difficulty is setting up the right equation when two objects are moving, or when the same person walks at different speeds, or when a boat fights a current. SSC GD tests your equation-building instinct, not your arithmetic.

Here's the analogy that makes relative speed click: imagine you're standing on a platform watching two trains. If both trains move in the same direction at 60 and 40 km/h, the faster train "gains" on the slower one at only 60 - 40 = 20 km/h from your perspective — it's as if the slower train is standing still and the faster one crawls at 20 km/h. If they move toward each other, they "close the gap" at 60 + 40 = 100 km/h. That's relative speed, and it governs every train-crossing and overtaking problem you'll see.

For boats and streams, the river current is the additional "push" or "resistance." A rower's own muscle gives him a still-water speed u. A downstream current of speed v adds to that (u + v), and an upstream current subtracts (u - v). You're always solving for u and v from the two given speeds.

Deep Dive

The Core Formula Variants

$S = \frac{D}{T}, \quad T = \frac{D}{S}, \quad D = S \times T$

When speed changes but distance stays the same:

$S_1 T_1 = S_2 T_2$

This is the single most useful identity for "walks at fraction of usual speed" problems. Memorise it in words: same distance, speed and time are inversely proportional.

Speed Change → Time Change Formula

If a person walks at $\frac{m}{n}$ of usual speed, the new time becomes $\frac{n}{m}$ of usual time. The change in time is:

$\Delta T = T_{\text{usual}} \times \left(\frac{n}{m} - 1\right)$

Example: walks at $\frac{3}{4}$ speed → new time = $\frac{4}{3}T$ → extra time = $\frac{1}{3}T$ . If $\frac{1}{3}T = \frac{1}{3}$ hr, then $T = 1$ hr.

Relative Speed

| Scenario | Relative Speed | |---|---| | Same direction | $|S_1 - S_2|$ | | Opposite direction | $S_1 + S_2$ |

Two-speed, same-distance, different-arrival-time problem (the classic train problem):

If a train is 10 min late at 50 km/h and 50 min late at 30 km/h, the difference in lateness is 40 min = $\frac{2}{3}$ hr. The distance D satisfies:

$\frac{D}{30} - \frac{D}{50} = \frac{2}{3}$

$D \times \frac{50-30}{30 \times 50} = \frac{2}{3}$

$D \times \frac{20}{1500} = \frac{2}{3} \implies D = \frac{2}{3} \times \frac{1500}{20} = 50 \text{ km}$

Look — the trap is using the absolute lateness (10 or 50), not the difference. The scheduled time cancels out when you subtract the two equations, and you're left with a clean equation.

Bus-Car / Two-Vehicle Problems

When Car speed = Bus speed + 25, and 500/Bus_speed - 500/Car_speed = 10, set Bus speed = x:

$\frac{500}{x} - \frac{500}{x+25} = 10$

$500(x+25) - 500x = 10x(x+25)$

$12500 = 10x^2 + 250x$

$x^2 + 25x - 1250 = 0 \implies (x+50)(x-25) = 0 \implies x = 25$

So Bus = 25 km/h, Car = 50 km/h.

Overtaking / Chase Problems

When a thief gets a head start of time $t$ at speed $S_T$ , the head start distance = $S_T \times t$ . The owner at speed $S_O$ covers this gap at relative speed $S_O - S_T$ .

Time to catch from owner's start = $\frac{S_T \times t}{S_O - S_T}$

Time from start of theft = Owner's catch time + $t$ .

Thief: 40 km/h, head start: 0.5 hr → head start distance = 20 km. Owner: 50 km/h → relative speed = 10 km/h. Owner's catch time = 20/10 = 2 hr from when owner starts. From start of theft = 2 + 0.5 = 2.5 hours.

Meeting Problems (Two People Walking Toward Each Other)

When A starts 3 min after B, and they eventually meet: use total distance covered by both equals 2 × (destination distance) - (remaining gap). The trick is to write total time for A and B separately, using A's known speed, then solve for B's speed from the meeting condition.

A's speed = 1 km in 18 min = $\frac{60}{18} = \frac{10}{3}$ km/h. A has walked 1 km at meeting → A's time = 18 min. B has walked 4.5 + (4.5 - 1) = 8 km (goes 4.5, returns 3.5 before meeting) in 18 + 3 = 21 min. B's speed = $\frac{8}{21} \times 60 = \frac{480}{21}$ ... let's verify: actually B walked 4.5 + (4.5-1) = 8 km in 21 min → $\frac{8 \times 60}{21}$ doesn't simplify cleanly; check: A walks 1 km in 18 min, total time elapsed for B = 21 min. Distance B covers = 4.5 + (4.5 - 1) = 8 km. Speed = $\frac{8}{21}$ km/min = $\frac{8 \times 60}{21}$ = not clean. Re-examine: meeting point is 1 km from destination, so A walked 4.5 - 1 = 3.5 km from start. A's time = 3.5 × 18 = 63 min. B's time = 63 + 3 = 66 min (B started 3 min earlier). B covered 4.5 + (4.5 - 1) = 8 km. B's speed = 8/66 km/min = 8×60/66 = 480/66 ≈ not 5. Correct reading: they meet after A walks 1 km from where they meet (1 km before destination). A walked (4.5 - 1) = 3.5 km. Time for A = 3.5 × 18 = 63 min. B started 3 min before A, so B ran for 66 min total. B went 4.5 km then came back (4.5 - 1) = 3.5 km = 8 km total. Speed of B = 8/66 × 60 = ~7.27 km/h... the answer is 5 km/h per the PYQ. The standard solution uses: A's speed = 1/18 km/min. They meet when A has walked 1 km (meeting point is 1 km before destination, since B is returning). Time for A to walk 3.5 km = 63 min. Check answer 5 km/h = 5/60 km/min. In 66 min B covers 66 × 5/60 = 5.5 km. But destination is 4.5 km, so B goes 4.5 km and returns 1 km = 5.5 km. This checks out perfectly.

Boats and Streams

$\text{Downstream speed} = u + v, \quad \text{Upstream speed} = u - v$

$u = \frac{(u+v) + (u-v)}{2} = \frac{\text{Downstream} + \text{Upstream}}{2}$

$v = \frac{(u+v) - (u-v)}{2} = \frac{\text{Downstream} - \text{Upstream}}{2}$

Upstream speed: 750 m in 600 s = 1.25 m/s = 4.5 km/h. Return (downstream): 750 m in 7.5 min = 450 s → 750/450 m/s = 5/3 m/s = 6 km/h. Still water speed = (4.5 + 6)/2 = 5.25 km/h.

Average Speed Trap

When distance is the same for both legs, average speed ≠ arithmetic mean of speeds. It equals the harmonic mean:

$\bar{S} = \frac{2 S_1 S_2}{S_1 + S_2}$

But for the runner problem — same distance every day, total distance / total time — that's a straightforward division. Total distance = 9 × 400 × 4 = 14,400 m. Total time = (88 + 96 + 89 + 87) = 360 min. Average = 14,400 / 360 = 40 m/min.

Memory Tricks & Shortcuts

patternFraction-Speed → Time Multiplier

When speed becomes p/q of usual, new time = q/p of usual. Extra time = (q/p - 1) × usual time. Set that equal to the given delay and solve in one step. Standard method: write two equations, solve system — 4 steps, ~40s. This shortcut: one multiplication — 10s.

Micro-example: speed = 5/4 usual → new time = 4/5 usual → saved time = (1 - 4/5) × T = T/5 = 30 min → T = 150 min = 2.5 hr.

patternTwo-Speed Lateness: Subtract the Equations

When a train/person arrives at two different lateness values at two speeds, NEVER work with absolute schedule. Always subtract: (Time at slow speed) - (Time at fast speed) = (Difference in lateness). This eliminates the unknown scheduled time and leaves a single clean equation. Standard method: introduce scheduled time as a third variable — 6 steps. This method: 3 steps, ~25s saved.

Micro-example: late 50 min at 30 km/h, late 10 min at 50 km/h → D/30 - D/50 = 40/60. Solve: D = 50 km.

patternChase: Head-Start Distance First

Don't set up simultaneous equations for overtaking. Calculate head-start distance = (thief's speed) × (head-start time). Then divide by relative speed to get catch time from owner's departure. Add head-start time for total time from theft.

Micro-example: 40 × 0.5 = 20 km head start. Relative speed = 50 - 40 = 10 km/h. Catch in 2 hr from owner's start → 2.5 hr from theft. Three arithmetic steps vs. two-variable system — saves ~30s.

patternStill Water Speed = Average of Downstream and Upstream

The formula u = (D + U)/2 where D is downstream speed and U is upstream speed is faster than solving two simultaneous equations from scratch. Convert distances and times to speeds first, then average them directly.

Micro-example: upstream = 4.5 km/h, downstream = 6 km/h → still water = (4.5 + 6)/2 = 5.25 km/h. One addition + one halving vs. two-equation elimination — saves ~20s.

eliminationAverage Speed for Constant Distance: Total D / Total T

When the same distance is covered every day/trip, do NOT use the harmonic mean formula — just divide total distance by total time. The harmonic mean is only for two-speed single-journey problems. Eliminate the formula, use direct division.

Micro-example: 9 laps × 400 m × 4 days = 14,400 m total. Total time = 360 min. Speed = 14,400/360 = 40 m/min. Avoids formula confusion in under 15s.

Fast-Solving Framework

Read the problem. Ask these three questions in sequence:

1. Is one person changing their own speed? (walks at 3/4 of usual rate) → Use the fraction-speed multiplier trick. One equation. Done.

2. Are two different speeds given for the same journey with different lateness? → Subtract to eliminate scheduled time. One equation. Solve for D.

3. Are two objects moving — either chasing or meeting? → Same direction: find head-start distance, divide by relative speed. → Opposite direction: add speeds, use total distance.

4. Boats and streams? → Convert all given data to speeds (km/h). Still water = (downstream + upstream)/2.

5. Average speed question? → Check if same distance every leg/day → Total D / Total T. If two different speeds for same single-journey distance → Harmonic mean = 2S₁S₂/(S₁+S₂).

If the problem gives you quadratic-looking numbers (500 km, 25 km/h difference), try factoring or smart substitution before grinding through the quadratic formula.