Time, Speed & Distance

The spine: $D = S \times T$

Every question in this topic is one equation in different clothes: $\text{Distance} = \text{Speed} \times \text{Time}, \qquad S = \frac{D}{T}, \qquad T = \frac{D}{S}$ Ask at every question: "which two of the three do I know?" Speed and time are inversely proportional for a fixed distance — if the speed drops to $\frac34$, the time becomes $\frac43$ of the old time. This inverse reflex solves a surprising number of questions on its own.

Unit conversion — the 5/18 rule

Train lengths are in metres and times in seconds, but speeds are usually quoted in km/h. Convert first, always: $1 \text{ km/h} = \frac{5}{18} \text{ m/s}, \qquad 1 \text{ m/s} = \frac{18}{5} \text{ km/h}$ Keep the friendly table by heart: 36 km/h = 10 m/s, 54 = 15, 72 = 20, 90 = 25, 108 = 30 m/s. To convert by hand, divide by 18 first when using $\frac5{18}$ (54 → 3 → 5×3 = 15).

Average speed — a division, not an average

$\text{Average speed} = \frac{\text{total distance}}{\text{total time}}$ It is the average of the speeds only when the times are equal. Distances are more often equal (out and back), and then the answer is the harmonic mean: $\bar{S} = \frac{2ab}{a + b}$ 40 km/h out, 60 km/h back → $\frac{2 \times 2400}{100} = 48$ km/h, not 50. The equal-distance average always sits below the arithmetic mean, and below the mid-value of the two speeds. Three equal legs at a, b, c give $\frac{3}{\frac1a + \frac1b + \frac1c}$ — e.g. 20, 30, 60 → $\frac{3}{\frac{3+2+1}{60}} = 30$ km/h. For legs of different lengths, go back to basics: add the distances, add the times, divide.

"Late by…" and "early by…" questions

The distance is the same in both runs, so equate the two time expressions. Walking at $\frac34$ of the usual speed makes the time $\frac43 t$; the lateness is $\frac43 t - t = \frac{t}{3}$. So "20 minutes late" → $t = 60$ minutes. With two speeds s₁ and s₂ and a stated gap between the arrival times: $\frac{d}{s_1} - \frac{d}{s_2} = \text{gap (in hours)}$ e.g. 5 min late at 4 km/h, 10 min early at 5 km/h → $\frac{d}{4} - \frac{d}{5} = \frac{15}{60} = \frac14$ → $d = 5$ km. Add the two gaps when one is late and the other early; subtract when both are late.

Percentage shortcuts

Speed 20% more → time $\frac{5}{6}$ of before (7 min early on a 42-min run, etc.). Speed 20% less → time $\frac54$ of before. Convert every "reduces his speed to ¾" into the time multiplier $\frac43$ before writing any equation.

Common traps

• Averaging the two speeds for equal distances (50 instead of 48).
• Dividing metres by km/h — no conversion done.
• Reading "speed is ¾ of usual" as "time is ¾ of usual" (it is 4/3).
• Mixing minutes and hours inside one equation — convert to hours (or fractions of an hour).

Quick revision

• $D = ST$; speed and time are inverse for fixed distance.
• km/h → m/s: × 5/18 (table: 36→10, 54→15, 72→20, 90→25).
• Equal distances: $\frac{2ab}{a+b}$; three equal legs: $3 \div (\frac1a+\frac1b+\frac1c)$.
• Late/early: $\frac{d}{s_1} - \frac{d}{s_2} =$ time gap in hours.
• Speed ×(p/q) ⇒ time ×(q/p).