Time, Speed & Distance

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high importance~2 Q in Tier 121 formulas⚡ 15 shortcuts5 subtopics

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Speed, Distance, Time & Unit Conversion

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⏱ 5 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

One relation drives everything:

$D = S \times T$ , so $S = \dfrac{D}{T}$ and $T = \dfrac{D}{S}$ .

Units: multiply km/h by $\dfrac{5}{18}$ to get m/s; multiply m/s by $\dfrac{18}{5}$ to get km/h. Friendly pairs: 36 km/h = 10 m/s, 54 = 15, 72 = 20, 90 = 25.

Average speed = total distance ÷ total time, never the average of the speeds. For the same distance out and back at $a$ and $b$ : $\dfrac{2ab}{a+b}$ .

One relation, three forms

Distance = Speed × Time. Know any two, find the third. Ask at the start of every question: which two do I have?

For a fixed distance, speed and time move opposite ways. Speed drops to $\dfrac{3}{4}$ → time grows to $\dfrac{4}{3}$ . That inverse reflex alone solves many questions before any equation is written.

Rule: Speed × $\dfrac{p}{q}$ turns time into × $\dfrac{q}{p}$ . Flip the fraction, never keep it.

The five-eighteen conversion

Lengths come in metres, times in seconds, speeds in km/h. Convert before dividing.

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. To convert 72 by hand: $72 \div 18 = 4$ , then $4 \times 5 = 20$ .

Watch: Dividing metres by a km/h speed is the most common avoidable error here.

Average speed is a division

\text{Average speed} = \frac{\text{total distance}}{\text{total time}}

It equals the average of the speeds ONLY when the times are equal. Out-and-back trips have equal DISTANCES, and then the answer is the harmonic mean:

\bar{S} = \frac{2ab}{a+b}

40 km/h out, 60 km/h back → $\dfrac{2 \times 40 \times 60}{100} = 48$ km/h, not 50. The equal-distance average always sits below the plain midpoint. Three equal legs at a, b, c give $3 \div \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$ : at 20, 30 and 60 that is $3 \div \frac{6}{120} = 30$ km/h.

Tip: Legs of different lengths? Add the distances, add the times, divide. No shortcut beats the definition.

Late by and early by

The distance is the same in both runs, so compare the two times.

Walking at $\dfrac{3}{4}$ of usual speed makes the time $\dfrac{4}{3}t$ ; the lateness is $\dfrac{t}{3}$ . "20 minutes late" → $t = 60$ minutes. With two speeds and a stated gap:

\dfrac{d}{s_1} - \frac{d}{s_2} = \text{gap in hours}

5 minutes late at 4 km/h and 10 minutes early at 5 km/h: $\dfrac{d}{4} - \frac{d}{5} = \frac{15}{60} = \frac{1}{4}$ , so $d = 5$ km. Add the two gaps when one run is late and the other early; subtract when both are late.

Speed times a fraction, time times the flip

Speed $\dfrac{4}{5}$ of usual → time $\dfrac{5}{4}$ of usual → $\dfrac{1}{4}$ late.
Speed $\dfrac{3}{4}$ of usual → time $\dfrac{4}{3}$ of usual → $\dfrac{1}{3}$ late.

Convert every "reduces his speed to ¾" into the time multiplier $\dfrac{4}{3}$ BEFORE writing the equation.

Careful: Reading "speed is ¾ of usual" as "time is ¾ of usual" is the classic trap. The time goes UP.

Per cent changes in speed

Speed 20% more → time $\dfrac{100}{120} = \frac{5}{6}$ of before. Speed 25% more → time $\dfrac{4}{5}$ . Speed 20% less → time $\dfrac{5}{4}$ .

A 20% faster ride that saves 10 minutes: the saving is $\dfrac{1}{6}$ of the usual time, so the usual time is 60 minutes. Convert the per cent to a fraction, then read the time change off it.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Direct distance-speed-time with conversion

How to spot it:

Two of distance, speed, time are given in mixed units; the third is asked, often with a km/h and m/s conversion.

S = \frac{D}{T}, \qquad \text{km/h} \times \frac{5}{18} = \text{m/s}

Method

Put all quantities in one unit system (metres and seconds, or km and hours).
Apply $D = S \times T$ in the direction you need.
Convert the answer to the unit the options use.

Why it works:

The formula is only consistent when distance, speed and time speak the same units.

Try this

A train covers 450 metres in 30 seconds. Its speed in km/h is:

Show solution

$\dfrac{450}{30} = 15$ m/s.
$15 \times \frac{18}{5} = 54$ km/h.

Answer

54 km/h

Type 2very common2 practice Q

Average speed over the legs of a journey

How to spot it:

A journey splits into legs — out and back at different speeds, or different speeds for different stretches; the average speed is asked.

\bar{S} = \frac{D_{total}}{T_{total}}, \quad \text{equal distances: } \frac{2ab}{a+b}

Method

Equal distances (a round trip): use $\dfrac{2ab}{a+b}$ .
Equal times: the plain average of the speeds.
Otherwise: compute each leg's time, then total distance ÷ total time.

Why it works:

Average speed is a mean over TIME — that is exactly what the harmonic mean does.

Try this

A car goes to a town at 40 km/h and returns along the same road at 60 km/h. Find the average speed for the round trip.

Show solution

$\dfrac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100}$ .
= $48$ km/h — not 50.

Answer

48 km/h

Type 3very common2 practice Q

Fraction of usual speed makes him late

How to spot it:

'Walking at 3/4 (or 4/5) of his usual speed he is 20 minutes late' — the usual time or the distance is asked.

\text{new time} = \frac{t}{f} \Rightarrow \text{late by} = \left(\frac{1}{f} - 1\right)t

Method

Speed fraction f of usual → time is $t/f$ .
Lateness = $t\left(\frac{1}{f} - 1\right)$ ; set it equal to the given minutes.
Solve for t; multiply by the usual speed if the distance is asked.

Why it works:

For the same distance, speed and time are inversely proportional.

Try this

Walking at $\frac{4}{5}$ of his usual speed, a man reaches his office 10 minutes late. Find his usual time.

Show solution

New time = $\dfrac{5}{4}t$ .
Lateness = $\dfrac{t}{4} = 10$ minutes.
$t = 40$ minutes.

Answer

40 minutes

Type 4common2 practice Q

Two fixed speeds, one late and one early

How to spot it:

'At s₁ km/h he is x minutes late; at s₂ km/h he is y minutes early' — the distance is asked.

\frac{d}{s_1} - \frac{d}{s_2} = \frac{x + y}{60}

Method

Both times are measured against the same unknown schedule, so subtract the two travel times.
The gap equals late + early, converted to hours.
Solve $d\left(\frac{1}{s_1} - \frac{1}{s_2}\right) = \text{gap}$ for d.

Why it works:

Writing both times against a common schedule cancels the schedule itself.

Try this

Walking at 4 km/h a student reaches school 5 minutes late; walking at 5 km/h he reaches 10 minutes early. Find the distance to school.

Show solution

$\dfrac{d}{4} - \frac{d}{5} = \frac{15}{60} = \frac{1}{4}$ .
$\dfrac{d}{20} = \frac{1}{4}$ .
$d = 5$ km.

Answer

5 km

Type 5common

Per cent change in speed saves time

How to spot it:

'Increasing his speed by 20% he arrives 10 minutes early' — the usual time is asked.

\text{speed } \times\frac{100+r}{100} \Rightarrow \text{time } \times\frac{100}{100+r}

Method

Convert the per cent to a time factor: +20% speed → time $\dfrac{5}{6}$ ; +25% → $\dfrac{4}{5}$ .
The saving is $1 - \text{factor}$ of the usual time.
Divide the given saving by that fraction.

Why it works:

A per cent change in speed flips into the reciprocal time factor.

Try this

A cyclist increases his speed by 20% and reaches college 10 minutes early. Find his usual time.

Show solution

Speed $\times\frac{6}{5}$ → time $\times\frac{5}{6}$ .
Saving = $\dfrac{1}{6}$ of usual time = 10 minutes.
Usual time = $60$ minutes.

Answer

60 minutes

Formula sheet

Basic relation

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

km/h to m/s

\text{km/h} \times \frac{5}{18} = \text{m/s}

Equal-distance average

\bar{S} = \frac{2ab}{a + b}

General average speed

\bar{S} = \frac{D_1 + D_2}{T_1 + T_2}

Shortcuts that save time

⚡ Convert first, always

Train lengths are in metres, times in seconds — reach m/s before anything else.

Example

Convert 90 km/h into m/s and find the time a 150 m train takes to cross a pole.

Show solution

$90 \times \frac{5}{18} = 25$ m/s.
Time = $\dfrac{150}{25} = 6$ seconds.

Answer

25 m/s; 6 seconds

⚡ Harmonic mean for round trips

Same distance out and back → 2ab/(a+b) in one line.

Example

A car goes to a town at 40 km/h and returns at 60 km/h. Find the average speed for the whole trip.

Show solution

$\dfrac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100}$ .
= $48$ km/h.

Answer

48 km/h

⚡ Early and late gaps are time equations

Both runs cover the SAME distance — equate or subtract the two time expressions.

Example

Walking at $\frac{3}{4}$ of his usual speed a man is 20 minutes late. Find his usual time.

Show solution

New time = $\dfrac{4}{3}t$ .
Lateness = $\dfrac{t}{3} = 20$ minutes.
$t = 60$ minutes.

Answer

60 minutes

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Averaging two speeds for equal DISTANCES.

Equal distances average by 2ab/(a+b); 40 and 60 give 48, not 50.

Mistake 02

Forgetting the 5/18 conversion and dividing metres by km/h.

Convert to m/s first: km/h × 5/18.

Mistake 03

Using ¾t as the new time when speed is ¾.

Speed down to ¾ makes time 4/3 of usual — flip the fraction.

Mistake 04

Mixing minutes and hours in one equation.

Convert everything to hours (or fractions of an hour) first.

Mistake 05

Averaging speeds of legs with different lengths.

Total distance ÷ total time, computing each leg's time separately.

Quick revision

Read this the night before the exam.

D = ST; for a fixed distance, speed and time are inverse.
km/h → m/s: × 5/18 (36→10, 54→15, 72→20, 90→25).
Equal distances: 2ab/(a+b); three equal legs: 3 ÷ (1/a + 1/b + 1/c).
Late/early: d/s₁ − d/s₂ = gap in hours; add opposite gaps.
Speed × p/q ⇒ time × q/p.
Speed 20% more ⇒ time 5/6; 25% more ⇒ 4/5.

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.

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