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AFCAT Numerical Ability · Speed, Time and Distance

The Basic Formula, Units and Journeys in Parts

Speed as distance over time, changing km/h into m/s, and journeys split into parts: going and coming back, two speeds on one trip, and rests on the way.

All Speed, Time and Distance notes

Why this matters

Every question asks for one of speed, distance and time from the other two, usually after a change of units. Marks go on minutes written as decimals and on km/h divided into metres. AFCAT set one question here in 2015 and one in 2019.

Concept 1 of 3: Speed, distance, time and their units

Speed is distance covered per unit of time. Know any two of speed, distance and time and the third follows. Most slips are in the units: a train's speed is given in km/h while its length is in metres.

Definition

  • speed=distancetime\text{speed} = \dfrac{\text{distance}}{\text{time}}, so distance=speed×time\text{distance} = \text{speed} \times \text{time} and time=distancespeed\text{time} = \dfrac{\text{distance}}{\text{speed}}.
  • km/h to m/s: multiply by 518\tfrac{5}{18}, because 1 km/h is 1,000 m in 3,600 s. m/s to km/h: multiply by 185\tfrac{18}{5}.
  • Minutes to hours: divide by 60. So 2 h 15 min is 2.25 h.
  • Two people walking off at right angles: the distance between them is the hypotenuse, a2+b2\sqrt{a^2 + b^2}, of the distances each has gone.

Speed and the unit change

speed=distancetime,1 km/h=518 m/s\text{speed} = \dfrac{\text{distance}}{\text{time}}, \qquad 1 \text{ km/h} = \tfrac{5}{18} \text{ m/s}

Worked example

A cyclist rides at 18 km/h for 2 h 40 min. How far does he go, and what is his speed in m/s?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

AFCAT · 2019 · AFCAT (II) 2019 — Numerical Ability · 04-201Easy

Example 1 · Speed, Time and Distance · Speed, Distance, Units and Journeys in Parts

Two friends started for a place, one by motorcycle and the other by train. The speed of the motorcycle is 30 km/h and that of the train is 24 km/h. The first one takes 6 h 12 min to reach the destination. Find the time taken by the second one.

Change the units before you divide

A 150 m train at 54 km/h: 150÷54150 \div 54 means nothing. Write 54 km/h as 15 m/s first, and the train passes a pole in 10 s.

Minutes are not decimal places

2 h 45 min is 2.75 h, not 2.45 h. Divide the minutes by 60.
Drill 5 more on speed, distance, time and their units

Concept 2 of 3: Going and coming back at two speeds

The road is the same length both ways. Call it dd, add the time out to the time back, and set the sum equal to the total time given.

Definition

  • Out at aa and back at bb in a total time TT: da+db=T\dfrac{d}{a} + \dfrac{d}{b} = T, so d=abTa+bd = \dfrac{abT}{a + b}.
  • Change the total time to hours first: 3 h 45 min is 3.75 h.
  • The same equation covers riding out and walking back, or going up a hill and coming down.

Distance from a round trip

da+db=T  ⇒  d=abTa+b\dfrac{d}{a} + \dfrac{d}{b} = T \;\Rightarrow\; d = \dfrac{abT}{a + b}

Worked example

A man rides to a town at 15 km/h and walks back at 5 km/h. The round trip takes 8 h. How far away is the town?
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

AFCAT · 2015 · AFCAT (I) 2015 — Numerical Ability · 04-222Moderate

Example 2 · Speed, Time and Distance · Speed, Distance, Units and Journeys in Parts

A man travelled from a point AA to BB at the rate of 25 km/h and walked back at the rate of 4 km/h. If the whole journey took 5 h 48 min, the distance between AA and BB is

The total time covers both legs

If the round trip takes 8 h, the 8 h is the ride out and the walk back together. Solving d15=8\tfrac{d}{15} = 8 uses the whole time on one leg.

Turn the minutes into hours

3 h 36 min is 3.6 h, not 3.36 h. Put the wrong figure into da+db=T\tfrac{d}{a} + \tfrac{d}{b} = T and every option is missed.
Drill 4 more on going and coming back at two speeds

Concept 3 of 3: One journey at two speeds, and rests on the way

When one journey is split between two speeds, call the distance on one part xx. The other part is what is left of the whole, and the two times add up to the total.

Definition

  • Distance DD in time TT, part at aa and the rest at bb: xa+D−xb=T\dfrac{x}{a} + \dfrac{D - x}{b} = T. This is a linear equation in xx.
  • Two equal halves: the time is D/2a+D/2b\dfrac{D/2}{a} + \dfrac{D/2}{b}.
  • Rests add time but no distance. A rest after every kilometre of a 5 km walk means 4 rests, because there is none after the last kilometre.
  • Walking one way and riding back takes T1T_1; walking both ways takes T2T_2. One walk is T22\tfrac{T_2}{2}, so one ride is T1−T22T_1 - \tfrac{T_2}{2}.

A journey in two parts

xa+D−xb=T\dfrac{x}{a} + \dfrac{D - x}{b} = T

Worked example

A man travels 60 km in 4 h, partly on foot at 5 km/h and partly by bus at 25 km/h. How far does he walk?
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Speed, Time and Distance · Speed, Distance, Units and Journeys in Parts

A man travelled a distance of 80 km in 7 h partly on foot at the rate of 8 km/h and partly on cycle at 16 km/h. The distance travelled on foot is

The two parts are x and what is left

Writing xa+xb=T\tfrac{x}{a} + \tfrac{x}{b} = T makes both parts the same length. The second part is D−xD - x.

No rest after the last stretch

A 5 km walk with a rest after every kilometre has 4 rests, not 5. The walker has arrived after the fifth.
Drill 6 more on one journey at two speeds, and rests on the way

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Watch out for (6)

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