AFCAT Numerical Ability · Speed, Time and Distance

Average Speed: Total Distance over Total Time

Average speed is the whole distance divided by the whole time, with quick forms for equal distances and equal times, fractions of a journey, and stoppages.

All Speed, Time and Distance notes

Why this matters

This is one of the biggest pages in the chapter and most of it is EASY, so marks are lost to one habit: taking the mean of the speeds. That is right only when the times are equal. Over equal distances the true average is lower, and the options are built on that gap.

Concept 1 of 5: Total distance over total time

Average speed is not the average of the speeds. It is the one steady speed that would cover the whole distance in the whole time.

Definition

$\text{average speed} = \dfrac{\text{total distance}}{\text{total time}}$ .
Given each part's distance and speed: work out each part's time first, then add.
Given each part's distance and time: add the distances, add the times, divide.
Times in minutes: add the minutes, then change the total to hours.

Average speed

\bar v = \dfrac{d_1 + d_2 + \dots}{t_1 + t_2 + \dots}

Worked example

A car goes 120 km at 60 km/h and then 60 km at 20 km/h. Find its average speed.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Speed, Time and Distance · Average Speed

A bus travels

150

km at

50

km/hr, then

200

km at

40

km/hr. What is the average speed for the entire journey?

Do not average the speeds

120 km at 60 km/h and 60 km at 20 km/h is 180 km in 5 h, 36 km/h. The mean of the two speeds, 40, counts the slow part as if it lasted no longer than the fast one.

Drill 38 more on total distance over total time

Concept 2 of 5: Speeds held for given times

When each speed is kept up for a stated time, each part's distance is speed times time. With equal times, the average is simply the mean of the speeds.

Definition

Each part: $d = v \times t$ . Then $\bar v = \dfrac{v_1 t_1 + v_2 t_2 + \dots}{t_1 + t_2 + \dots}$ , a mean weighted by time.
Equal times: $\bar v = \dfrac{v_1 + v_2 + \dots + v_n}{n}$ .
Minutes to hours first: 45 min is 0.75 h.

Average over given times

\bar v = \dfrac{v_1 t_1 + v_2 t_2}{t_1 + t_2}

Worked example

A truck runs for 1 h at 35 km/h and then for 3 h at 55 km/h. Find its average speed.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Speed, Time and Distance · Average Speed

A car travels for

2

hours at

50

km/hr, then for

3

hours at

40

km/hr, and finally for

1

hour at

60

km/hr. What is the average speed?

Weight by time

The plain mean of the speeds is right only when the times are equal. 1 h at 35 km/h and 3 h at 55 km/h gives 50 km/h, not the mean 45, because the faster speed lasted three times as long.

Drill 10 more on speeds held for given times

Concept 3 of 5: Equal distances: there and back, or equal stretches

Over equal distances the slower speed lasts longer, so it pulls the average below the mean of the speeds. The distance itself cancels and need not be known.

Definition

Two equal distances at $a$ and $b$ , such as there and back: $\bar v = \dfrac{2ab}{a + b}$ .
Three equal distances: $\bar v = \dfrac{3}{\tfrac1a + \tfrac1b + \tfrac1c}$ .
If the distance is given anyway, total distance over total time gives the same answer.
The average always lies between the speeds and below their mean.

Average over two equal distances

\bar v = \dfrac{2ab}{a + b}

Worked example

A man drives to a town at 30 km/h and back at 70 km/h. Find his average speed.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 3 · Speed, Time and Distance · Average Speed

The speed of a train going from Nagpur to Allahabad is 100 km/h, while its speed is 150 km/h when coming back from Allahabad to Nagpur. The average speed of the train during the whole journey is

The mean is too high

Out at 30 km/h and back at 70 km/h averages 42 km/h, not 50. The return is quick, so the slow leg fills most of the time.

Work out a changed speed first

'Out at 24 km/h and back 50% faster' means 36 km/h back. Find the second speed before using

\tfrac{2ab}{a + b}

: here 28.8 km/h.

Drill 20 more on equal distances: there and back, or equal stretches

Concept 4 of 5: Fractions of the journey, and working back from an average

When a journey is split by fractions, pick a total distance that every fraction divides. When the average is given instead, run the same equation backwards for the missing speed.

Definition

A fraction $p$ of the way at $a$ and the rest, $q = 1 - p$ , at $b$ : $\bar v = \dfrac{1}{\tfrac pa + \tfrac qb}$ .
In practice, take a convenient total (a multiple of every denominator), find each part's time and divide.
Given the average of two equal halves and one speed: solve $\dfrac{2ab}{a + b} = \bar v$ for the other.
Speeds such as $x$ and $3x$ for equal halves: the formula gives a multiple of $x$ , set it equal to the average.

Average over fractions of the way

\bar v = \dfrac{1}{\tfrac pa + \tfrac qb} \quad (p + q = 1)

Worked example

A cyclist rides a third of the way at 10 km/h and the rest at 20 km/h. Find his average speed.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Speed, Time and Distance · Average Speed

A person covers

2/5

of distance at

40

km/hr and remaining at

60

km/hr. What is the average speed?

Fractions of distance weight the times, not the speeds

A third of the way at 10 km/h and two thirds at 20 km/h is not

\tfrac13 \times 10 + \tfrac23 \times 20 \approx 16.7

km/h. Each part took 1 h, so the answer is 15 km/h.

Drill 4 more on fractions of the journey, and working back from an average

Concept 5 of 5: Stoppages and breaks

A stop adds time but no distance. Counting its stops, a bus covers less in an hour than its running speed; the shortfall is the distance it would have run while standing still.

Definition

Average including a break $= \dfrac{\text{distance}}{\text{moving time} + \text{break}}$ . Speed while moving $= \dfrac{\text{distance}}{\text{moving time}}$ .
Minutes stopped per hour: the distance lost each hour, $v_1 - v_2$ , at the running speed $v_1$ .
'Without stops it would take 4 h, with stops 5 h' means 1 h of stops.

Minutes stopped per hour

\text{stop per hour} = \dfrac{v_1 - v_2}{v_1} \times 60 \text{ min}

Worked example

Without stops a train runs at 60 km/h; with stops it averages 48 km/h. How many minutes does it stop each hour?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionEasy

Example 5 · Speed, Time and Distance · Average Speed

Excluding stoppages, the speed of a bus is 54 km/hr and including stoppages, it is 45 km/hr. For how many minutes does the bus stop per hour?

Divide by the running speed

The 12 km lost each hour would have been run at 60 km/h, so it is 12 minutes. Dividing by the 48 km/h average gives 15 minutes, which is wrong.

Include the break, or leave it out, as asked

'Average speed including the break' puts the break in the time. 'Speed while driving' leaves it out. The options usually carry both.

Drill 5 more on stoppages and breaks

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 (5)

Total distance over total time
Average speed
$\bar v = \dfrac{d_1 + d_2 + \dots}{t_1 + t_2 + \dots}$
Speeds held for given times
Average over given times
$\bar v = \dfrac{v_1 t_1 + v_2 t_2}{t_1 + t_2}$
Equal distances: there and back, or equal stretches
Average over two equal distances
$\bar v = \dfrac{2ab}{a + b}$
Fractions of the journey, and working back from an average
Average over fractions of the way
$\bar v = \dfrac{1}{\tfrac pa + \tfrac qb} \quad (p + q = 1)$
Stoppages and breaks
Minutes stopped per hour
$\text{stop per hour} = \dfrac{v_1 - v_2}{v_1} \times 60 \text{ min}$

Watch out for (7)

Do not average the speeds
→ Total distance over total time
Weight by time
→ Speeds held for given times
The mean is too high
→ Equal distances: there and back, or equal stretches
Work out a changed speed first
→ Equal distances: there and back, or equal stretches
Fractions of distance weight the times, not the speeds
→ Fractions of the journey, and working back from an average
Divide by the running speed
→ Stoppages and breaks
Include the break, or leave it out, as asked
→ Stoppages and breaks

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