Speed, time and distance for AFCAT: average speed, relative speed and late–early questions

Speed, time and distance is the parent of two other AFCAT topics: trains, boats and streams are this chapter with a twist. So the effort you put in here counts three times.

Every question uses distance = speed × time. What separates a 40-second solution from a three-minute one is knowing four supporting ideas: converting units, using proportion instead of equations, the correct formula for average speed, and relative speed. This guide explains each, with the reason it works, and then solves the question types that come up.

Idea 1: units

Keep all three quantities in matching units: km, hours and km/h, or metres, seconds and m/s.

• km/h to m/s: multiply by 5/18. 72 km/h = 72 × 5/18 = 20 m/s.
• m/s to km/h: multiply by 18/5. 15 m/s = 15 × 18/5 = 54 km/h.

Why 5/18: 1 km/h is 1,000 metres in 3,600 seconds, and 1,000/3,600 = 5/18.

Useful pairs to know: 18 km/h = 5 m/s, 36 km/h = 10 m/s, 54 km/h = 15 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s.

Idea 2: proportion saves equations

From distance = speed × time:

• Same time: distance is proportional to speed.
• Same distance: time is inversely proportional to speed. If the speed ratio is a : b, the time ratio is b : a.

So if a person travels at 3/4 of the usual speed, the journey takes 4/3 of the usual time. The extra time is 1/3 of the usual time, and a "late by 20 minutes" question solves in one line: usual time = 60 minutes.

Idea 3: average speed

Average speed = total distance ÷ total time. It is not the average of the speeds.

For two equal distances at speeds a and b:

Average speed = 2ab ÷ (a + b)

Why: take each distance as d. Total time = d/a + d/b = d(a + b)/ab. Total distance 2d divided by that time gives 2ab/(a + b). You spend longer at the slower speed, so the average is pulled towards it.

At 40 and 60 km/h over equal distances, the average is 2 × 40 × 60 ÷ 100 = 48 km/h, not 50.

If the two parts take equal times instead, the average speed is (a + b)/2.

Idea 4: relative speed

When two objects move, what matters is how fast the gap between them changes.

Time to meet or overtake = the gap to close ÷ relative speed.

Late and early

A candidate who arrives late at one speed and early at another has given you the difference between two travel times. Write

d/(slower speed) − d/(faster speed) = total time difference

in hours. "10 minutes late" and "5 minutes early" are 15 minutes apart, which is 1/4 hour.

Stoppages

If a bus averages x km/h without stoppages and y km/h including them, it stops for

(x − y) ÷ x × 60 minutes per hour

Why: in each hour it covers only y km, which at its moving speed x takes y/x of an hour. The rest of the hour, (x − y)/x, is spent stopped.

Worked examples

Example 1 (average speed). A car goes from P to Q at 40 km/h and returns at 60 km/h. Find the average speed for the round trip.

• Equal distances, so 2 × 40 × 60 ÷ (40 + 60) = 4,800 ÷ 100 = 48 km/h.

Example 2 (meeting). Two cars 360 km apart drive towards each other at 50 km/h and 70 km/h. When and where do they meet?

• Relative speed = 120 km/h, so they meet after 360 ÷ 120 = 3 hours.
• The slower car has covered 50 × 3 = 150 km from its start.

Example 3 (late and early). Walking at 4 km/h, an aspirant reaches the exam centre 10 minutes late. At 5 km/h, they are 5 minutes early. Find the distance.

• The time difference is 15 minutes = 1/4 hour.
• d/4 − d/5 = 1/4, so d/20 = 1/4 and d = 5 km.
• Check: at 4 km/h it takes 75 minutes and at 5 km/h 60 minutes. The correct time is 65 minutes.

Example 4 (proportion). Walking at 3/4 of the usual speed, a student reaches 20 minutes late. Find the usual time.

• Time becomes 4/3 of the usual, so the extra 1/3 of the usual time = 20 minutes.
• Usual time = 60 minutes.

Example 5 (overtaking). A cyclist leaves a point at 10 km/h. Thirty minutes later, a second cyclist leaves the same point at 15 km/h in the same direction. When and where does the second catch up?

• In 30 minutes the first cyclist is 10 × 0.5 = 5 km ahead.
• Relative speed = 15 − 10 = 5 km/h, so the gap closes in 1 hour after the second starts.
• They meet 15 × 1 = 15 km from the start.

Example 6 (stoppages). Without stoppages a bus averages 60 km/h; with stoppages, 45 km/h. How many minutes per hour does it stop?

• (60 − 45) ÷ 60 × 60 = 15 minutes.

Common mistakes

• Mixed units. A speed in km/h with a time in minutes gives nonsense. Convert first.
• Adding speeds in the same direction. Same direction subtracts; opposite adds.
• Forgetting the head start in overtaking questions: the gap exists before the chaser starts.
• Treating "late" and "early" as the same side. Late and early add up to the time difference; two "late" figures subtract.

Practice set

1. Convert 54 km/h to m/s.
2. A car covers 300 km in 5 hours. Find its speed.
3. Find the average speed for equal distances at 30 km/h and 50 km/h.
4. Two people 18 km apart walk towards each other at 4 km/h and 5 km/h. When do they meet?
5. Convert 25 m/s to km/h.
6. Travelling at 5/6 of the usual speed, a person reaches 10 minutes late. Find the usual time.
7. A person goes to work at 30 km/h and returns at 20 km/h. The round trip takes 5 hours. How far is the workplace?
8. A bus averages 54 km/h without stoppages and 45 km/h with them. How many minutes per hour does it stop?

Answers:

1. 15 m/s. 54 × 5/18.
2. 60 km/h. 300 ÷ 5.
3. 37.5 km/h. 2 × 30 × 50 ÷ 80 = 3,000 ÷ 80.
4. 2 hours. 18 ÷ (4 + 5).
5. 90 km/h. 25 × 18/5.
6. 50 minutes. Time becomes 6/5 of usual, so 1/5 of usual = 10 minutes.
7. 60 km. d/30 + d/20 = 5, so 5d/60 = 5 and d = 60.
8. 10 minutes. (54 − 45) ÷ 54 × 60 = 9/54 × 60.

What to do next

• Memorise the five km/h and m/s pairs so conversion becomes automatic.
• Practise ten average-speed and ten relative-speed questions, checking units each time.
• Move on to trains, boats and streams, then revise time and work, which uses the same rate × time logic.