Speed, Time and Distance: Formulas and Unit Conversion

Every question in this chapter rests on one relation between three quantities:

Distance = Speed × Time, so Speed = Distance ÷ Time and Time = Distance ÷ Speed

The formula only works if the units match. Speed in km/h needs distance in kilometres and time in hours; speed in m/s needs metres and seconds. Most wrong answers come from mixing units, not from the formula.

Worked example 1: time for a journey

A bus covers 240 km at 60 km/h. Time = 240 ÷ 60 = 4 hours.

Worked example 2: minutes into hours

A walker at 6 km/h takes 2 hours 30 minutes. That is 2.5 hours, so the distance is 6 × 2.5 = 15 km.

One kilometre is 1,000 metres and one hour is 3,600 seconds, so 1 km/h = 1,000 ÷ 3,600 = 5/18 m/s.

Convert	Multiply by	Example
km/h to m/s	5/18	36 km/h = 10 m/s
m/s to km/h	18/5	10 m/s = 36 km/h

Worked example 3: 72 km/h

72 × 5/18 = 20 m/s.

In the other direction, 15 m/s × 18/5 = 54 km/h. A useful habit is to remember a few pairs: 18 km/h = 5 m/s, 36 = 10, 54 = 15 and 72 = 20.

When the distance is fixed, doubling the speed halves the time. More generally, if the speeds are in the ratio a : b, the times are in the ratio b : a. This removes the need to know the distance at all.

Worked example 4: speed ratio 3 : 4

Two people walk the same road at speeds in the ratio 3 : 4. The faster one takes 30 minutes less. Their times are in the ratio 4 : 3, a difference of 1 part, so 1 part is 30 minutes. The slower one takes 4 × 30 = 120 minutes.

The other two proportions are direct. For a fixed time, distance is proportional to speed; for a fixed speed, distance is proportional to time.

Worked example 5: the same time

Two cars leave together and drive for 3 hours at 50 km/h and 70 km/h. In the same time the distances are in the ratio 5 : 7, and the faster car ends up 3 × (70 − 50) = 60 km ahead.

A special case appears often: "at a fraction of his usual speed, he is late by t minutes". At speed p/q of usual, the time becomes q/p of usual, and the extra time is (q − p)/p of the usual time.

Worked example 6: three-quarters of the usual speed

At 3/4 of her usual speed a woman is 20 minutes late. Her time becomes 4/3 of usual, so the extra is 1/3 of the usual time. Usual time = 20 × 3 = 60 minutes.

"At speed s₁ he is late; at speed s₂ he is early" means the two journey times differ by the late minutes plus the early minutes. Since the distance is the same:

Distance = (s₁ × s₂ × time difference) ÷ (s₂ − s₁)

Worked example 7: 40 and 50 km/h

At 40 km/h a man is 10 minutes late, and at 50 km/h he is 5 minutes early. The times differ by 15 minutes = 1/4 hour. Distance = 40 × 50 × 1/4 ÷ 10 = 50 km. Check: 50/40 h = 75 min and 50/50 h = 60 min, a 15-minute gap.

If both speeds make him late, by different amounts, the time difference is the larger delay minus the smaller one.

A vehicle that stops on the way has two speeds: its running speed without stops, and its lower average speed including stops. The distance lost each hour, run at full speed, gives the minutes spent stopped.

Worked example 8: 60 and 45 km/h

Without stops a bus runs at 60 km/h; with stops it averages 45 km/h. Each hour it loses 15 km, which at 60 km/h takes 15 ÷ 60 × 60 = 15 minutes. So it stops for 15 minutes every hour.

✘ Writing 4 hours 30 minutes as 4.3 hours. It is 4.5 hours.
✘ Multiplying by 18/5 when converting km/h to m/s. It is 5/18.
✘ Using the speed ratio instead of its inverse for times over the same distance.
✘ In late-and-early questions, subtracting the minutes instead of adding them.

These basics feed into train problems, boats and streams and the average speed formula. The formula list is in the Quant cheat code.

Q1. What is 90 km/h in metres per second? easy

(a) 324 m/s (b) 25 m/s (c) 15 m/s (d) 30 m/s

✅ Answer: (b) 25 m/s

💡 Explanation: Multiply by 5/18 to convert km/h to m/s: 90 × 5 ÷ 18 = 25 m/s.

Q2. A car covers 360 km in 4 hours 30 minutes. What is its speed? easy

(a) 90 km/h (b) 72 km/h (c) 75 km/h (d) 80 km/h

✅ Answer: (d) 80 km/h

💡 Explanation: 4 hours 30 minutes is 4.5 hours. Speed = 360 ÷ 4.5 = 80 km/h. Using 4.3 hours for 4 h 30 min is the usual slip.

Q3. Walking at 4/5 of his usual speed, a man reaches office 15 minutes late. What is his usual time for the journey? medium

(a) 60 minutes (b) 75 minutes (c) 45 minutes (d) 48 minutes

✅ Answer: (a) 60 minutes

💡 Explanation: At 4/5 of the speed he takes 5/4 of the time, which is 1/4 of the usual time extra. That extra is 15 minutes, so the usual time is 4 × 15 = 60 minutes.

Q4. At 30 km/h a student is 20 minutes late; at 40 km/h he is 10 minutes early. What is the distance? medium

(a) 40 km (b) 50 km (c) 60 km (d) 72 km

✅ Answer: (c) 60 km

💡 Explanation: The two times differ by 20 + 10 = 30 minutes = 1/2 hour. Distance = product of speeds × time difference ÷ difference of speeds = 30 × 40 × 1/2 ÷ 10 = 60 km.

Q5. Without stoppages a bus runs at 54 km/h; with stoppages its average is 45 km/h. For how many minutes per hour does it stop? medium

(a) 9 minutes (b) 12 minutes (c) 15 minutes (d) 10 minutes

✅ Answer: (d) 10 minutes

💡 Explanation: In each hour the bus loses 54 − 45 = 9 km of running. At 54 km/h, 9 km takes 9 ÷ 54 × 60 = 10 minutes, so it stops 10 minutes per hour.

Q6. Two cyclists cover the same route at speeds in the ratio 5 : 6. The slower takes 12 minutes longer. How long does the slower cyclist take? hard

(a) 72 minutes (b) 60 minutes (c) 66 minutes (d) 84 minutes

✅ Answer: (a) 72 minutes

💡 Explanation: For the same distance, times are in the inverse ratio 6 : 5. The difference of 1 part is 12 minutes, so the slower cyclist takes 6 × 12 = 72 minutes.

Why does 5/18 convert km/h to m/s?

One kilometre is 1,000 metres and one hour is 3,600 seconds. So 1 km/h = 1,000 ÷ 3,600 m/s, which simplifies to 5/18 m/s. Going the other way, multiply by 18/5.

How do I write 4 hours 30 minutes in the formula?

Convert minutes to a fraction of an hour by dividing by 60. Thirty minutes is 0.5 hours, so the time is 4.5 hours. Never write it as 4.30, which would mean 4 hours and 18 minutes.

All problems and solutions on this page are original. No outside facts are relied on; every numeric answer is recomputed by software before publishing.

How this page was prepared. Drafted with AI assistance (Claude (Anthropic AI model)) from the sources listed above. 15 calculations on this page were recomputed by software and matched. Automated checks also covered structure, internal links and duplicate text. Software checks cannot confirm facts. Reviewed by praveenkumar on 2026-10-23 against the sources listed above.

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