Speed, distance, time
v = d / t
The v = d / t triangle, all three ways: average speed from distance and duration, distance from speed and duration, travel time from distance and speed. The duration is entered in hours, minutes and seconds, the way you know it; the speed is returned both in km/h and in m/s (exact factor 3.6) and as a pace per kilometre, the runner’s reading. The note explains the classic round-trip trap: the average speed is not the average of the speeds.
50 km/h
- Calculation
- v = 100000 m / 7200 s = 13.889 m/s
- Pace
- 1 min 12 s / km
The average speed is 50 km/h, that is 13.889 m/s.
This is an average speed: total distance over total time, stops included if they are in the duration. It says nothing about the instantaneous speed, the one on the dial.
The average speed of a round trip is not the average of the speeds: 60 km/h out and 40 km/h back make 48 km/h, not 50, because the slow leg takes more time. Always go back to total distance over total time.
The factor between km/h and m/s is exactly 3.6: an hour has 3600 seconds and a kilometre 1000 metres. Dividing km/h by 3.6 gives m/s; multiplying m/s by 3.6 gives km/h.
Scientific dossier
What the tool computes, what it assumes, where it stops being valid, and where its data comes from.
Method & formulasv = d / t
v = d / t
d = v × t
t = d / v
v (m/s) = v (km/h) / 3.6
pace (min/km) = time / distance
One relation, three readings. The factor 3.6 between km/h and m/s is exact: 3600 seconds per hour, 1000 metres per kilometre. Pace is the inverse of speed, per kilometre: 12 km/h and 5 min/km are the same thing.
- Average speed
- · total distance divided by total time, stops included if they are in the duration.
- Instantaneous speed
- · the one on the dial, at a given instant. This tool says nothing about it: it only sees totals.
- Pace
- · the time per kilometre, the inverse of speed. The runner’s natural quantity: you aim for "5 min per km", not "12 km/h".
Validity domainThe tool computes averages over a trip: it assumes nothing about what happens between departure and arrival, neither constant speed nor a straight line.
The tool computes averages over a trip: it assumes nothing about what happens between departure and arrival, neither constant speed nor a straight line. The predicted travel time (t = d / v), however, assumes the stated speed is held on average over the whole distance, which a real trip with stops does not do: the result is a floor, not a promise.
The round-trip trapDrive out at 60 km/h, come back the same way at 40 km/h: the average is not 50 km/h but 48.
Drive out at 60 km/h, come back the same way at 40 km/h: the average is not 50 km/h but 48. The slow leg takes more time than the fast one, and the time-weighted average of the speeds leans towards the slow one: that is the harmonic mean, 2 / (1/60 + 1/40) = 48. The only calculation that never lies is going back to the totals: total distance, total time, one division. That is exactly what the tool does, and that is why it asks for the duration rather than two speeds.
Reading the resultThe speed is returned in both units because both serve: km/h for the road, m/s for physics, where every formula (kinetic energy, momentum, free fall) expects metres per second.
The speed is returned in both units because both serve: km/h for the road, m/s for physics, where every formula (kinetic energy, momentum, free fall) expects metres per second. The pace per kilometre is the same information seen from the runner’s side. The travel time is broken into hours, minutes and seconds: 2.5 h reads 2 h 30 min.