Average speed for a round trip — total distance over total time

The column on the left holds the one-way distance and the speeds out and back. On the right, from the top, are the total distance, the average speed and the total time. Average speed divides the total distance above it by the total time below it.

The bottom card, Mean of the two speeds, is there for comparison: it adds the two speeds and divides by 2. Changing any number updates the four cards on the right together.

Average speed is total distance ÷ total time

A speed is the distance covered divided by the time it took; in units, km ÷ hours = km/h. The average speed of a round trip is the same division: the distance there and back, divided by the time the whole trip took.

Here the total distance is 60 × 2 = 120 km and the total time is 3 hours, so the average speed is 120 ÷ 3 = 40 km/h. At a steady 40 km/h the whole way, the trip would take the same 3 hours.

The mean of the two speeds, (60 + 30) ÷ 2, is 45 km/h. That calculation does not account for the time spent at each speed, which is why it does not give the average speed of the trip.

The slow way back takes twice as long

The two speeds do not count equally, and comparing the time spent on each leg shows why. Board 2 splits the total time into the trip out and the trip back.

The inputs on the left are the same three as on Board 1. The two cards on the right are the time out and the time back.

• Out, at 60 km/h — 60 ÷ 60 = 1 hour
• Back, at 30 km/h — 60 ÷ 30 = 2 hours

The distance is 60 km both ways, but the slower way back takes twice as long. Of the 3 hours on the road, 2 are spent at 30 km/h, so the average speed ends up below 45 km/h, closer to the slower speed.

Together the two legs take 1 + 2 = 3 hours, Board 1's Total time. That card simply puts these two divisions into one formula.

Equal times give the plain average

Averaging the two speeds directly does give the average speed when the same time is spent at each. Board 3 takes a time and a speed for each of two parts of a journey.

On the left are the time and speed for part 1 and part 2. Total distance, top right, adds up time × speed for the two parts. Total time, bottom right, adds the two times. Average speed in the middle is total distance ÷ total time, as on Board 1.

As set up, the board has 1 hour at 60 km/h and 1 hour at 30 km/h. That is 60 + 30 = 90 km in 2 hours, an average of 45 km/h — the same as Board 1's Mean of the two speeds. When the times are equal, adding the speeds and dividing by 2 does give the average speed.

With Part 2 time set to 2, the total becomes 60 + 60 = 120 km in 3 hours, an average of 40 km/h: the round trip again, because 2 hours at 30 km/h covers exactly the 60 km back.

So the round trip's 40 km/h is an average in which 60 km/h counts for 1 hour and 30 km/h for 2 hours: (60 × 1 + 30 × 2) ÷ 3 = 40. An average where each value counts in proportion to how much of it there is — here, how long it lasted — is called a weighted average.

Equal distances give the harmonic mean

When the two legs are the same distance, as on a round trip, the average speed can also be written as 2 × speed out × speed back ÷ (speed out + speed back). This kind of average is called the harmonic mean. For 60 and 30: 2 × 60 × 30 ÷ 90 = 40.

The one-way distance does not appear in that formula. Both the total distance and the total time are proportional to it, so it cancels out in the division — and changing the distance leaves the average speed where it was.

Other numbers

On Board 1, keeping 60 km each way and changing the speeds out and back:

With the same speed both ways, the two numbers agree. The further apart the speeds, the more the average speed leans towards the slower one. Swapping the speeds out and back changes nothing: each speed still covers the same distance, so it is still 40 km/h.

At 10 km/h, the way back alone takes 6 hours (measured on Board 2). Six of the 7 hours are spent at 10 km/h, and the average drops to about 17.14 km/h.

With a one-way distance of 100 km, still 60 km/h out and 30 km/h back, the trip is 200 km in 5 hours. The average speed is 40 km/h, the same as for 60 km.

With the one-way distance back at 60 km and 30 km/h out, the way out alone takes 60 ÷ 30 = 2 hours. Then no speed on the way back can bring the round trip up to an average of 60 km/h: 120 km at 60 km/h leaves only 2 hours for the whole trip, and the way out has already used them. On Board 1, 300 km/h back gives about 54.55 km/h, and even the input's maximum of 1000 km/h gives about 58.25 km/h.

With Part 1 time set to 2 and Part 2 time to 1 on Board 3, the trip is 2 hours at 60 km/h and 1 hour at 30 km/h: 150 km in 3 hours, 50 km/h — closer to the speed kept up for longer.

What this calculation covers, and what it does not

The boards treat each leg as driven at one steady speed. Real trips speed up and slow down for traffic lights and queues. Even so, with the total distance and the total time known, the average speed is still total distance ÷ total time; how the speed varied along the way does not enter into it.

Whether breaks count as part of the total time does change the answer. On Board 3, with part 1 set to an hour at 60 km/h and part 2 to half an hour at a speed of 0 — a 30-minute stop — the trip is 60 km in 1.5 hours, an average of 40 km/h, while the speed on the move stays at 60 km/h. Which of the two applies depends on what the number is for.

Read next — Concentration — finding it, diluting and mixing It calculates concentrations, including the result of mixing two salt solutions: the salt and the total weight are added up separately, and the result is the plain average of the two only for equal weights.

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