Knowing two speeds alone is not enough to determine the average speed for the entire journey. You also need to know how long you travelled at each speed, or over what distance. If you travel the same distance out and back, the slower segment takes up more time, so it has a greater impact on the overall average.
The round-trip average speed calculator (in Hungarian) is designed for outbound and return journeys of equal length. It handles stops separately and can also calculate the one-way distance from a known total time. The following examples are mathematical models using constant segment speeds; on a real route, traffic and speed limits impose further constraints.
The basic formula is the same for every segment
Average speed = total distance travelled / total time elapsed.
This definition is also found in the OpenStax physics material on speed. You include every segment travelled in the total distance. Stops are included in the elapsed time depending on whether you are looking for the average of the entire journey or the average while moving.
If you travel 150 kilometres in 3 hours from departure to arrival, the overall average is 50 km/h. This figure is not changed by the number of faster or slower segments in between. A detailed breakdown of the route is only necessary when you still need to calculate the total time.
For a segment with varying speed, you can use its measured average if it is the quotient of the segment's total distance and the corresponding time. An instantaneous or peak speed is not suitable for this. Record whether the time basis for the segment average includes stops: waiting time already included must not be added a second time in the summary.
The units for time and speed must be consistent. For speed in km/h, calculate time in hours. 90 minutes is 1.5 hours; dividing directly by 90 minutes would give a unit of km/minute, which is a different numerical value.
Equal distance: why 48 for 60 and 40 km/h?
Suppose the one-way distance is 120 km. Outbound, the constant speed is 60 km/h, so the journey takes 120 / 60 = 2 hours. On the return, at 40 km/h, it takes 120 / 40 = 3 hours. The total distance is 240 km, the time is 5 hours, so the average is 48 km/h.
The arithmetic mean of the two speeds would be 50. This differs here because 40 km/h was valid for three hours, while 60 km/h was valid for two. Weighting by time gives the same result: (60 × 2 + 40 × 3) / (2 + 3) = 48.
For two segments of equal length without stops, the shorter formula is 2 × v₁ × v₂ / (v₁ + v₂). In the example, 2 × 60 × 40 / 100 = 48. This is the harmonic mean of the two positive speeds. Always state the conditions of the formula, as it cannot be applied unchanged to unequal distances.
The one-way distance cancels out in the quotient here. A 12 + 12 kilometre route and a 120 + 120 kilometre route give the same average without stops at the two given speeds. The duration naturally changes by a factor of ten.
Equal duration: when is 50 km/h correct?
Now you travel for half an hour at 60 km/h, then for half an hour at 40 km/h. The first segment is 30 km, the second 20 km. The total distance is 50 km, the total time is 1 hours. The average is 50 km/h, which is indeed the arithmetic mean of the two speeds.
In this example, the time for the segments is equal, but the distances differ. If every segment has the same duration, you can simply average the speeds. For differing durations, a time-weighted average is required.
Ten minutes at 60 km/h means 10 kilometres. Fifty minutes at 40 km/h means 33.333… kilometres. The overall average for the hour is thus 43.333… km/h. The pair of speeds remains unchanged, but the differing time ratio yields a new result. Therefore, keep the duration of the segments when calculating the average for a workout or route.
Stops: two meaningful, different averages
Let's return to the 120 kilometre round trip. The moving time is 5 hours, with a total of 30 minutes of stops. The total elapsed time is 5.5 hours. The average while moving remains 48 km/h, but the average calculated from departure to arrival is 240 / 5.5 = 43.636… km/h.
The calculator provides both results. For journey time planning, start with the data that includes stops, because the actual arrival time affects the meeting or connection time. In a performance log, record which average is included so that different days are comparable.
The automatic pausing of a stopwatch app can alter the displayed basis. If one device excludes waiting time and the other includes it, they may show different averages for the same route. With the time difference calculator, you can separately check the elapsed time from the departure and arrival times.
Calculating distance back from total time
You know the two constant segment speeds: 60 and 40 km/h. The round trip, including stops, lasted 330 minutes, of which 30 minutes were breaks. The moving time is 300 minutes, i.e. 5 hours. Let the unknown one-way distance be d.
The time equation is d / 60 + d / 40 = 5. The coefficient of the two time quotients is 1/60 + 1/40 = 1/24, so d / 24 = 5. From this, d = 120 km. In the calculator's time mode, you get the same result by entering the total time and the stop.
If the stop time is equal to or longer than the total time, no positive distance can result from the model. In this case, the inputs must be corrected. Subtract the break once from the 330 minutes; entering 30 minutes again alongside the already cleaned 300 minutes of moving time would result in an incorrect distance.
Is the targeted overall average achievable at all?
For a theoretical tour, the distance out and back is 10 km each. The constant outbound speed is 4 km/h, and you plan a total of half an hour of stops. The outbound journey alone uses 10 / 4 = 2.5 hours. We now look for the return speed from the desired overall average.
If 4 km/h is the target average including stops, the total 20 km allows 20 / 4 = 5 hours. This leaves 5 − 2.5 − 0.5 = 2 hours for the return. This requires a return speed of 10 / 2 = 5 km/h.
If you raise the target average to 5 km/h, the total time can only be 20 / 5 = 4 hours. This leaves 4 − 2.5 − 0.5 = 1 hours for the return, so 10 km/h would be needed. Raising the target average by 25% in this situation requires twice the return speed.
| Target overall average, including stops | Allowed total time | Time remaining for return | Required return speed |
|---|---|---|---|
| 4 km/h | 5 hours | 2 hours | 5 km/h |
| 5 km/h | 4 hours | 1 hours | 10 km/h |
Measure the plan against the pace you can actually sustain. If you can only count on a maximum of 5 km/h on the return, the overall average in this example will be at most 4 km/h. To reach a higher target, you must change the route, the departure plan, or the available time.
There is also a purely mathematical limit. The outbound journey and the break already occupy 3 hours; any return journey of positive length adds further time to this. The overall average is therefore less than 20 / 3 ≈ 6.6667 km/h, no matter what finite return speed you enter. Reaching this limit would require zero return time.
Generally, first calculate the quotient of total distance / target average. Subtract the known journey time and stops. A positive, finite return speed can only be calculated from a positive remainder. In the calculator, you can check the average by entering the resulting speed back into the distance mode; there is no separate target speed planning field.
Unequal distances and multiple segments
In the theoretical example, you travel a 200 km segment at 50 km/h, then a 100 km segment at 100 km/h. The times are 4 and 1 hours. The total 300 km corresponds to 5 hours, so the average is 60 km/h. The short formula derived for two speeds over equal distances would give a different result here.
Work in the same way for multiple segments: calculate the time for each, sum the distances and the times, and finally divide the two sums. The unit converter helps if the input distances or speeds are in different units.
The OpenStax chapter introducing equations of motion provides further examples of the relationship distance = speed × time. In the DG round-trip tool, the one-way distance is common to both segments, so check the 200 and 100 kilometre example with a separate summary.
In a current, relate both speeds to the bank
In your own ideal boat model, the constant speed relative to the water is 10 km/h, and the river flows uniformly at 2 km/h. The boat travels ten kilometres downstream parallel to the current, then returns over the same section. The speed relative to the bank is 12 km/h downstream and 8 km/h upstream. The addition is derived in the OpenStax chapter on relative motion.
The total journey time is 10/12 + 10/8 ≈ 2.0833 hours. The average speed calculated with the unrounded time is 20 / (10/12 + 10/8) = 9.6 km/h. Thus, the increase of two km/h in one direction and the same decrease suffered in the other do not restore the 10 km/h average: the slower direction takes longer.
Generally, for a speed v relative to the water and a current u, for two equal distances the formula is v_avg = (v² − u²)/v = v − u²/v, provided that 0 ≤ u < v, and there are no stops. If u reaches v, the boat no longer makes progress upstream in the model; in this case, no finite round trip can be constructed. The example is a speed and time calculation; it does not size the real boat's performance or the safety of a journey.
Distance and displacement: zero can also be correct
If you return to the starting point, your displacement is zero. Your average speed calculated from the total distance is still positive. In physics, the quotient of displacement/time belongs to the average velocity vector; it is worth distinguishing this from the value of distance travelled/time.
The calculator shows the average of the distance actually travelled. As a check, without stops, the result should fall between the two positive segment speeds. With stops, it can be lower than the lower moving speed. With these two quick checks, many unit and time management errors can be detected before the result is used.
Final knowledge test
Choose one answer for each question. Each correct answer earns 1 point; an incorrect answer, a skipped question or more than one selected answer earns 0 points. You can score 0–3 points in total. Read the solutions after answering.
1. You travel 120 km at 60 km/h, and the same distance at 40 km/h, without stops. What is the average speed for the entire journey?
- A) 48 km/h.
- B) 40 km/h.
- C) 50 km/h.
2. You walked 10 km at 4 km/h, then rested for half an hour. What return speed is needed for the 20 km journey to have an average of 5 km/h, including the rest?
- A) 10 km/h.
- B) 5 km/h.
- C) 6 km/h.
3. The boat's speed relative to the water is 10 km/h, and the current is 2 km/h. What is the average speed relative to the bank for the 10–10 km round trip, without stops?
- A) 9.6 km/h.
- B) 10 km/h.
- C) 12 km/h.
Solutions and scoring
1. A. The two times are 2 and 3 hours. The total distance of 240 km corresponds to 5 hours, so 240/5 = 48 km/h. For revision: Equal distance: why 48 for 60 and 40 km/h?.
2. A. The target allows a total of 20/5 = 4 hours. Of this, 2.5 hours of walking and 0.5 hours of rest have passed, so 10 hours remain for 1 km. For revision: Is the targeted overall average achievable at all?.
3. A. Relative to the bank, the two speeds are 12 and 8 km/h. The total time is 10/12 + 10/8 hours, so the average for the 20 km journey is 9.6 km/h. For revision: In a current, relate both speeds to the bank.
Each correct answer is worth 1 point; an incorrect or skipped answer is worth 0 points. The total score can be 0–3.
- 0 points: work through the sections marked in the three solutions again, and write down the initial conditions of the examples.
- 1 point: derive the correct answer for the two missed questions; separately note which condition or calculation step was omitted.
- 2 points: return to the section linked to the single missed question, then check the result again by changing one input.
- 3 points: you correctly applied the three relationships in this task set. Try one of the article's checks with your own data, recording the conditions and units as well.
Translated and adapted from the Hungarian article: Hungarian original.