Module 3: Kinematics · Theory · Prerequisite: K1
A journey has two averages, and the one you want depends on what you were asking.
Mahavakya
An average compresses a history into a single number. What survives the compression depends entirely on what you divided by.
On this page
Learning objectives
After this reading you should be able to:
- Compute average speed and average velocity for a journey with a reversal in it.
- Say why the arithmetic mean of two speeds is usually the wrong answer.
- Handle the equal-time and equal-distance cases without memorising two formulas.
- Ask, of any averaging problem, what is being accumulated and what is being held fixed.
- State the relation between average speed and the magnitude of average velocity, and when they are equal.
1. Two averages, not one
K1 established that a journey has two lengths — the distance travelled and the magnitude of the displacement. Divide each by the time taken and you get two different averages.
average speed = s / Δt a scalar
average velocity = Δr / Δt a vector
Neither is more real than the other. Average speed asks how hard the journey worked; average velocity asks how much progress it made. They answer different questions and disagree whenever the motion reverses.
Worked example
A cyclist rides 900 m north in 3 minutes, rests for 1 minute, then rides 300 m south in 2 minutes. Find the average speed and the average velocity.
Total time is 6 min = 360 s. Distance is 900 + 300 = 1200 m. Displacement is 900 − 300 = 600 m north.
average speed = 1200 / 360 = 3.33 m/s
average velocity = 600 / 360 = 1.67 m/s north
The rest minute appears in both denominators and in neither numerator. It lowers both averages without contributing to either distance or displacement — which is exactly right, because the cyclist was there and was not moving.
2. Why you cannot average speeds
A car covers the first half of a journey at 20 km/h and the second half at 60 km/h. What is the average speed?
Almost every student answers 40. It is the arithmetic mean, it looks obviously right, and it is wrong — or rather, it is the answer to a question that was not asked.
The trouble is the phrase “the first half”. Half of what? Half the time and half the distance are different journeys, and they give different answers with the same two speeds.
Same speeds, same total distance, different averages.
The reason is visible in the picture. When the times are equal, the car spends as long going fast as going slow. When the distances are equal, it spends far longer at the slow speed — two hours against forty minutes — so the slow speed dominates the average.
Averaging speeds directly is almost always wrong, because an average speed is a total distance divided by a total time — and neither total is obtained by averaging the parts.
3. Equal times
A body travels at v1 for a time t, then at v2 for the same time t. The distances are v1t and v2t, and the total time is 2t:
vavg = (v1t + v2t) / 2t
= (v1 + v2) / 2 the arithmetic mean
So the naive answer is right — when the times are equal. That is the one case in which it is, and it is also the case students assume without checking. With n equal intervals the result generalises to the ordinary mean of the n speeds.
4. Equal distances
Now the same body covers a distance d at v1 and then another d at v2. This time the times are unequal — d/v1 and d/v2:
vavg = 2d / (d/v1 + d/v2)
= 2v1v2 / (v1 + v2) the harmonic mean
Notice that d cancels. The answer does not depend on how long the halves were — only on the two speeds, which is why the result is worth deriving once rather than memorising.
With the figure’s numbers: 2(20)(60)/(80) = 2400/80 = 30 km/h, as drawn.
The harmonic mean is always the smaller of the two. That is not a coincidence — equal distances means more time spent at the lower speed, and time is what the average is divided by. If your equal-distance answer comes out larger than the equal-time answer, you have made an arithmetic error.
5. The question that decides
You now have two formulas, which is two more than you need. Neither is worth memorising, because both come from the same definition and one question tells you which applies.
What is being accumulated, and what is being held fixed?
Average speed accumulates distance and divides by time. So find the total distance, find the total time, and divide. Everything else is arithmetic.
Applied to the two cases: if the times are equal, the times are what you know, and the distances follow from v = s/t. If the distances are equal, the distances are what you know, and the times follow. In both you are doing the same thing — assembling the two totals — and the formulas are just the results of having done it.
That question will keep working. It applies in K8 when the equations of motion are derived, in K9 when acceleration varies, and in K13 when velocities are measured from different frames. Whenever an average appears, ask what is being accumulated and what is fixed.
6. Stopping counts
A bus travels 3 km at 30 km/h, stops for 6 minutes, then travels 3 km at 45 km/h. Find its average speed.
The temptation is to average the two speeds, or to apply the harmonic mean and be done. Both ignore the stop, and the stop is doing real work in the answer.
first leg: 3/30 h = 6 min stop: 6 min second leg: 3/45 h = 4 min
total distance 6 km, total time 16 min = 4/15 h
vavg = 6 ÷ (4/15) = 22.5 km/h
Lower than either speed the bus ever travelled at, and correctly so: for six of the sixteen minutes it was not travelling at all. The average is over the whole interval, and the whole interval includes the standing still.
7. The bound between them
K1 showed that s ≥ |Δr| for any journey. Divide both sides by the same Δt and the inequality survives:
average speed ≥ |average velocity|
Equality only when the motion is in a straight line without reversal — the same condition as before, because it is the same inequality.
This is worth carrying as a check. If you compute an average speed of 4 m/s and an average velocity of 7 m/s, you have made a mistake, and you know it without re-reading the question.
One consequence catches students out. A body that returns to where it started has zero displacement, so its average velocity is zero however fast it went. Not small — zero. Meanwhile its average speed may be considerable. Both statements describe the same journey.
Contemplation
Two cars complete the same route in the same time. Both have an average speed of 60 km/h. The first held sixty the whole way. The second alternated between twenty and a hundred, braking and accelerating for the entire journey.
One number describes both, and the number is honest. It is also nearly silent about what happened. An average is a compression, and compression means discarding — usefully, deliberately, and at a cost that does not announce itself.
That is worth remembering the next time an average seems to settle a question. It settles the question it was constructed to answer. Whether that was your question is a separate matter, and no amount of computing the average more carefully will tell you.
Common misconceptions
1. Average speed is the average of the speeds
Only when the time intervals are equal. Average speed is total distance over total time, and neither total comes from averaging.
2. Average velocity is the average of the velocities
Same error with a vector. It is net displacement over total time, and the intermediate velocities do not appear.
3. Rest periods can be ignored
They add to the total time and nothing to the distance, so they lower the average. The bus above averages 22.5 km/h having never travelled slower than 30.
4. vavg = (u + v)/2 is a general formula
It holds only for constant acceleration, where the velocity changes uniformly, and you will meet it properly in K8. Applied to a journey in two stages at different constant speeds it is the equal-time case, which is a different situation with the same algebra. Applied to anything else it is wrong.
5. Zero average velocity means the body did not move
It means the body ended where it began. A runner completing a lap has zero average velocity and a substantial average speed.
Reflection
- Under what circumstances would the equal-time and equal-distance averages come out the same?
- A journey’s average speed is 50 km/h. What is the largest its average velocity could be, and the smallest?
- Why does the distance d cancel in the harmonic mean but the time t not cancel in the arithmetic mean — or does it?
Key takeaways
- Average speed is total distance over total time; average velocity is net displacement over total time.
- Averaging the speeds gives the right answer only when the times are equal.
- Equal times give the arithmetic mean; equal distances give the harmonic mean, which is always the smaller.
- Rather than memorising both, ask what is being accumulated and what is held fixed.
- Time spent at rest counts in the denominator and not the numerator.
- Average speed ≥ |average velocity|, with equality only for straight unreversed motion.
What comes next
Every average on this page describes an interval. None of them says how fast the body was going at any particular moment — the cyclist averaging 3.33 m/s was, at various instants, doing rather more and rather less, and nothing here can recover that.
K4 shrinks the interval until the question becomes answerable.
Prerequisite: K1 · Reading: 14 min · Practice: 25 min · Difficulty: Beginner