Average Speed vs Average Velocity Calculator
Average Speed vs Average Velocity Calculator
Average speed is total distance over total time. Average velocity is net displacement over total time. The headline here is the ratio of the two — one when the journey is a straight run, exactly zero when you end up where you started, and a verdict on how much your path doubled back.
A two-leg journey: speed, velocity and the ratio between them
You cycle 300 m along a straight road in 60 s, then turn round and cycle 100 m back in 40 s. Find the average speed, the average velocity, and the ratio of displacement to distance.
Same denominator, different numerator — and the ratio that compares them
- total distance
- the length of the path actually travelled, in m. Always positive, and it adds up over every leg whichever way the leg went.
- net displacement
- the straight-line change in position from start to finish, in m, with a sign. A return journey has a net displacement of zero.
- total time
- the whole elapsed time, in s, including any stationary intervals. Both averages use it, which is why the ratio between them is purely a ratio of path to displacement.
- ratio
- displacement ÷ distance, the headline figure. 1 for a straight run forwards, 0 for a journey that returns to its start, −1 for one that ends up straight backwards. Dimensionless.
- time-weighted
- both averages weight each leg by its duration, not by its length. That is why the plain mean of two leg speeds is generally not the average speed.
Worked example
You cycle 300 m along a straight road in 60 s, then turn round and cycle 100 m back in 40 s. Find the average speed, the average velocity, and the ratio of displacement to distance.
Total distance: add every leg. 300 + 100 = 400 m. The direction of a leg never matters for distance; it is a length of path.
Net displacement: take the directions into account. Calling the outward direction positive, 300 − 100 = 200 m. You finish 200 m from where you started.
Total time. 60 + 40 = 100 s, for both averages.
The two averages. Average speed = 400/100 = 4.00 m/s. Average velocity = 200/100 = 2.00 m/s in the outward direction. The speed is exactly twice the magnitude of the velocity here.
The headline ratio. displacement ÷ distance = 200/400 = 0.500: half of the path you covered turned into net progress.
And the trap. The leg speeds are 300/60 = 5.00 m/s and 100/40 = 2.50 m/s, whose plain mean is 3.75 m/s. The average speed is 4.00 m/s, not 3.75 m/s, because the faster leg lasted longer and both averages weight by time. Averaging the speeds is wrong unless the legs took equal times.
One journey, four numbers that are all called an average of something
| Quantity | Definition | This journey | When is it the answer? |
|---|---|---|---|
| Average speed | total distance ÷ total time | 400/100 = 4.00 m/s | Whenever the question says speed, or asks how fast you were going on average. |
| Average velocity | net displacement ÷ total time | 200/100 = 2.00 m/s | Whenever the question says velocity, or asks for a direction as well as a size. |
| Mean of the leg speeds | (5.00 + 2.50)÷2 | 3.75 m/s — not an average speed | Only when every leg took the same time. Otherwise it is nothing. |
| Displacement ÷ distance | 200 ÷ 400 | 0.500 | When you want to know how direct the journey was. It is the ratio of the first two. |
Four journeys, and what happens to each average
| Journey | Distance | Displacement | Average speed | Average velocity | Ratio |
|---|---|---|---|---|---|
| Straight run, 400 m in 100 s | 400 m | 400 m | 4.00 m/s | 4.00 m/s forwards | 1.000 |
| 300 m out, 100 m back, 100 s | 400 m | 200 m | 4.00 m/s | 2.00 m/s forwards | 0.500 |
| 200 m out, 200 m back, 100 s | 400 m | 0 m | 4.00 m/s | exactly zero | 0.000 |
| 100 m out, 300 m back, 100 s | 400 m | −200 m | 4.00 m/s | 2.00 m/s backwards | −0.500 |
| 0 m out, 400 m back, 100 s | 400 m | −400 m | 4.00 m/s | 4.00 m/s backwards | −1.000 |
| 380 m out, 20 m back, 100 s | 400 m | 360 m | 4.00 m/s | 3.60 m/s forwards | 0.900 |
Why the denominator is the same, and the mean-of-speeds trap
The two averages differ in one place only: the numerator. Both divide by the total elapsed time. Average speed puts the length of the path on top; average velocity puts the net change in position on top. Everything else about them follows from that single difference, including why the speed can never be smaller than the magnitude of the velocity — a path between two points is never shorter than the straight line between them.
The consequences are worth spelling out. Average speed is a scalar and is always positive. Average velocity is a vector: it has a sign in one dimension and a direction in two or three, and it can be negative or zero. A journey that returns to its starting point has a net displacement of exactly zero and therefore an average velocity of exactly zero, however hard you worked — and the average speed over the same journey is not zero at all. If a question about a return trip has the answer “zero”, that is usually the whole point of the question rather than a sign that you have gone wrong.
The trap is averaging the speeds. Both averages weight each leg by its duration, so a plain mean of two leg speeds is the right answer only when the two legs took equal times. In the worked example the legs are 5.00 m/s for 60 s and 2.50 m/s for 40 s; the plain mean is 3.75 m/s and the actual average speed is 4.00 m/s, because the faster leg lasted longer. The classic version of this is the journey out at one speed and back at another over the same distance, where the average speed is the harmonic mean rather than the arithmetic one: 60 km/h out and 40 km/h back over the same road averages 48 km/h, not 50, because the slower leg occupies more of the time. The figures under the answer print the plain mean alongside the real average speed so you can watch them disagree.
Do not forget the stationary time. Both denominators are the total elapsed time, which includes every second spent waiting at a junction or stopped at a station. A train that covers 120 km in two hours of running and then stands for one hour has an average speed of 40 km/h over the three hours, not 60. Questions are written specifically to see whether the stationary interval was counted, and omitting it is one of the few errors here that cannot be spotted by a sanity check on the final number.
The link to the rest of kinematics is the displacement-to-distance ratio that this page leads with. It is exactly the average velocity divided by the average speed, it runs from −1 to 1, and it measures how direct a journey was independently of how fast it was. It is also the same distinction that decides whether the signed area under a velocity–time graph answers the question — the motion-graph page works through that — and the same distinction that separates the displacement s returned by the equations of motion from the distance travelled when the motion reverses inside the interval.
In two or three dimensions the same definitions hold with the word displacement doing more work. The average velocity is the straight-line vector from the start point to the finish point divided by the total time, whatever shape the path between them took; the average speed is the length of that path divided by the same time. So a car that drives 3 km east and then 4 km north in half an hour has travelled 7 km, giving an average speed of 14 km/h, but its displacement is the 5 km hypotenuse, giving an average velocity of 10 km/h on a bearing east of north. Neither number is wrong and they answer different questions; the ratio 5/7 is what this page’s headline would report.
A note on instantaneous versus average, because the words matter in a mark scheme. Everything on this page is an average over a stated interval, and an average tells you nothing about what happened inside it: a journey with an average speed of 4 m/s may have been a steady 4 m/s or a sprint followed by a rest. The instantaneous speed is the limit of the average as the interval shrinks to nothing, which on a displacement–time graph is the gradient of the tangent rather than of a chord. If a question says “at t = 6 s” it wants an instantaneous value and an average will not do; if it says “over the first 6 s” it wants an average and a tangent will not do.
One more idea that follows for free. For uniformly accelerated motion, and only then, the average velocity is also (u + v)/2 — the mean of the start and end velocities. That is why the equation s = (u + v)t/2 exists and why it needs no acceleration in it. It is a special case and not a definition: for any non-uniform acceleration, (u + v)/2 is not the average velocity, and using it is the same category of error as averaging the leg speeds above.
Where the two averages matter outside the classroom. A delivery driver’s average speed decides the fuel and the hours; the average velocity decides nothing, because the van comes back to the depot and the average velocity is zero. A runner on a 400 m track has a lap-average velocity of exactly zero and a lap-average speed that wins races. A migrating bird’s average velocity over a season is the thing that gets it to its destination, and its average speed is the thing that costs it energy. In each case the quantity worth measuring is the one that matches the question being asked, and the only way to get that wrong is not to notice there are two of them.
Frequently asked questions
What is the difference between average speed and average velocity?
The denominator is the same and the numerator is not. Average speed is the total distance travelled divided by the total time; average velocity is the net displacement divided by the total time. Speed is a scalar and always positive; velocity is a vector and carries a sign or a direction. They are equal in magnitude only when the motion never reverses, and the average speed is never the smaller of the two.
Can the average velocity be zero while the average speed is not?
Yes, and it is the standard illustration. Any journey that finishes where it started has a net displacement of exactly zero, so its average velocity is exactly zero — while its average speed is the total path length over the total time and is not zero. Walk 200 m out and 200 m back in 100 s and the average speed is 4 m/s while the average velocity is 0. If a question gives the answer zero, that is the answer.
Why can I not just average the two leg speeds?
Because both averages weight each leg by its duration, not by its length, and a plain mean weights them equally. 300 m at 5.00 m/s then 100 m at 2.50 m/s gives an average speed of 4.00 m/s, not the 3.75 m/s a plain mean suggests, because the faster leg lasted 60 s against the slower leg’s 40 s. Averaging the speeds is correct only when the legs took equal times, which is rare and is usually the thing the question is testing.
What about out at 60 and back at 40 over the same distance?
That is the harmonic-mean case, and the answer is 48 km/h rather than 50. Over the same distance the slower leg takes longer, so it occupies more of the total time and pulls the average down. The general result for two equal distances is 2v1v2/(v1 + v2), which for 60 and 40 is 4800/100 = 48. Enter the two legs as equal distances with their own times above and the calculator shows it.
Does time spent stationary count?
Yes, in both denominators. The total time is the whole elapsed time from start to finish, including waiting. A train covering 120 km in two hours of motion and then standing for an hour has an average speed of 40 km/h across the three hours. This is one of the most frequently set versions of the question and the omission is invisible in the final number, so there is nothing to catch it except reading the question carefully.
Is the average velocity the same as (u + v)/2?
Only for uniformly accelerated motion. When the acceleration is constant, the average velocity does equal the mean of the initial and final velocities, and that is exactly why s = (u + v)t/2 is a valid equation of motion. For any other kind of motion the mean of the end velocities is not the average velocity — a journey that starts and ends at rest has (u + v)/2 = 0 and may well have covered a hundred metres in between.
How does this work in two dimensions?
Identically, with displacement meaning the straight-line vector from start to finish. Drive 3 km east then 4 km north in half an hour: the distance is 7 km, so the average speed is 14 km/h, while the displacement is the 5 km hypotenuse, so the average velocity is 10 km/h on a bearing east of north. The ratio of displacement to distance is 5/7 = 0.714. Nothing about the definitions changes; only the arithmetic for the displacement gets a Pythagoras step.
Related calculators
References
- NCERT, Physics: Textbook for Class XI, Part I, Chapter 2 “Motion in a Straight Line”, §2.4 “Kinematic Equations for Uniformly Accelerated Motion”. Derives the three equations writing v0 for the initial velocity and x for the displacement, and takes g as 9.8 m s−2 (10 m s−2 in Example 2.3).
- OCR, Data, Formulae and Relationships Booklet for AS and A Level GCE Physics A (H156 and H556), 8 pages. Lists the acceleration of free fall as g = 9.81 m s−2 alongside the thirteen other physical constants the papers assume.
- BIPM, The International System of Units (SI), 9th edition (2019). Definitions of the metre and the second, and the convention that the radian is the coherent SI unit of plane angle with dimension one — which is why every trigonometric formula in mechanics is written in radians.