Velocity Calculator

average velocity = net displacement ÷ time, and final velocity = 2 × average velocity − initial velocity

Δx

The net displacement — not the distance walked. It is where the object ended up relative to where it started, in metres, kilometres, feet or miles, and it can be negative, which is how the direction gets into the answer. Go 60 m out and 60 m back and this field is zero, which is why the average velocity is zero even though you travelled 120 m

t

How long the journey took, in seconds, minutes or hours. It is the divisor, so it has to be positive; a zero here would be a journey of no duration, and the average velocity would not be defined. Nothing about the sign of the displacement changes this

v₀

The speed the object already had when the measurement started, in metres per second, kilometres per hour, miles per hour or feet per second. Leave it at zero for a standing start. It enters the answer only through the final velocity: a journey that begins at 10 m/s and averages 15 ends at 20, whereas the same journey from a standing start would end at 30

v̄

The average velocity, the main row, in metres per second with kilometres per hour underneath. It is a straight division of the two fields above and it carries their sign. Note what it is not: it is not the average speed. Speed averages the distance covered; velocity averages the displacement, so a there-and-back trip has an average velocity of zero while its average speed is not zero at all

v

The final velocity, from the relation that under constant acceleration the average sits exactly halfway between the start and the end. It is a model assumption rather than a fact about every journey: a steady cruise gives a final velocity equal to the starting one, which is correct, while a journey with stops and starts gives a number that no instantaneous speedometer reading ever showed

Use this page when you know how far something moved and how long it took and you want the velocity rather than the speed: a car trip with a known start and end point, a lift travelling between two floors, a sprinter's 100 m split, or a drone returning to where it took off. The negative sign is the reason to prefer it over a plain distance-over-time division, because it is what makes a there-and-back journey come out at zero instead of at the average of a path length. The final velocity row is the second reason: if the motion is roughly constant acceleration, one subtraction from the average tells you how fast the object was going at the end, which is what you need before you can work out its momentum or its kinetic energy.

1. The defaults: 100 m in 5 s from a standing start

   1. Net displacement 100 m, time 5 s, starting from rest
   2. Average velocity: Δx ÷ t = 100 ÷ 5 = 20 m/s
   3. In kilometres per hour: 20 × 3.6 = 72 km/h
   4. Final velocity: 2 × 20 − 0 = 40 m/s
   5. In kilometres per hour: 40 × 3.6 = 144 km/h

   From a standing start the final velocity is exactly twice the average, and that factor of two is worth keeping: it follows from the average sitting halfway between the start and the end, so if the start is zero the end has to be twice the middle. Here it means a sprinter who covers 100 m in 5 s averages 20 m/s and crosses the line at 40 m/s, which is 144 km/h — faster than any human has run, and the reason this is a worked example rather than a record. The arithmetic is exact and the model is the part that fails.

2. Already moving: 150 m in 10 s, starting at 10 m/s

   1. Net displacement 150 m, time 10 s, initial velocity 10 m/s
   2. Average velocity: 150 ÷ 10 = 15 m/s, which is 54 km/h
   3. Final velocity: 2 × 15 − 10 = 20 m/s
   4. In kilometres per hour: 20 × 3.6 = 72 km/h
   5. Check the assumption: constant acceleration from 10 to 20 m/s over 10 s is 1 m/s², and 10 × 10 + ½ × 1 × 10² = 150 m, which is the displacement we started with

   This is the row that catches an implementation which ignores the starting velocity: drop that field and the final velocity comes out at 30 m/s instead of 20, while the first two rows stay exactly right. The check in the last step is worth doing once, because it shows the three quantities are not independent — pick a displacement, a time and a starting velocity and you have also chosen the acceleration, whether or not you meant to. That is why the final velocity row is only as good as the constant-acceleration assumption.

3. There and back: the average velocity is zero

   1. Net displacement 0 m, time 5 s, initial velocity 10 m/s
   2. Average velocity: 0 ÷ 5 = 0 m/s, and the kilometres-per-hour row is 0 as well
   3. Final velocity: 2 × 0 − 10 = −10 m/s
   4. In kilometres per hour: −10 × 3.6 = −36 km/h
   5. You ended where you started, so the displacement is nothing — and the final velocity is negative because you had to turn round to get there

   This is the case that separates velocity from speed, and it is the reason this page exists next to the speed, distance and time one. Walk 60 m out and 60 m back in 12 s and you averaged 10 m/s of speed and 0 m/s of velocity; both statements are true and they answer different questions. The negative final velocity is the same idea seen from the other end: under the constant-acceleration assumption the object had to be moving backwards at the end in order to arrive at zero displacement from a forward start. A page that took the absolute value of the displacement would report 0 and 10 here, and it would look entirely reasonable.

4. The other direction: 60 m backwards in 6 s

   1. Net displacement −60 m, time 6 s, starting from rest
   2. Average velocity: −60 ÷ 6 = −10 m/s, which is −36 km/h
   3. Final velocity: 2 × (−10) − 0 = −20 m/s, which is −72 km/h
   4. Check the assumption: constant acceleration to reach −20 m/s in 6 s is −20 ÷ 6 = −3.333… m/s², and ½ × (−3.333…) × 6² = −60 m

   The sign is doing real work here rather than decorating the answer: without it, this journey and a 60 m walk forwards would be indistinguishable, and the final velocity row could not be negative at all. It is the same convention as the one used for forces pointing backwards, and it is the reason this page asks for net displacement rather than distance. If you are only ever interested in how fast and never in which way, the speed, distance and time calculator is the right page and this one will keep offering you a minus sign you do not want.

The final velocity row assumes constant acceleration, because it relies on the average sitting halfway between the starting and ending velocities. A journey that is mostly steady with a stop in the middle still produces a number there, and that number is not a speed the object ever had. The displacement is net, so a route that wanders is described only by where it ended up. Speeds are treated as ordinary arithmetic at the top of the input range, where they are a fraction of the speed of light.

What is the average velocity formula?

Average velocity is the net displacement divided by the time taken, v̄ = Δx ÷ t. Cover 100 m in 5 s and it is 20 m/s, which is 72 km/h. The word to notice is net: it is where the object ended up relative to the start, not the distance it travelled, which is why the answer can be zero for a journey that clearly involved moving.

What is the difference between average velocity and average speed?

Velocity divides the displacement; speed divides the distance covered. Walk 60 m out and 60 m back in 12 s and you travelled 120 m, so the average speed is 10 m/s, while your displacement is zero, so the average velocity is 0 m/s. Neither answer is wrong — they answer different questions, and this page gives the second one because direction matters whenever the motion reverses.

Why can the average velocity be negative?

Because it carries the direction of the net displacement. Enter −60 m over 6 s and the average velocity is −10 m/s, which means 10 m/s in the direction you have chosen to call negative. This is the same convention that lets a force pointing backwards be written with a minus sign, and it is why the page asks for net displacement rather than distance: a distance cannot be negative, so it cannot express a direction.

How is the final velocity worked out?

From the average, by v = 2v̄ − v₀. Under constant acceleration the average velocity sits exactly halfway between the starting and finishing velocities, so the finish is twice the average minus the start. Averaging 15 m/s from a 10 m/s start means finishing at 20 m/s. From a standing start the final velocity is simply twice the average, which is why 100 m in 5 s implies 40 m/s at the line.

Can I use this page for a journey with stops?

For the average velocity, yes — that only needs the displacement and the time, and it does not care how the motion was distributed. For the final velocity, no: the doubling relation assumes constant acceleration, and a journey with a stop in the middle has a final velocity that this relation cannot recover. The average comes out right either way, which is the useful half to trust on a real trip.

• Time, Velocity, and Speed (College Physics 2e, §2.3) — average velocity as displacement over time, its sign, and how it differs from average speed — OpenStax
• Velocity — the distinction between velocity and speed, and the average and instantaneous forms — Wikipedia
• NIST Guide for the Use of the International System of Units (SI), Appendix B.8 — Factors for Units Listed Alphabetically: the kilometre-per-hour row (1 km/h = 0.277 777 8 m/s) that this page's second and fourth rows are converted with — NIST