What is the displacement formula?

s = ut + ½at², one of the standard kinematics equations for constant acceleration, where u is the initial velocity, a the acceleration and t the time. From rest at 2 m/s² for 5 seconds that is 0 + ½ × 2 × 25 = 25 m, and the final velocity is v = u + at = 10 m/s. The ½ is the reason the distance grows with the square of the time: twice the time from rest is four times the distance.

What is the difference between displacement and distance travelled?

Displacement is the straight line from where you started to where you finished, with a direction; distance travelled is how far you actually went. Braking from 10 m/s at 2 m/s² for 10 seconds takes you 25 m out and 25 m back: the displacement is 0 m and the distance travelled is 50 m. Both are correct, and that is why this page prints both — a driver looking at an odometer wants the 50, and a physicist drawing a diagram wants the 0.

How do I calculate stopping distance?

Use s = ut + ½at² with a negative, and stop the calculation when the final velocity reaches zero. From 70 mph (31.29 m/s) at 1 g the time to stop is 3.19 s and the distance is 49.9 m, about 164 ft. Two things are missing from that figure in real driving: the reaction time before the brakes go on — one second at 70 mph is another 31 m — and a deceleration of 1 g, which only a dry road gives you.

Is average velocity the same as (u + v) ÷ 2?

Yes, whenever the acceleration is constant — and it is also displacement divided by time, which gives the same number. From rest to 26.82 m/s in 5 seconds, both routes give 13.41 m/s. The two definitions part company the moment the object reverses: in the example that returns to its start, (u + v) ÷ 2 is 0 as well, but that is a coincidence of symmetry, and in general it is the displacement divided by time that stays meaningful.

What does a negative displacement mean?

That the object finished on the other side of the starting point, measured along whichever direction you decided was positive. It is a direction, not a mistake. The same goes for the velocity and the acceleration: in a braking problem the initial velocity is positive and the acceleration is negative, and a reader who enters both as positive numbers will get a steadily accelerating car instead of a stopping one, with no error message to warn them.

What happens if the acceleration is zero?

The formula collapses to s = ut and this page becomes the ordinary speed times time calculation, which is the job of the speed distance time page. That is a legitimate input here rather than an error — it is the boundary case where the two pages agree — and the final velocity simply equals the initial velocity. The interesting cases on this page all have a non-zero acceleration, which is exactly what the other page cannot express.

Displacement Calculator

Result

25.00 m

Displacement

Distance travelled: 25.00 m
Displacement (ft): 82.02 ft
Final velocity: 10.00 m/s
Final velocity (km/h): 36.00 km/h
Average velocity: 5.00 m/s

Displacement calculator: how far an object ends up from where it started when it is accelerating steadily, plus how far it actually travelled getting there. The formula is s = ut + ½at², with the final velocity from v = u + at. The two distance outputs agree whenever the object keeps going the same way, and they come apart the moment it turns round — which is the whole reason the page reports both. The defaults are a standing start at 2 m/s² for 5 seconds: 25 m of displacement and 25 m of travel, a final speed of 10 m/s, and an average of 5 m/s. The worked examples take the same equation through the case that confuses everyone, where the object brakes to a stop and then rolls back, so it ends up exactly where it began with 150 ft on the odometer. The reference table below is that story in table form, one row per second, and it is the clearest way to see displacement and distance diverge. This page is one-dimensional on purpose: the sign of the velocity and the sign of the acceleration carry the direction, and a two-dimensional problem is this calculation done twice, once per axis.

Displacement, distance and final velocity second by second

Time (s)	Displacement (ft)	Distance travelled (ft)	Final velocity (ft/s)
0	0	0	30
2	48	48	18
4	72	72	6
6	72	78	-6
8	48	102	-18
10	0	150	-30

One object, braking from 30 ft/s at 6 ft/s², sampled every two seconds for ten seconds — and the clearest available picture of the difference this page is about. The displacement column climbs to 72 ft at 4 seconds, is still 72 ft at 6 seconds because the object turned round in between, and is back to 0 at the end. The distance column only ever goes up, reaching 150 ft. The velocity column runs 30, 18, 6, -6, -18, -30, so it passes through zero exactly halfway between the 4 s and the 6 s rows, and that crossing is the moment the two distance columns stop agreeing. Read the middle two columns together: they match for the first four seconds and never match again.

Formula

displacement = u × t + ½ × a × t² final velocity = u + a × t average velocity = displacement ÷ t

u: Initial velocity in metres per second, signed — the field also takes km/h, mph and ft/s. Negative means it is already moving the other way when the clock starts, and this page deliberately allows it
a: Acceleration in m/s², signed. Negative is braking, which is the most common use of this page; the field also takes ft/s² and multiples of standard gravity, so 1 g of braking is 9.80665 m/s²
t: Time in seconds — the field also takes minutes and hours. It has to be positive: average velocity divides by it, so zero is rejected rather than reported as an infinite speed
s: Displacement, ut + ½at², in metres and feet. It is the straight-line distance from the start, with a sign, not the length of the path
v: Final velocity, u + at, in m/s and km/h. It comes from its own formula rather than from the displacement: displacement is the area under the velocity graph and final velocity is its height, and the two are independent of each other

Use this page for any motion where the acceleration is steady: braking distances, take-off and landing runs, a ball thrown up and falling back, a train pulling out of a station, a car merging onto a motorway, a lift's travel time, a sprinter's first seconds. It answers two questions that a speed-distance-time calculation cannot, because that one assumes the speed is constant. The first is how far something goes before it stops, given the deceleration — that is the stopping distance, and it is the single most useful number on this page for a driver. The second is whether the object ever changes direction, because that is where displacement and distance stop agreeing, and a reader who has only been given the displacement will think a zero is a bug. Note what the acceleration has to come from: this page takes it as a given, and the friction coefficient a road can supply is what sets it, so the honest chain for a braking distance runs from the road surface through the friction page and into this one.

Worked examples

A car doing 30 ft/s decelerating at 6 ft/s² for 10 seconds
1. 30 ft/s = 9.144 m/s and 6 ft/s² = 1.8288 m/s² (the fields take SI units)
2. Displacement: 9.144 × 10 + ½ × (-1.8288) × 100 = 91.44 - 91.44 = 0 m
3. Distance travelled: it stops at 5 s and comes back, so 45.72 m out and 45.72 m back... in fact 22.86 m each way, 45.72 m in total
4. Final velocity: 9.144 + (-1.8288) × 10 = -9.144 m/s, so it is moving backwards at the same speed it started
5. Average velocity: 0 ÷ 10 = 0 m/s
This is the example the page exists for. The displacement is zero and the odometer says 150 ft, and both are right — the object is back where it started. Anyone who asked this page one question and got 0 would reasonably conclude it had failed, which is why distance travelled is a separate row rather than a footnote. The final velocity is also worth reading: it is negative, and it has the same magnitude as the starting speed, because a symmetric deceleration over twice the stopping time returns the object to its start. The average velocity row then says something that surprises people — 0 m/s, which is true and useless, and shows why averaging the speeds you pass through is not the same as averaging over the trip.
A car accelerating from rest to 60 mph in 5 seconds
1. 60 mph = 26.8224 m/s, so the acceleration needed is 26.8224 ÷ 5 = 5.36448 m/s²
2. Displacement: 0 × 5 + ½ × 5.36448 × 25 = 67.06 m
3. Distance travelled: the same 67.06 m, because the car never reverses
4. Final velocity: 5.36448 × 5 = 26.82 m/s, which is 96.56 km/h
5. Average velocity: 67.06 ÷ 5 = 13.41 m/s, which is exactly half the final speed
220 ft is a realistic figure for a quick car, and 5.36 m/s² is about 0.55 g of acceleration, which is at the top of what a front-wheel-drive family car can do from a standstill. The last step is the useful identity: from rest, the average speed is exactly half the final speed, so the distance is just the final speed times half the time. That shortcut is only valid when the acceleration is constant and the starting speed is zero, and it is the fastest way to sanity-check any answer this page gives you.
Stopping distance from 70 mph at 1 g of braking
1. 70 mph = 31.2928 m/s; 1 g of braking is 9.80665 m/s²
2. Time to stop: 31.2928 ÷ 9.80665 = 3.19 s
3. Displacement: 31.2928 × 3.191 + ½ × (-9.80665) × 3.191² = 99.86 - 49.93 = 49.93 m
4. Final velocity: 0 m/s — the car is stopped
5. Average velocity: 49.93 ÷ 3.191 = 15.65 m/s, half the initial speed
Just under 50 metres, or 164 ft, and it is the number a driver should carry: 1 g is about the best a dry road gives an ordinary car, and it is what an emergency stop actually achieves. Read it next to the reaction time, which this page does not model — a driver who takes one second to move their foot has already travelled 31 m before the braking starts, so the real distance from 70 mph is closer to 80 m. It also shows why speed is so expensive: the stopping distance goes with the square of the speed, so 70 mph needs about twice the room of 50 mph, and the last few mph before an impact are the ones that decide how bad it is.

Limitations

Everything here assumes the acceleration is constant, and almost nothing in the real world is: a car's brakes are not perfectly progressive, drag rises with the square of speed, and a wheel that locks behaves differently from one that does not. The page is one-dimensional, so it describes motion along a line; anything that curves — a corner, a projectile, a ball thrown at an angle — is this calculation done separately on each axis, and the total is the vector sum rather than the number this page prints. The sign convention matters and is easy to get wrong: the initial velocity and the acceleration must both be measured along the same direction, and a braking car with a positive initial velocity needs a negative acceleration, not a smaller positive one. It assumes the object is a point, so it says nothing about wheelbase, suspension travel or the fact that a real vehicle pitches forward as it brakes. It gives no help with where the acceleration comes from: the friction a road can supply depends on the surface and the tyres, and the friction page is the one that turns a road surface into an m/s². And the distance-travelled output assumes the object reverses at most once, which is all a constant acceleration can produce — anything that oscillates is outside this model.

Frequently asked questions

What is the displacement formula?: s = ut + ½at², one of the standard kinematics equations for constant acceleration, where u is the initial velocity, a the acceleration and t the time. From rest at 2 m/s² for 5 seconds that is 0 + ½ × 2 × 25 = 25 m, and the final velocity is v = u + at = 10 m/s. The ½ is the reason the distance grows with the square of the time: twice the time from rest is four times the distance.
What is the difference between displacement and distance travelled?: Displacement is the straight line from where you started to where you finished, with a direction; distance travelled is how far you actually went. Braking from 10 m/s at 2 m/s² for 10 seconds takes you 25 m out and 25 m back: the displacement is 0 m and the distance travelled is 50 m. Both are correct, and that is why this page prints both — a driver looking at an odometer wants the 50, and a physicist drawing a diagram wants the 0.
How do I calculate stopping distance?: Use s = ut + ½at² with a negative, and stop the calculation when the final velocity reaches zero. From 70 mph (31.29 m/s) at 1 g the time to stop is 3.19 s and the distance is 49.9 m, about 164 ft. Two things are missing from that figure in real driving: the reaction time before the brakes go on — one second at 70 mph is another 31 m — and a deceleration of 1 g, which only a dry road gives you.
Is average velocity the same as (u + v) ÷ 2?: Yes, whenever the acceleration is constant — and it is also displacement divided by time, which gives the same number. From rest to 26.82 m/s in 5 seconds, both routes give 13.41 m/s. The two definitions part company the moment the object reverses: in the example that returns to its start, (u + v) ÷ 2 is 0 as well, but that is a coincidence of symmetry, and in general it is the displacement divided by time that stays meaningful.
What does a negative displacement mean?: That the object finished on the other side of the starting point, measured along whichever direction you decided was positive. It is a direction, not a mistake. The same goes for the velocity and the acceleration: in a braking problem the initial velocity is positive and the acceleration is negative, and a reader who enters both as positive numbers will get a steadily accelerating car instead of a stopping one, with no error message to warn them.
What happens if the acceleration is zero?: The formula collapses to s = ut and this page becomes the ordinary speed times time calculation, which is the job of the speed distance time page. That is a legitimate input here rather than an error — it is the boundary case where the two pages agree — and the final velocity simply equals the initial velocity. The interesting cases on this page all have a non-zero acceleration, which is exactly what the other page cannot express.

References

Displacement (College Physics 2e, §2.1) — displacement as a signed straight-line quantity, and how it differs from the distance walked — OpenStax
Equations of motion — the kinematic equations for constant acceleration, the sign conventions, and the standard braking-distance derivation — Wikipedia
NIST Guide to the SI — the metre, the second and the exact mile and foot factors used to convert the speeds in the examples above — National Institute of Standards and Technology

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