The Displacement Calculator finds net change in position under constant acceleration from average velocity and time, from the pair of end velocities, or from initial velocity, acceleration and time. Enter the known values for the matching mode, and the tool returns metres (or another length unit) with each substitution written out.
Displacement answers how far the object ends from where it started, along a chosen axis. That is a different question from how much path it traced, and the gap between those two answers is the teaching point this page owns.
Calculate displacement from velocity and time
Displacement equals average velocity multiplied by elapsed time when acceleration is constant. That single product is the area under the velocity-time graph for the interval. Enter the average velocity (or enough data to form it) and the duration, and the tool returns the signed change in position in the unit selected for length.
s = v̄ tWith v̄ = 8 m/s and t = 3 s:
s = 8 × 3 = 24 m
Those figures match the average of u = 5 m/s and v = 11 m/s over three seconds. If velocity is constant, average velocity is that constant value and the same product applies without needing u and v separately. Variable acceleration breaks the shortcut: the true displacement is still the integral of velocity, but it is no longer ½(u + v)t.
Calculate displacement from initial and final velocity
When both end velocities and the time are known, displacement is the average of those velocities times the interval. Acceleration drops out of the formula, which is why this form is the right choice when a is missing. The arithmetic is identical to multiplying average velocity by time; only the inputs differ.
s = ½(u + v)tRearrangements:
u = 2s/t − v
v = 2s/t − u
t = 2s/(u + v)Substitute u = 5 m/s, v = 11 m/s, t = 3 s:
s = ½(5 + 11) × 3 = ½ × 16 × 3 = 24 m
If u + v = 0 and s is not zero, time cannot be found from this rearrangement: the object has equal and opposite end velocities, and the form that needs their sum fails. Use another equation that keeps acceleration or a known average when that happens.
Calculate displacement from acceleration and time
Initial velocity, acceleration and time fix displacement through a relation that includes a term in t². Final velocity is not required. Free-fall problems usually arrive in this form: drop from rest, set a to g, and solve for how far the object falls in a given time.
s = ut + ½at²Rearrangements for u and a:
u = (s − ½at²)/t
a = 2(s − ut)/t²Solving for t needs the quadratic formula and is covered in its own section below. A cart with u = 5 m/s, a = 2 m/s² and t = 3 s:
s = (5 × 3) + ½(2)(3²) = 15 + ½(2)(9) = 15 + 9 = 24 m
Same 24 m as the average-velocity route, which is a useful consistency check when more than three values are known.
Distinguish displacement from distance
Displacement is a vector from start point to end point. Distance is the scalar length of the path taken. Both can use metres, and both can share a numeric value on a one-way straight trip, which is why the words get treated as synonyms until a problem turns around.
Walk five metres east and then five metres west. Distance travelled is 10 m. Displacement is 0 m, because the end point coincides with the start. Average speed for the trip is positive; average velocity is zero. A ball thrown straight up and caught at the same height has the same pattern: large path length, zero net displacement for the round trip.
| Quantity | Type | Measures |
|---|---|---|
| Distance | Scalar | Path length |
| Displacement | Vector | Straight change from start to finish |
Exam questions that ask for "how far" without naming displacement or distance are ambiguous on purpose. Read whether the answer can be negative. If a signed metre value is expected, the quantity is displacement. Speed-versus-velocity detail lives on the Velocity Calculator; the vector-versus-scalar contrast for position change is what this page owns.
Find displacement for u = 0, a = 9.81, t = 3
An object dropped from rest falls under standard gravity. Taking down as positive, u = 0, a = 9.80665 m/s² and t = 3 s. Prose often rounds g to 9.81; the calculation below uses the exact defined value 9.80665 so the result matches the engine fixture.
1. List what is known. u = 0 m/s, a = 9.80665 m/s², t = 3 s. Final velocity is not required for displacement.
2. Apply the equation. s = ut + ½at²
3. Substitute. s = (0 × 3) + ½ × 9.80665 × 3² s = 0 + ½ × 9.80665 × 9 s = ½ × 88.25985
4. Solve. s = 44.129925 m, reported as 44.13 m to four significant figures.
Air resistance is ignored. Light objects and long falls diverge from this figure. Local g also varies slightly with latitude and altitude; coursework treats 9.80665 (or 9.81) as constant.
Solve for time using the quadratic formula
When displacement, initial velocity and acceleration are known, time satisfies a quadratic because t appears as t and as t² in s = ut + ½at². Two roots can be real. Both positive roots can be physical: a projectile passes a given height once rising and once falling.
Rearranged for t when a ≠ 0:
½a t² + u t − s = 0
t = [−u ± √(u² + 2as)] / aFixture with u = 0, a = 2 m/s², s = 100 m:
t = [0 ± √(0 + 2×2×100)] / 2 = ±√400 / 2
Positive root: t = 20 / 2 = 10 s
Negative root: t = −10 s, discarded as a time before the motion began. When a = 0 the quadratic path is not used; time is simply s / u if u ≠ 0. The Kinematics Calculator reports both positive roots and labels which pass they correspond to on a vertical throw.
Read the sign of a displacement
Positive displacement means the end point lies in the positive direction from the start point you chose. Negative displacement means the finish lies the other way along the same axis. Returning toward the origin reduces the magnitude of displacement even while distance along the path keeps rising with every step.
A runner who jogs to +40 m and back to +10 m has a final displacement of +10 m for the whole outing, not 40 m and not 70 m. The 70 m figure is distance. Sign errors usually come from flipping the axis halfway through: if up was positive for the launch velocity, it must stay positive when gravity is entered as −9.80665 m/s².
Zero displacement does not mean the object stayed still. It means start and finish coincide. Any closed loop in space has zero net displacement and a positive path length.
Frequently asked questions
Displacement problems trip people up when path length and net change in position are confused, or when a quadratic in time returns two roots. The answers below open by restating each question, then give the formula, sign rule or free-fall figure that matches the engine fixtures used on this page.
What is the formula for displacement with average velocity?
Displacement is average velocity times time: s = v̄ t. Under constant acceleration, v̄ = (u + v) / 2, so the same result is s = ½(u + v)t.
How do you calculate displacement with acceleration?
Use s = ut + ½at² when initial velocity, acceleration and time are known. For a drop from rest lasting 3 s at g = 9.80665 m/s², s = ½ × 9.80665 × 9 ≈ 44.13 m.
What is the difference between displacement and distance?
Displacement is the signed straight change from start to finish. Distance is how much path was travelled. A round trip can pile up distance while finishing with zero displacement.
Can displacement be negative?
Yes. Negative means the finish lies opposite the chosen positive direction. Magnitude is still in metres; the minus sign is directional information, not a claim that length is less than zero in an absolute sense.
Why does solving for time give two answers?
Because the displacement equation is quadratic in t. A ball thrown upward crosses any height below the peak twice. Both positive times are real. Negative times are dropped.
Does free fall use g = 9.8 or 9.80665?
Standard gravity is defined as exactly 9.80665 m/s². Textbooks often round to 9.81 or 9.8. Pick one value and stay consistent. This tool uses 9.80665 in calculation and may display 44.13 m for a three-second drop from rest.
How is displacement read from a velocity-time graph?
The area between the velocity graph and the time axis equals displacement. Under constant acceleration that area is a trapezium with parallel sides u and v, which recovers ½(u + v)t.
What if the object changes direction during the interval?
Net displacement still equals the signed area. Motion one way contributes positive area; motion the other way contributes negative area. Distance would add the absolute areas instead.
When should you use s = vt instead of s = ut + ½at²?
Use s = v̄ t (often written s = vt when v is already an average or a constant) when you already have average or constant velocity. Use s = ut + ½at² when you have u, a and t but not the final speed.
Is displacement the same in every reference frame?
Changing the origin shifts absolute coordinates, but the difference between end and start (the displacement for an interval) is unchanged by a fixed shift of origin. Rotating or accelerating frames are a different discussion outside constant-acceleration SUVAT along one axis.
Summary
Net change in position follows from s = v̄ t, from s = ½(u + v)t, and from s = ut + ½at². A three-second free fall from rest at g = 9.80665 m/s² gives s ≈ 44.13 m.
Distance and displacement share a unit and diverge as soon as the path doubles back. Time solved from displacement is quadratic when acceleration is nonzero, so two positive roots can both be physical on a vertical throw. Sign follows the axis declared at the start of the problem.