Uniform Acceleration Displacement Calculator s = v₀t + ½at²

The uniform-acceleration displacement calculator solves s = v0 t + \tfrac12 a t^2: given an initial velocity v0, a constant acceleration a and a time t, it finds the displacement s over that interval; it also runs in reverse, giving the initial velocity or the acceleration from a displacement and a time—how far a stone falls in 3 seconds, how far a car travels braking from 100 km/h, and what average acceleration covers 100 m in 8 seconds from rest.

Uniform Acceleration Displacement Calculator s = v₀t + ½at²

About this calculator

The uniform-acceleration displacement calculator solves $s = v_0 t + \tfrac12 a t^2$ : given an initial velocity $v_0$ , a constant acceleration $a$ and a time $t$ , it finds the displacement $s$ over that interval; it also runs in reverse, giving the initial velocity or the acceleration from a displacement and a time—how far a stone falls in 3 seconds, how far a car travels braking from 100 km/h, and what average acceleration covers 100 m in 8 seconds from rest.

It is the second of the four basic kinematic equations, paired with the final velocity calculator ( $v = v_0 + at$ ): that one says how the velocity changes, this one how far the motion goes. The two equations carry the same physics—constant acceleration—and differ only in which quantity is asked for. Displacement is the area under the straight line of a v–t graph; this geometric reading is the main thread of the page and the less obvious connection to the trapezoid area.

What it does not do: it does not solve for time from a displacement (that is a quadratic equation, and the hand calculation is given below); it does not compute the final velocity (use the paired calculator); it does not handle acceleration that changes with time; it is one-dimensional only—projectile problems need the horizontal and vertical directions treated separately; and it does not convert units.

Formula and variables

s = v_0\,t + \tfrac{1}{2}\,a\,t^2

The inverse forms (two implemented by the engine):

v_0 = \frac{s - \tfrac12 a t^2}{t},\qquad a = \frac{2\,(s - v_0 t)}{t^2}

Symbol	Meaning	Calculator field	Unit	Range and notes
$v_0$	initial velocity (at $t = 0$ )	v₀ (m/s)	m/s	any real number; default 0. A minus sign means opposite to the positive direction
$a$	acceleration (constant)	a (m/s²)	m/s²	any real number; default 9.8. Negative when opposite to $v_0$
$t$	elapsed time	t (s)	s	$t \ge 0$ ; no default, empty on open
$s$	displacement (signed distance from the start at time $t$ )	Displacement s (m)	m	output; an input when solving in reverse. Signed

Dimensional check: $\text{m/s} \times \text{s} = \text{m}$ and $\text{m/s}^2 \times \text{s}^2 = \text{m}$ ; both terms are lengths. The $\tfrac12$ comes from a triangle's area (see Principles).

How to use this calculator

Open the panelthis page corresponds to “Math tools → Formulas → Displacement (uniform accel.)” (/#/mathtools/formula?calculator=calculator.phys-displacement). On open, “v₀” = 0 and “a” = 9.8 are already filled in and “t” is empty; the time field stays empty while “Displacement s” carries a “calculate” marker and shows 0—the engine does not solve for time from a displacement, and an empty $t$ participates as 0, giving $s = 0$ . This is not a result.
Find the displacemententer the time in seconds and the displacement updates immediately. For deceleration enter a negative $a$ ; the displacement can be negative (ending below the starting point).
Solve for accelerationclear “a”, enter $v_0$ , $t$ and $s$ , and the acceleration field switches to “calculate” and shows $2(s - v_0 t)/t^2$ .
Solve for initial velocityclear “v₀”, enter $a$ , $t$ and $s$ , and the initial-velocity field shows $(s - \tfrac12 at^2)/t$ .
Find the timenot supported by the engine. With $t$ cleared the field stays empty. By hand: $\tfrac12 a t^2 + v_0 t - s = 0$ , so $t = \dfrac{-v_0 + \sqrt{v_0^2 + 2as}}{a}$ (take the positive root); or first solve for $t$ from $v$ in the final velocity calculator and come back to this page.
Read the resultvalues keep at most 8 decimal places with trailing zeros removed.
Copy the formulathe copy button at the right of the panel title puts the LaTeX s = v_0 t + \tfrac{1}{2} a t^2 on the clipboard.

Worked examples

All four examples were recomputed by the engine's phys-displacement compute routine; the display convention is at most 8 decimal places with trailing zeros removed, in units m/s, m/s², s and m.

Example 1: the 1 : 4 : 9 of free fall

Inputs: $v_0 = 0$ , $a = 9.8$ , $t = 1$ . $s = 0 + \tfrac12 \times 9.8 \times 1 = 4.9$ . Interface shows: displacement s 4.9.

Change the time to 2 and 3: the interface shows 19.6 and 44.1 in turn. The ratio $4.9 : 19.6 : 44.1 = 1 : 4 : 9$ —the distance fallen is proportional to the square of the time. The distance added in each successive second is 4.9, 14.7 and 24.5, in the ratio 1 : 3 : 5: the “odd-number rule” Galileo inferred from inclined-plane experiments in 1638, and his evidence that speed is proportional to time (not to distance).

Example 2: the sliding distance of a car braking from 100 km/h

A car travelling at 100 km/h ( $27.78\ \text{m/s}$ ) decelerates at $-5\ \text{m/s}^2$ . First solve for the stopping time in the final velocity calculator, giving 5.55555556 s; then return to this page: $v_0 = 27.77777778$ , $a = -5$ , $t = 5.55555556$ . $s = 27.78 \times 5.56 - \tfrac12 \times 5 \times 5.56^2 = 154.32 - 77.16 = 77.16$ . Interface shows: displacement s 77.16049384.

Check: the time-free formula $v^2 = v_0^2 + 2as$ gives $s = 27.78^2 / (2 \times 5) = 77.1604938$ , in agreement; the average velocity $\tfrac12 \times 27.78 = 13.89\ \text{m/s}$ times 5.56 s is also 77.16. Braking distance is proportional to the square of the speed: double the speed and the distance quadruples—the physical basis of speed limits.

Figure 1: displacement is the area under the v–t graph (left: the braking triangle of Example 2); the s–t graph is a parabola (right: free fall from Example 1, 1 : 4 : 9)

Example 3: solving for acceleration and initial velocity from a displacement

Solving for acceleration: a car starts from rest and covers 100 m in 8 seconds; what is the average acceleration? Clear $a$ , enter $v_0 = 0$ , $t = 8$ , $s = 100$ : $a = 2 \times 100 / 64 = 3.125$ . Interface shows: acceleration a 3.125.

Compare with $a = 3.47222222$ for “0 to 100 km/h in 8 seconds” in the final velocity calculator: over the same 8 seconds that car covers $\tfrac12 \times 3.47222222 \times 64 = 111.11$ m. Enter $v_0 = 0$ , $a = 3.47222222$ , $t = 8$ and the interface shows displacement 111.11111104—the final 04 is a rounding tail from substituting the 8-decimal acceleration back in, not an error.

Solving for initial velocity: an object decelerating at $a = -5$ travels 50 m in 2 seconds; how fast was it to begin with? Clear $v_0$ , enter $a = -5$ , $t = 2$ , $s = 50$ : $v_0 = (50 + 10)/2 = 30$ . Interface shows: initial velocity v₀ 30.

Example 4: a vertical throw and a train

Vertical throw (upward positive, $a = -9.8$ ): a ball thrown at 19.6 m/s has $s = 39.2 - 19.6 = 19.6$ at $t = 2$ . Interface shows: displacement s 19.6—the highest point is 19.6 m up (where $v = 0$ ; see Example 3 of the final velocity calculator). At $t = 4$ , $s = 78.4 - 78.4 = 0$ . Interface shows: 0—the ball is back at the height of the throw. The displacement is zero, but the ball's distance travelled is 39.2 m: displacement is the signed net change, distance the total length of the path.

Train: a train travelling at 20 m/s accelerates at 0.5 m/s² for 1 minute. $v_0 = 20$ , $a = 0.5$ , $t = 60$ : $s = 1200 + 900 = 2100$ . Interface shows: displacement s 2100. The constant-velocity part (1200 m) and the acceleration part (900 m) appear as the two terms.

The inputs of Example 1 can be entered directly into the panel to reproduce it; change the time to 1, 2 and 4 and watch the displacement grow as 1 : 4 : 9 : 16; change the acceleration to 1.62 (the Moon) and see that the same 3 seconds covers only 7.29 m.

Principle and derivation

Displacement is the area under the v–t graph

For constant velocity, displacement $= vt$ , the area of a rectangle on a v–t graph. Under uniform acceleration the velocity changes, but time can be cut into many small intervals, each approximately constant-velocity, and the total displacement becomes the area of the trapezoid below the line. The line rises from $v_0$ to $v_0 + at$ , so the trapezoid area $= \tfrac12 (v_0 + v_0 + at)\,t = v_0 t + \tfrac12 a t^2$ —this is where the $\tfrac12$ comes from. The braking triangle on the left of Figure 1 is the special case of $v_0 = 27.78$ falling to 0: area $\tfrac12 \times 27.78 \times 5.56$ . The trapezoid area calculator computes exactly the same thing.

Understanding it through average velocity

With constant acceleration the average velocity is exactly the arithmetic mean of the initial and final velocities, $\bar v = \tfrac12 (v_0 + v)$ , and the displacement is $= \bar v\, t$ . Substituting $v = v_0 + at$ returns $s = v_0 t + \tfrac12 at^2$ . This is also the fourth kinematic equation, $s = \tfrac12 (v_0 + v)\,t$ . It holds only for uniform acceleration—with variable acceleration the average velocity is no longer the midpoint of the endpoints.

Eliminating time: $v^2 = v_0^2 + 2as$

Solve $v = v_0 + at$ for $t = (v - v_0)/a$ , substitute into the displacement formula and simplify to obtain the time-free $v^2 = v_0^2 + 2as$ . Example 2 used it to check the braking distance. Multiplying both sides by $\tfrac12 m$ gives $\tfrac12 m v^2 - \tfrac12 m v_0^2 = mas = Fs$ —the work–energy theorem: the work done by the net force equals the change in kinetic energy. The kinetic energy calculator takes over from there.

The s–t graph is a parabola

$s$ is a quadratic function of $t$ , so its graph is a parabola (right of Figure 1). With $v_0 = 0$ the vertex is at the origin and the direction of opening is set by the sign of $a$ . The tangent slope at any point of the parabola is the velocity at that instant: at $t = 0$ the slope is $v_0$ . For a vertical throw the s–t graph is a downward-opening parabola whose vertex is the highest point, and whose second crossing of the horizontal axis is the moment the ball returns to the height of the throw ( $t = 4$ in Example 4).

Why solving for time is a quadratic equation

Given $s$ , $v_0$ and $a$ , finding $t$ means solving $\tfrac12 a t^2 + v_0 t - s = 0$ . It may have two positive roots (a ball thrown upward passes the same height twice, rising and falling) or none (the ball never reaches that height). The engine deliberately does not return a single answer in the field, to avoid silently discarding a root; calculate by hand as in step 5 of “How to use” and consider both roots.

Galileo's inclined plane

Free fall is too fast—in 1638 there was no clock able to measure it. Galileo rolled balls down a smooth inclined plane, timed them with a water clock, and found that distance is proportional to the square of time and that the distances covered in successive equal intervals are in the ratio 1 : 3 : 5 : 7, with the ratios unchanged when the inclination changed. From this he inferred the same for a vertical fall (a 90° incline). The $g$ in $s = \tfrac12 g t^2$ was not measured accurately until after 1638.

Assumptions

Constant acceleration: $a$ is unchanged over the whole interval $t$ . Real vehicle acceleration varies with speed, so the result is an equivalent average.
Straight-line, one-dimensional motion: displacement, velocity and acceleration are collinear, with direction carried by signs.
Motion continues up to $t$ : the formula does not stop when the object stops (see the note after Example 4).
Drag is ignored: the free-fall examples exclude air resistance.
SI units: m/s, m/s², s, m; no conversion is performed.
Interface conventions: an unfilled $t$ participates as 0, so the displacement shows 0 when the panel opens; in the inverse modes an empty denominator $t$ falls back to 1.

Scope and limitations

Can calculate: $s$ from $v_0$ , $a$ and $t$ ; $a$ or $v_0$ from $s$ , $t$ and the other quantity; negative values and negative displacements are supported.
Approximate only: variable acceleration can only be approximated piecewise; taking $g = 9.8$ carries a value difference of about 0.1%.
Cannot calculate: time from a displacement (a quadratic, by hand); final velocity (use the paired calculator); two-dimensional projectiles (per direction); distance travelled (the sum of absolute displacements, needing to know about any reversal).
Does not: convert units; decide whether the motion has already stopped.
Numerics: substituting 8-decimal results back leaves a rounding tail in the last digit (111.11111104 in Example 3).

Common pitfalls

Forgetting the ½: $s = at^2$ makes 3 seconds of free fall 88.2 m, twice the true value. The $\tfrac12$ comes from the triangle's area and is not optional.
Taking displacement for distance: a vertical throw returning to the start has zero displacement but non-zero distance (Example 4).
Calculating past the stop: for $t$ beyond the stopping time the formula drives the car backwards (note after Example 4).
Entering km/h directly: entering 100 km/h as 100 turns the braking distance into 1000 m. Convert to 27.78 m/s first.
Forgetting the minus sign on deceleration: the braking $a$ is $-5$ ; $+5$ makes the car go ever further.
Using the final velocity in place of the initial: the $v_0$ in the formula is the velocity at the start of the interval.
Squaring something other than time: in $\tfrac12 a t^2$ only $t$ is squared.

Typical use cases

Teaching

The 1 : 4 : 9 of free fall and the odd-number rule; the area method on v–t graphs; displacement versus distance; the symmetry of a vertical throw; the origin of the work–energy theorem.

Traffic safety

Braking distance for a given speed and deceleration (combined with the constant-velocity reaction distance $= v \times t_\text{reaction}$ ); an intuitive account of how speed limits relate to the square law.

Engineering

Travel of lift, conveyor and feed drive ramps; positioning lengths for cranes and machine tools; inferring the acceleration needed from a known travel and time.

Sport and everyday life

Entry time and speed estimates for diving heights; the highest point of a throw; estimating well or building height from a fall time (allowing for the sound travelling back).

FAQ

Why is the displacement 0 and the time empty when the panel opens?

The time has not been filled in and the engine treats $t$ as 0; the engine does not solve for time from a displacement, so the time field stays empty. Enter a time and it is normal.

How do I find the time from a displacement?

Solve the quadratic $\tfrac12 a t^2 + v_0 t - s = 0$ : $t = \dfrac{-v_0 + \sqrt{v_0^2 + 2as}}{a}$ ; with $v_0 = 0$ it simplifies to $t = \sqrt{2s/a}$ . Or first solve for $t$ from $v$ in the final velocity calculator.

What is the difference between displacement and distance?

Displacement is the signed straight-line distance from the start to the end; distance is the total length of the path. They are equal for straight-line motion without reversal; after a reversal the displacement starts to shrink while the distance keeps growing.

Why is braking distance proportional to the square of the speed?

Put $v = 0$ in $v^2 = v_0^2 + 2as$ : $s = v_0^2 / (2|a|)$ . Double the speed and $v_0^2$ quadruples.

Does a thrown ball travel the same distance up and down?

Without drag, yes: the s–t parabola is symmetric about the highest point. With drag the downward leg is slower and takes longer.

Can it be used when air resistance is large?

No. Drag grows with speed, so the acceleration is no longer constant. Raindrops, parachutes and feathers are not uniformly accelerated.

References and further reading

The phys-displacement definition in the CalcX engine source src/data/formulasScience.ts (compute(known, solveFor) for s / v₀ / a); every figure on this page was recomputed by that engine.
Wikipedia, Equations of motion — Constant translational acceleration in a straight line （访问日期：2026-09-09）—the complete table of five quantities and four equations, with derivations.
Wikipedia, Displacement (geometry) （访问日期：2026-09-09）—the definitions of displacement and distance.
Galileo Galilei, Dialogues Concerning Two New Sciences (1638), Third Day, translated by Crew & de Salvio （访问日期：2026-09-09）—the original account of the odd-number rule and the inclined-plane experiments.
Wikipedia, Braking distance （访问日期：2026-09-09）—the braking-distance formula, reaction distance and typical decelerations.
ISO 80000-3:2019, Quantities and units — Part 3: Space and time—definitions of the quantities and units of displacement, velocity and acceleration.

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Sources & review

Reviewed by: CalcX 编辑组
Updated: 2026-09-09

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