Physics · Kinematics

Uniformly accelerated motion: position–time equation

Q: When may you use s = ½at² without the v₀t term?

Use it when $v_{0} = 0$ and a is constant over the interval. Starting from rest alone is insufficient, for example with changing propulsion acceleration. For $v_{0} =\frac{ 20 m}{s}$ and $a = -\frac{5 m}{s^{2}}$, displacement to rest after 4 s is 40 m. The sign of a depends on the chosen axis. Check the initial state in the selected reference frame: release from rest is frame dependent. An object released from a moving vehicle may already have initial velocity relative to the ground. Check the acceleration assumption separately. For an initially stationary object with $a = 0$, displacement remains zero at all times.

Q: What do you do when the time is not given?

For constant a, use $v^2=v_0^2+2as$ with signed displacement s. For $v_{0} =\frac{ 20 m}{s}$, $a = -\frac{4 m}{s^{2}}$ and $v = 0$ this gives $s = 50 m$. The square $v^{2}$ alone does not determine direction; the sign of v must match the motion. At $a = 0$ this relation cannot determine displacement. Check the result by substituting it back and checking units: both sides have units $\frac{m^{2}}{s^{2}}$. Do not extend a braking model past rest without justification. A reversal uses the same signed quantities but must be distinguished from ordinary stopping. The model interval decides whether a calculated state is physically relevant.

Q: How are the distance-time and velocity-time laws related?

Differentiating $x(t)=x_0+v_0t+\frac12at^2$ gives $v(t)=v_0+at$. Signed area in a velocity–time plot gives displacement: regions below the time axis count negatively. Distance adds their magnitudes. The slope of the v-t graph is a; the slope of the x-t graph is v. For $v_{0} = -\frac{6 m}{s}$ and $a = +\frac{2 m}{s^{2}}$, the first three seconds contribute −9 m of displacement and the next two contribute +4 m. Total displacement is −5 m but total distance is 13 m. This illustrates why treating all graph area as unsigned gives a different quantity.

Q: Why does the distance quadruple when the time doubles?

Starting from rest with constant a gives s(t) = ½at², so $s(2t) = 4s(t)$. With $v_{0} \ne{} 0$, the linear term v₀t doubles while the acceleration term quadruples; the blanket fourfold rule no longer applies. A reversal also requires distinguishing displacement and distance. For an original comparison, $v_{0} =\frac{ 3 m}{s}$ and $a =\frac{ 2 m}{s^{2}}$ give displacement of 4 m at one second and 10 m at two seconds. The ratio is 2.5 rather than four. Starting from rest with the same acceleration instead gives 1 m and 4 m, matching the square law.

The position–time equation gives signed displacement $s = x - x_{0}$ at constant acceleration. Only a start from rest makes s purely proportional to $t^{2}$ .

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Formula

Calculate

s = \frac{1}{2} a t^{2} + v_{0} t

s in metres [m] · a in m/s² · t in seconds [s] · v₀ in m/s

Variables & units

Variables & units
Symbol	Meaning	Unit
$s$	Signed displacement x − x₀	m
$a$	Constant acceleration component	m/s²
$t$	Elapsed time since the start	s
$v₀$	Signed initial velocity	m/s

LaTeX

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

QUANTA STUDYquanta-study.de

Original v-t line with positive v₀ and positive constant a. The marked trapezoid gives displacement s=Δx and, in this case, distance travelled. — At constant a, v(t) is a straight line. Signed area gives s=Δx. In the illustrated case v is positive, so displacement equals distance travelled.

Displacement calculator for constant acceleration

Find displacement, initial velocity or acceleration for motion along a straight line with constant acceleration.

Find

Initial velocity v₀

Velocity at the start, with sign, e.g. 36 km/h; 0 when starting from rest.

Acceleration a

Constant acceleration with sign, e.g. 2 m/s²; when braking it points against the motion.

m/s²

Time t

Time elapsed since the start, never negative, e.g. 5 s.

Filled with example values. Type your own and the result updates straight away.

Result: Displacement Δx (signed)

Δx = 75 m

With v₀ = 36 km/h, a = 2 m/s² and t = 5 s, Δx = 75 m.

Working · rearrange for Δx, substitute, evaluate

1. Rearrange for Δx
$\Delta x=v_0t+\frac12at^2$
2. Substitute values (in SI units)
- v₀ = 36 km/h = 10 m/s
$\Delta x=10\cdot 5+\frac12\cdot 2\cdot 5^2=75$
3. Result
Δx = 75 m

Quick tips

From rest (v₀ = 0) this becomes Δx = ½ · a · t²: twice the time, four times the distance.
Braking to a stop: first find the stopping time t = |v₀| ÷ |a| and use only that. A longer time would make the model carry on backwards.

Assumptions and limits

Fixed axis, constant signed acceleration and elapsed time t ≥ 0. Δx is displacement, not total path length. No inverse time target: it may have multiple nonnegative solutions.

Derivation & background

Integrating $v(t)=v_0+at$ gives $x(t)=x_0+v_0t+\frac12at^2$ . The symbol s denotes x − x₀. The time-free relation $v^2=v_0^2+2as$ uses the same displacement. For free fall, the acceleration component is +g or −g according to the chosen axis.

Exam blueprint

Validity range

Valid for a constant acceleration component along the chosen axis. s denotes signed displacement, not total distance during a reversal. Varying a requires integration or suitable separate intervals.

Derivation steps

Integrate constant acceleration twice, retaining initial values and signs.

1Choose an axis and start time $t = 0$ ; specify x₀ and v₀ at that start.
2For constant a, $\frac{dv}{dt} = a$ throughout the interval.
3Integrate from 0 to t: v(t) − v₀ = at, giving $v(t)=v_0+at$ .
4Displacement is signed area: x(t) − x₀ = ∫₀ᵗv(τ)dτ.
5Substitute and integrate: s = x − x₀ = v₀t + ½at².
6Check the derivative and initial value. For distance during a reversal, add absolute segment displacements.

Rearrangements

Time from distance (for v₀ = 0)

t = \sqrt{\frac{2s}{a}}

Require release from rest, constant a and $a \ne{} 0$ . Here s is signed displacement; $\frac{2s}{a}$ must be nonnegative. Choose the nonnegative root for elapsed time.

Acceleration from distance and time $(v_{0} = 0)$

a = \frac{2s}{t^2}

This is how driving tests and inclined-plane experiments are evaluated.

Time-free equation

v^2 = v_0^2 + 2as

Uses signed displacement and constant a. Squared velocity determines magnitude only; its sign must follow from the motion phase.

Task variant

A car accelerates from rest over 100 m at $a =\frac{ 2 m}{s^{2}}$ . How long does it take?

$t = \sqrt{(\frac{2s}{a})} = \sqrt{(\frac{200}{2})} = \sqrt{100} = 10 s$ .

A sled covers 45 m from rest in 3 s. Find a.

$a =\frac{ 2s}{t^{2}} =\frac{ 2\cdot{}45}{9} =\frac{ 10 m}{s^{2}}$ .

Common mistakes

Forgetting the factor ½ or not squaring t.

The acceleration contribution contains ½ and $t^{2}$ . Doubling time quadruples displacement only for $v_{0} = 0$ and constant a; reversal additionally requires separating distance and displacement.

Dropping the v₀t term although an initial velocity is given.

Only for a start from rest does s = ½at² hold.

Mixing speeds in $\frac{km}{h}$ with distances in metres.

Divide $\frac{km}{h}$ by 3.6 before substituting.

Exam context

Standard in acceleration, braking and overtaking problems, often combined with $v = a\cdot{}t + v_{0}$ or the time-free equation.

Formula cluster

Kinematics of straight-line motion

Distance law, velocity law and force law interlock.

v = a·t + v₀

Velocity-Time Law

h = ½gt²

Free Fall

ΣF_ext = F_res = ma

Newton's Second Law

Worked example

A car starts from rest $(v_{0} = 0)$ with $a =\frac{ 2 m}{s^{2}}$ . After $t = 5 s$ : $s = \frac{1}{2} \times{} 2 \times{} 5^{2} = 25 m$ .

Applications

Acceleration and braking distance calculations, accident analysis, runway design, driver assistance systems

How do you calculate with Uniformly Accelerated Motion (Position–Time Equation)?

Here is how to work through a typical Uniformly Accelerated Motion (Position–Time Equation) (s = ½at² + v₀t) task step by step:

1
Task
A car accelerates from rest over 100 m at $a =\frac{ 2 m}{s^{2}}$ . How long does it take?
Solution path
$t = \sqrt{(\frac{2s}{a})} = \sqrt{(\frac{200}{2})} = \sqrt{100} = 10 s$ .
2
Task
A sled covers 45 m from rest in 3 s. Find a.
Solution path
$a =\frac{ 2s}{t^{2}} =\frac{ 2\cdot{}45}{9} =\frac{ 10 m}{s^{2}}$ .

Understand and apply the equation

With constant acceleration, a cart covers larger or smaller distances in later seconds, depending on its motion direction. The position–time equation combines these movements. First a simple area explains the factor one half; initial conditions, reversal and stopping then show how to use the equation.

When is the equation for a start from rest sufficient?

The short form $s=\frac12at^2$ requires constant acceleration and a start from rest, so $v_0=0$ . A nonzero initial velocity requires retaining its initial contribution. A cart starts from rest and gains 2 m/s each second. During the first second velocity rises from 0 to 2 m/s, averaging 1 m/s and covering 1 m. During the second it rises from 2 to 4 m/s, averaging 3 m/s and adding another 3 m. After two seconds the total is 4 m. Equal time intervals therefore do not mean equal distances under acceleration.[1]

Prerequisite: v is the rate of position change; a is the rate of velocity change. Both refer to the same axis.
Distinguish initial position x₀, final position x and displacement s = x − x₀. Its magnitude equals distance only when there is no reversal.
Rectangle area is base times height; triangle area is half as much. These facts explain the equation without calculus.

What is the displacement–time equation?

For constant acceleration, the displacement–time equation is $s=v_0t+\frac12at^2$ ; final position is $x=x_0+s$ . Choose a positive axis and define t = 0. Initial position x₀, velocity component v₀ and acceleration component a refer to that axis. The expression ½at² + v₀t, often labelled s, gives displacement x − x₀, which can be negative. Total distance is nonnegative and must be calculated in separate segments when direction reverses.[1][2]

\begin{aligned}x(t)&=x_0+v_0t+\frac12at^2\\s(t):&=x(t)-x_0\end{aligned}

The short form s = v₀t + ½at² uses s for displacement.

Three initial quantities fix the parabola: x₀ shifts it vertically, v₀ fixes its initial slope and a determines curvature. At a = 0 it becomes a straight line. For nonzero a, displacement is not generally proportional to t² because v₀t can contribute substantially.

Integrating twice while preserving initial values

Under constant a, the velocity graph is a straight line. Split its signed area from zero to t into an initial contribution and a change contribution. The rectangle has height v₀ and width t. The triangle represents the additional change at and has half the corresponding rectangle area. Both contributions retain their signs, so the decomposition also works for negative a or reversal.[1]

\begin{aligned}s&=s_{\mathrm{initial}}+s_{\mathrm{change}}\\&=v_0t+\frac12t(at)\\&=v_0t+\frac12at^2\\x&=x_0+s\end{aligned}

Initial velocity produces the linear contribution; its steady change produces the quadratic contribution.[1]

If you know calculus, you can evaluate the same area directly. Integrate a to obtain velocity, then integrate v to obtain position. The following lines provide a second justification of the same equation; the area argument is sufficient for the first exercises.

v(t)-v_0=\int_0^t a\,\mathrm d\tau=at

First integral: velocity changes linearly.

x(t)-x_0=\int_0^t(v_0+a\tau)\,\mathrm d\tau=v_0t+\frac12at^2

Second integral: the factor ½ comes from integrating τ.

Differentiate to check: dx/dt = v₀ + at and d²x/dt² = a. Also check x(0) = x₀. If the starting time is nonzero, replace t throughout by elapsed time Δt = t − t₀ and use initial values at t₀.

Why the factor one half appears

QUANTA STUDYquanta-study.de

With v₀ = 2 m/s and a = 2 m/s², the rectangle gives 6 m and the triangle 9 m displacement by 3 s: 15 m total.[1]

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How do you calculate acceleration and possible time roots?

\begin{aligned}[v_0t]&=\frac{\mathrm m}{\mathrm s}\mathrm s=\mathrm m\\{}[at^2]&=\frac{\mathrm m}{\mathrm s^2}\mathrm s^2=\mathrm m\end{aligned}

Both terms are lengths; divide km/h by 3.6 first.

\begin{aligned}a&=\frac{2(s-v_0t)}{t^2}\quad(t\ne0)\\v^2&=v_0^2+2as\end{aligned}

The time-free equation uses the same signed displacement s.

t=\frac{-v_0\pm\sqrt{v_0^2+2as}}{a}\quad(a\ne0)

Check real roots, t ≥ 0 and the interval over which the model applies.

A position can be reached twice during a reversal. With v₀ = 6 m/s, a = −2 m/s² and s = 8 m, the roots are t = 2 s and t = 4 s: outbound and return passages. A braking model that ends at rest instead allows only times up to t = 3 s.

The quadratic formula gives time roots once the displacement–time relation is written as a quadratic equation. Time direction, the initial state and the model’s valid interval determine which roots describe the observed motion.

Example: a cart reverses direction

Negative initial velocity

Given: x₀ = 0 m, v₀ = −2 m/s, a = +1.5 m/s², t = 4 s.
Find: Find final position, final velocity and total distance.

Approach: first find when v = 0. Calculate displacements before and after that instant separately; their magnitudes give distance.
$\begin{aligned}x(t)&=x_0+v_0t+\frac12at^2\\v(t)&=v_0+at\\t_*&=-\frac{v_0}{a}\end{aligned}$
Substitution: use the specified initial values with their signs.
$\begin{aligned}x(4)&=-2\cdot4+\tfrac12\cdot1.5\cdot4^2=4\,\mathrm m\\{} v(4)&=4\,\mathrm{m/s}\end{aligned}$
Reversal occurs at v = 0; the cart moves left until then.
$\begin{aligned}t_*&=\frac{2}{1.5}=\frac43\,\mathrm s\\{} x(t_*)&=-\frac43\,\mathrm m\end{aligned}$
Add the two segments as positive lengths.
$L=\left|{-\frac43}-0\right|+\left|4-\left(-\frac43\right)\right|=\frac{20}{3}\,\mathrm m$

Result: Final position +4 m, final velocity +4 m/s, distance 6.67 m.

Check: Distance exceeds |s| = 4 m. The magnitude of average velocity is 1 m/s; average speed is 1.67 m/s.

Example: stopping distance without extending past the stop

In an inertial frame and for constant mass, braking acceleration follows from net force and mass through Newton’s second law. In the following example this acceleration is assumed constant until rest, where the braking model ends.[5][1]

Braking from 72 km/h

Given: v₀ = 72 km/h, a = −4 m/s², constant deceleration until rest.
Find: Calculate braking time and distance excluding reaction time.

Approach: stopping ends the braking phase at v = 0. Find that time before substituting it into the displacement equation.
$\begin{aligned}0&=v_0+at_{\mathrm{stop}}\\t_{\mathrm{stop}}&=-\frac{v_0}{a}\\s_{\mathrm{stop}}&=v_0t_{\mathrm{stop}}+\frac12at_{\mathrm{stop}}^2\end{aligned}$
Units: convert initial speed to metres per second before substitution.
$v_0=\frac{72}{3.6}=20\,\mathrm{m/s}$
v = 0 gives the endpoint of the braking model.
$t_{\mathrm{stop}}=\frac{0-20}{-4}=5\,\mathrm s$
Insert the stopping time into displacement.
$s=20\cdot5+\tfrac12(-4)5^2=50\,\mathrm m$

Result: Stopping time 5 s, braking distance 50 m.

Check: The time-free check gives 0² = 20² + 2·(−4)·50. Extending to 6 s would predict negative velocity, which is no longer simple stopping.

How do you read areas and slopes in motion graphs?

In a velocity graph, displacement is signed area. A part below the time axis counts negatively. To find distance travelled, add magnitudes of the separate areas instead. In a position graph, a falling curve means motion in the negative direction; its minimum can mark reversal. Position and distance therefore need not share the same graph. Under constant acceleration, the velocity–time law determines the straight line whose area you evaluate here.[1][2]

\begin{aligned}\bar v&=\frac{s}{t}=\frac{v_0+v(t)}2\\L&=\int_0^t|v(\tau)|\,\mathrm d\tau\end{aligned}

The endpoint average holds for constant a; it is not generally average speed.

A straight trend across measured velocity–time points can support approximately constant a. Connecting only two points establishes an average acceleration. Starting, braking and rest can be modelled in separate intervals, carrying the final position of each into the next.

When the position–time model must change

For time-dependent acceleration, integrate a(t) rather than inserting an arbitrary instantaneous value into ½at². Drag, changing propulsion or curved motion require an appropriate model. In multiple dimensions, the equations can apply component by component when each acceleration component used is constant. For ideal free fall near Earth’s surface, acceleration is constant when changes in g are neglected and gravity is the only force.[4][1][6]

A real stopping distance contains additional phases. At 20 m/s, an assumed 0.8 s reaction time adds 16 m before the braking calculation above: 66 m in this model. These example assumptions do not replace a specific vehicle or safety assessment.

In projectile motion without air resistance, the motion separates into components: horizontal velocity is constant and vertical gravitational acceleration is constant. The spatial trajectory is nevertheless not a straight-line motion.

Five exercises with solutions

Calculate · Foundation

An object starts from rest at x₀ = 3 m with a = 2 m/s². Where is it after 4 s?

Show hint

Calculate displacement separately from initial position.

Learning goal: Combine initial position and displacement to find final position.

Show solution and calculation

Approach: first calculate displacement from the initial conditions.
Starting from rest eliminates v₀t, giving s = 16 m after four seconds.
Only then add initial position: x = 19 m.

Check: initial position shifts the entire position graph by 3 m without changing distance or velocity.

\begin{aligned}s&=v_0t+\frac12at^2\\&=\frac12(2\,\mathrm{m/s^2})(4\,\mathrm s)^2=16\,\mathrm m\\x&=x_0+s=3\,\mathrm m+16\,\mathrm m=19\,\mathrm m\end{aligned}

[1]

Rearrange · Application

v₀ = 5 m/s, a = 2 m/s², target displacement 24 m. Find the positive time.

Show hint

Solve t² + 5t − 24 = 0.

Learning goal: Select the physically appropriate time root.

Show solution and calculation

Approach: initial velocity is nonzero, so retain the linear term.
In SI units the equation is t² + 5t − 24 = 0, which factors as (t − 3)(t + 8) = 0.
Time after the start is 3 s; −8 s lies outside the interval.

Check: 15 m from initial motion plus 9 m from acceleration gives the required 24 m.

\begin{aligned}s&=v_0t+\frac12at^2\\24\,\mathrm m&=(5\,\mathrm{m/s})t+(1\,\mathrm{m/s^2})t^2\\t&=3\,\mathrm s\qquad\lor\qquad t=-8\,\mathrm s\\s(3\,\mathrm s)&=15\,\mathrm m+9\,\mathrm m=24\,\mathrm m\end{aligned}

[1]

Transfer · Application

v₀ = 10 m/s, constant a = −2 m/s². Find the distance to rest.

Show hint

Use v² = v₀² + 2as with v = 0.

Learning goal: Find stopping distance from the time-free equation.

Show solution and calculation

Approach: set velocity to zero at rest.
Stopping time is 5 s.
Velocity decreases uniformly from 10 to 0 m/s, so its average is 5 m/s and stopping distance is 25 m.

Check: the time-free equation gives the same 25 m using negative a. The model is not continued beyond rest.

\begin{aligned}t_{\mathrm{stop}}&=-\frac{v_0}{a}=-\frac{10}{-2}\,\mathrm s=5\,\mathrm s\\s&=v_0t+\frac12at^2\\&=10(5)\,\mathrm m+\frac12(-2)(5)^2\,\mathrm m=25\,\mathrm m\end{aligned}

[1]

Interpret · Application

One-dimensional motion is $x(t)=(1\,\mathrm{m/s^2})t^2-(4\,\mathrm{m/s})t$ for 0≤t≤4 s. Find initial and final positions and the turning point. Determine displacement and distance travelled.

Show hint

$v=\frac{dx}{dt}$ vanishes at reversal; split distance into two legs.

Learning goal: Distinguish displacement from path length using a position parabola.

Show solution and calculation

Approach: x(0)=x(4 s)=0, so displacement is zero. v=2t−4 in SI units vanishes at 2 s, where x=−4 m.
The body travels 4 m left and 4 m back, totaling 8 m.

Check: positive curvature gives constant a=2 m/s² despite initial negative motion. Final position alone does not give total distance.

\begin{aligned}\Delta x&=x(4)-x(0)=0\\L&=|x(2)-x(0)|+|x(4)-x(2)|=8\,\mathrm m\end{aligned}

[1][2]

Error analysis · Application

A particle starts from rest at x₀=5 m and accelerates at constant a=2 m/s² along +x. At x=9 m a solution sets $v^2=2ax=36\,\mathrm{m^2/s^2}$ . Correct it and check using elapsed time.

Show hint

The equation contains x−x₀ rather than absolute position.

Learning goal: Use displacement rather than absolute position in the time-free equation.

Show solution and calculation

Approach: displacement is 4 m, so $v^2=2a(x-x_0)=16\,\mathrm{m^2/s^2}$ and v=+4 m/s.

Check: $x-x_0=\frac12at^2$ gives t=2 s and v=at=4 m/s. The incorrect result depends on an arbitrary coordinate origin; changing that origin must not change predicted physical velocity.

\begin{aligned}v&=\sqrt{2a(x-x_0)}=\sqrt{2\cdot2\cdot4}\,\mathrm{m\,s^{-1}}=4\,\mathrm{m\,s^{-1}}\\t&=\sqrt{\frac{2(x-x_0)}a}=2\,\mathrm s\end{aligned}

[1]

What to take away

Position combines initial position, initial motion and the acceleration contribution. Displacement is purely proportional to t² only from rest.
Negative displacement is allowed. During reversal, total distance exceeds displacement magnitude.
For stopping, first find the instant of rest. With multiple time roots, identify which belongs to the actual motion phase.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Stated the displacement–time equation and start-from-rest condition directly; made existing rearrangements and graphs accessible through learning questions. Added contextual links for the velocity law, braking-force model and free fall.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added prerequisites and an intuitive introduction; clarified the derivation, worked-example approaches and exercise solutions, and added a summary.
October 2, 2026Linked time roots to the quadratic formula and component motion to projectile motion.

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Quanta. Uniformly Accelerated Motion (Position–Time Equation). Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/gleichmaessig-beschleunigte-bewegung

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Frequently asked questions about Uniformly Accelerated Motion (Position–Time Equation)

How do you calculate the distance under uniform acceleration?+

s = v₀t + ½at² gives signed displacement at constant a. With $v_{0} = 0$ , $a =\frac{ 2 m}{s^{2}}$ and $t = 5 s$ it is +25 m. Add x₀ to find final position. Without reversal, distance is |s|; with reversal, add absolute segment displacements. Convert all quantities to consistent SI units first. As a check, substitute $t = 0$ : displacement must be zero. Differentiation must return v₀ + at. A negative s is not an arithmetic error; it indicates a final position opposite the positive axis. The equation alone also excludes any reaction distance before braking begins.

When may you use s = ½at² without the v₀t term?+

Use it when $v_{0} = 0$ and a is constant over the interval. Starting from rest alone is insufficient, for example with changing propulsion acceleration. For $v_{0} =\frac{ 20 m}{s}$ and $a = -\frac{5 m}{s^{2}}$ , displacement to rest after 4 s is 40 m. The sign of a depends on the chosen axis. Check the initial state in the selected reference frame: release from rest is frame dependent. An object released from a moving vehicle may already have initial velocity relative to the ground. Check the acceleration assumption separately. For an initially stationary object with $a = 0$ , displacement remains zero at all times.

What do you do when the time is not given?+

For constant a, use $v^2=v_0^2+2as$ with signed displacement s. For $v_{0} =\frac{ 20 m}{s}$ , $a = -\frac{4 m}{s^{2}}$ and $v = 0$ this gives $s = 50 m$ . The square $v^{2}$ alone does not determine direction; the sign of v must match the motion. At $a = 0$ this relation cannot determine displacement. Check the result by substituting it back and checking units: both sides have units $\frac{m^{2}}{s^{2}}$ . Do not extend a braking model past rest without justification. A reversal uses the same signed quantities but must be distinguished from ordinary stopping. The model interval decides whether a calculated state is physically relevant.

How are the distance-time and velocity-time laws related?+

Differentiating $x(t)=x_0+v_0t+\frac12at^2$ gives $v(t)=v_0+at$ . Signed area in a velocity–time plot gives displacement: regions below the time axis count negatively. Distance adds their magnitudes. The slope of the v-t graph is a; the slope of the x-t graph is v. For $v_{0} = -\frac{6 m}{s}$ and $a = +\frac{2 m}{s^{2}}$ , the first three seconds contribute −9 m of displacement and the next two contribute +4 m. Total displacement is −5 m but total distance is 13 m. This illustrates why treating all graph area as unsigned gives a different quantity.

Why does the distance quadruple when the time doubles?+

Starting from rest with constant a gives s(t) = ½at², so $s(2t) = 4s(t)$ . With $v_{0} \ne{} 0$ , the linear term v₀t doubles while the acceleration term quadruples; the blanket fourfold rule no longer applies. A reversal also requires distinguishing displacement and distance. For an original comparison, $v_{0} =\frac{ 3 m}{s}$ and $a =\frac{ 2 m}{s^{2}}$ give displacement of 4 m at one second and 10 m at two seconds. The ratio is 2.5 rather than four. Starting from rest with the same acceleration instead gives 1 m and 4 m, matching the square law.

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Which formula describes Uniformly Accelerated Motion (Position–Time Equation)?

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How do you rearrange Uniformly Accelerated Motion (Position–Time Equation) for Time from distance (for v₀ = 0)?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §3.4 Motion with Constant Acceleration

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E_{kin} = \frac{1}{2} m v^2

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Uniformly accelerated motion: position–time equation

The position–time equation gives signed displacement

s = x - x_{0}

at constant acceleration. Only a start from rest makes s purely proportional to

t^{2}

BasicExam-relevantLower secondaryUpper secondary

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Symbol

Meaning

Unit

s

Signed displacement x − x₀

a

Constant acceleration component

m/s²

t

Elapsed time since the start

v₀

Signed initial velocity

m/s

Derivation & background

Exam blueprint

Validity range

Valid for a constant acceleration component along the chosen axis. s denotes signed displacement, not total distance during a reversal. Varying a requires integration or suitable separate intervals.

Derivation steps

Integrate constant acceleration twice, retaining initial values and signs.

1Choose an axis and start time $t = 0$ ; specify x₀ and v₀ at that start.
2For constant a, $\frac{dv}{dt} = a$ throughout the interval.
3Integrate from 0 to t: v(t) − v₀ = at, giving $v(t)=v_0+at$ .
4Displacement is signed area: x(t) − x₀ = ∫₀ᵗv(τ)dτ.
5Substitute and integrate: s = x − x₀ = v₀t + ½at².
6Check the derivative and initial value. For distance during a reversal, add absolute segment displacements.

How do you calculate with Uniformly Accelerated Motion (Position–Time Equation)?

Here is how to work through a typical Uniformly Accelerated Motion (Position–Time Equation) (s = ½at² + v₀t) task step by step:

Task

A car accelerates from rest over 100 m at $a =\frac{ 2 m}{s^{2}}$ . How long does it take?

Solution path

$t = \sqrt{(\frac{2s}{a})} = \sqrt{(\frac{200}{2})} = \sqrt{100} = 10 s$ .

Task

A sled covers 45 m from rest in 3 s. Find a.

Solution path

$a =\frac{ 2s}{t^{2}} =\frac{ 2\cdot{}45}{9} =\frac{ 10 m}{s^{2}}$ .

Understand and apply the equation

When is the equation for a start from rest sufficient?

Prerequisite: v is the rate of position change; a is the rate of velocity change. Both refer to the same axis.
Distinguish initial position x₀, final position x and displacement s = x − x₀. Its magnitude equals distance only when there is no reversal.
Rectangle area is base times height; triangle area is half as much. These facts explain the equation without calculus.

What is the displacement–time equation?

\begin{aligned}x(t)&=x_0+v_0t+\frac12at^2\\s(t):&=x(t)-x_0\end{aligned}

The short form s = v₀t + ½at² uses s for displacement.

Integrating twice while preserving initial values

\begin{aligned}s&=s_{\mathrm{initial}}+s_{\mathrm{change}}\\&=v_0t+\frac12t(at)\\&=v_0t+\frac12at^2\\x&=x_0+s\end{aligned}

Initial velocity produces the linear contribution; its steady change produces the quadratic contribution.[1]

v(t)-v_0=\int_0^t a\,\mathrm d\tau=at

First integral: velocity changes linearly.

x(t)-x_0=\int_0^t(v_0+a\tau)\,\mathrm d\tau=v_0t+\frac12at^2

Second integral: the factor ½ comes from integrating τ.

Why the factor one half appears

QUANTA STUDYquanta-study.de

With v₀ = 2 m/s and a = 2 m/s², the rectangle gives 6 m and the triangle 9 m displacement by 3 s: 15 m total.[1]

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How do you calculate acceleration and possible time roots?

\begin{aligned}[v_0t]&=\frac{\mathrm m}{\mathrm s}\mathrm s=\mathrm m\\{}[at^2]&=\frac{\mathrm m}{\mathrm s^2}\mathrm s^2=\mathrm m\end{aligned}

Both terms are lengths; divide km/h by 3.6 first.

\begin{aligned}a&=\frac{2(s-v_0t)}{t^2}\quad(t\ne0)\\v^2&=v_0^2+2as\end{aligned}

The time-free equation uses the same signed displacement s.

t=\frac{-v_0\pm\sqrt{v_0^2+2as}}{a}\quad(a\ne0)

Check real roots, t ≥ 0 and the interval over which the model applies.

Example: a cart reverses direction

Negative initial velocity

Given: x₀ = 0 m, v₀ = −2 m/s, a = +1.5 m/s², t = 4 s.
Find: Find final position, final velocity and total distance.

Approach: first find when v = 0. Calculate displacements before and after that instant separately; their magnitudes give distance.
$\begin{aligned}x(t)&=x_0+v_0t+\frac12at^2\\v(t)&=v_0+at\\t_*&=-\frac{v_0}{a}\end{aligned}$
Substitution: use the specified initial values with their signs.
$\begin{aligned}x(4)&=-2\cdot4+\tfrac12\cdot1.5\cdot4^2=4\,\mathrm m\\{} v(4)&=4\,\mathrm{m/s}\end{aligned}$
Reversal occurs at v = 0; the cart moves left until then.
$\begin{aligned}t_*&=\frac{2}{1.5}=\frac43\,\mathrm s\\{} x(t_*)&=-\frac43\,\mathrm m\end{aligned}$
Add the two segments as positive lengths.
$L=\left|{-\frac43}-0\right|+\left|4-\left(-\frac43\right)\right|=\frac{20}{3}\,\mathrm m$

Result: Final position +4 m, final velocity +4 m/s, distance 6.67 m.

Check: Distance exceeds |s| = 4 m. The magnitude of average velocity is 1 m/s; average speed is 1.67 m/s.

Example: stopping distance without extending past the stop

Braking from 72 km/h

Given: v₀ = 72 km/h, a = −4 m/s², constant deceleration until rest.
Find: Calculate braking time and distance excluding reaction time.

Approach: stopping ends the braking phase at v = 0. Find that time before substituting it into the displacement equation.
$\begin{aligned}0&=v_0+at_{\mathrm{stop}}\\t_{\mathrm{stop}}&=-\frac{v_0}{a}\\s_{\mathrm{stop}}&=v_0t_{\mathrm{stop}}+\frac12at_{\mathrm{stop}}^2\end{aligned}$
Units: convert initial speed to metres per second before substitution.
$v_0=\frac{72}{3.6}=20\,\mathrm{m/s}$
v = 0 gives the endpoint of the braking model.
$t_{\mathrm{stop}}=\frac{0-20}{-4}=5\,\mathrm s$
Insert the stopping time into displacement.
$s=20\cdot5+\tfrac12(-4)5^2=50\,\mathrm m$

Result: Stopping time 5 s, braking distance 50 m.

Check: The time-free check gives 0² = 20² + 2·(−4)·50. Extending to 6 s would predict negative velocity, which is no longer simple stopping.

How do you read areas and slopes in motion graphs?

\begin{aligned}\bar v&=\frac{s}{t}=\frac{v_0+v(t)}2\\L&=\int_0^t|v(\tau)|\,\mathrm d\tau\end{aligned}

The endpoint average holds for constant a; it is not generally average speed.

When the position–time model must change

Five exercises with solutions

Calculate · Foundation

An object starts from rest at x₀ = 3 m with a = 2 m/s². Where is it after 4 s?

Show hint

Calculate displacement separately from initial position.

Learning goal: Combine initial position and displacement to find final position.

Show solution and calculation

Approach: first calculate displacement from the initial conditions.
Starting from rest eliminates v₀t, giving s = 16 m after four seconds.
Only then add initial position: x = 19 m.

Check: initial position shifts the entire position graph by 3 m without changing distance or velocity.

\begin{aligned}s&=v_0t+\frac12at^2\\&=\frac12(2\,\mathrm{m/s^2})(4\,\mathrm s)^2=16\,\mathrm m\\x&=x_0+s=3\,\mathrm m+16\,\mathrm m=19\,\mathrm m\end{aligned}

[1]

Rearrange · Application

v₀ = 5 m/s, a = 2 m/s², target displacement 24 m. Find the positive time.

Show hint

Solve t² + 5t − 24 = 0.

Learning goal: Select the physically appropriate time root.

Show solution and calculation

Approach: initial velocity is nonzero, so retain the linear term.
In SI units the equation is t² + 5t − 24 = 0, which factors as (t − 3)(t + 8) = 0.
Time after the start is 3 s; −8 s lies outside the interval.

Check: 15 m from initial motion plus 9 m from acceleration gives the required 24 m.

\begin{aligned}s&=v_0t+\frac12at^2\\24\,\mathrm m&=(5\,\mathrm{m/s})t+(1\,\mathrm{m/s^2})t^2\\t&=3\,\mathrm s\qquad\lor\qquad t=-8\,\mathrm s\\s(3\,\mathrm s)&=15\,\mathrm m+9\,\mathrm m=24\,\mathrm m\end{aligned}

[1]

Transfer · Application

v₀ = 10 m/s, constant a = −2 m/s². Find the distance to rest.

Show hint

Use v² = v₀² + 2as with v = 0.

Learning goal: Find stopping distance from the time-free equation.

Show solution and calculation

Approach: set velocity to zero at rest.
Stopping time is 5 s.
Velocity decreases uniformly from 10 to 0 m/s, so its average is 5 m/s and stopping distance is 25 m.

Check: the time-free equation gives the same 25 m using negative a. The model is not continued beyond rest.

\begin{aligned}t_{\mathrm{stop}}&=-\frac{v_0}{a}=-\frac{10}{-2}\,\mathrm s=5\,\mathrm s\\s&=v_0t+\frac12at^2\\&=10(5)\,\mathrm m+\frac12(-2)(5)^2\,\mathrm m=25\,\mathrm m\end{aligned}

[1]

Interpret · Application

One-dimensional motion is $x(t)=(1\,\mathrm{m/s^2})t^2-(4\,\mathrm{m/s})t$ for 0≤t≤4 s. Find initial and final positions and the turning point. Determine displacement and distance travelled.

Show hint

$v=\frac{dx}{dt}$ vanishes at reversal; split distance into two legs.

Learning goal: Distinguish displacement from path length using a position parabola.

Show solution and calculation

Approach: x(0)=x(4 s)=0, so displacement is zero. v=2t−4 in SI units vanishes at 2 s, where x=−4 m.
The body travels 4 m left and 4 m back, totaling 8 m.

Check: positive curvature gives constant a=2 m/s² despite initial negative motion. Final position alone does not give total distance.

\begin{aligned}\Delta x&=x(4)-x(0)=0\\L&=|x(2)-x(0)|+|x(4)-x(2)|=8\,\mathrm m\end{aligned}

[1][2]

Error analysis · Application

A particle starts from rest at x₀=5 m and accelerates at constant a=2 m/s² along +x. At x=9 m a solution sets $v^2=2ax=36\,\mathrm{m^2/s^2}$ . Correct it and check using elapsed time.

Show hint

The equation contains x−x₀ rather than absolute position.

Learning goal: Use displacement rather than absolute position in the time-free equation.

Show solution and calculation

Approach: displacement is 4 m, so $v^2=2a(x-x_0)=16\,\mathrm{m^2/s^2}$ and v=+4 m/s.

Check: $x-x_0=\frac12at^2$ gives t=2 s and v=at=4 m/s. The incorrect result depends on an arbitrary coordinate origin; changing that origin must not change predicted physical velocity.

\begin{aligned}v&=\sqrt{2a(x-x_0)}=\sqrt{2\cdot2\cdot4}\,\mathrm{m\,s^{-1}}=4\,\mathrm{m\,s^{-1}}\\t&=\sqrt{\frac{2(x-x_0)}a}=2\,\mathrm s\end{aligned}

[1]

What to take away

Position combines initial position, initial motion and the acceleration contribution. Displacement is purely proportional to t² only from rest.
Negative displacement is allowed. During reversal, total distance exceeds displacement magnitude.
For stopping, first find the instant of rest. With multiple time roots, identify which belongs to the actual motion phase.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Stated the displacement–time equation and start-from-rest condition directly; made existing rearrangements and graphs accessible through learning questions. Added contextual links for the velocity law, braking-force model and free fall.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added prerequisites and an intuitive introduction; clarified the derivation, worked-example approaches and exercise solutions, and added a summary.
October 2, 2026Linked time roots to the quadratic formula and component motion to projectile motion.

Cite this reference page

Quanta. Uniformly Accelerated Motion (Position–Time Equation). Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/gleichmaessig-beschleunigte-bewegung

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Frequently asked questions about Uniformly Accelerated Motion (Position–Time Equation)

How do you calculate the distance under uniform acceleration?+

When may you use s = ½at² without the v₀t term?+

What do you do when the time is not given?+

How are the distance-time and velocity-time laws related?+

Why does the distance quadruple when the time doubles?+

Retain Uniformly Accelerated Motion (Position–Time Equation) for exams

Create a curated FSRS exam set for s = ½at² + v₀t: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Uniformly accelerated motion: position–time equation

Variables & units

Displacement calculator for constant acceleration

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Kinematics of straight-line motion

Worked example

Applications

How do you calculate with Uniformly Accelerated Motion (Position–Time Equation)?

When is the equation for a start from rest sufficient?

What is the displacement–time equation?

Integrating twice while preserving initial values

Why the factor one half appears

How do you calculate acceleration and possible time roots?

Example: a cart reverses direction

Negative initial velocity

Example: stopping distance without extending past the stop

Braking from 72 km/h

How do you read areas and slopes in motion graphs?

When the position–time model must change

Five exercises with solutions

An object starts from rest at x0 = 3 m with a = 2 m/s2. Where is it after 4 s?

v0 = 5 m/s, a = 2 m/s2, target displacement 24 m. Find the positive time.

v0 = 10 m/s, constant a = −2 m/s2. Find the distance to rest.

One-dimensional motion is x(t)=(1 m/s2)t2−(4 m/s)tx(t)=(1\,\mathrm{m/s^2})t^2-(4\,\mathrm{m/s})tx(t)=(1m/s2)t2−(4m/s)t for 0≤t≤4 s. Find initial and final positions and the turning point. Determine displacement and distance travelled.

A particle starts from rest at x0=5 m and accelerates at constant a=2 m/s2 along +x. At x=9 m a solution sets v2=2ax=36 m2/s2v^2=2ax=36\,\mathrm{m^2/s^2}v2=2ax=36m2/s2. Correct it and check using elapsed time.

What to take away

Sources for this explanation

Frequently asked questions about Uniformly Accelerated Motion (Position–Time Equation)

Subject references and further reading

Related formulas

More formulas: Mechanics

Retain Uniformly Accelerated Motion (Position–Time Equation) for exams

Uniformly accelerated motion: position–time equation

Variables & units

Displacement calculator for constant acceleration

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Kinematics of straight-line motion

Worked example

Applications

How do you calculate with Uniformly Accelerated Motion (Position–Time Equation)?

When is the equation for a start from rest sufficient?

What is the displacement–time equation?

Integrating twice while preserving initial values

Why the factor one half appears

How do you calculate acceleration and possible time roots?

Example: a cart reverses direction

Negative initial velocity

Example: stopping distance without extending past the stop

Braking from 72 km/h

How do you read areas and slopes in motion graphs?

When the position–time model must change

Five exercises with solutions

An object starts from rest at x0 = 3 m with a = 2 m/s2. Where is it after 4 s?

v0 = 5 m/s, a = 2 m/s2, target displacement 24 m. Find the positive time.

v0 = 10 m/s, constant a = −2 m/s2. Find the distance to rest.

One-dimensional motion is x(t)=(1 m/s2)t2−(4 m/s)tx(t)=(1\,\mathrm{m/s^2})t^2-(4\,\mathrm{m/s})tx(t)=(1m/s2)t2−(4m/s)t for 0≤t≤4 s. Find initial and final positions and the turning point. Determine displacement and distance travelled.

A particle starts from rest at x0=5 m and accelerates at constant a=2 m/s2 along +x. At x=9 m a solution sets v2=2ax=36 m2/s2v^2=2ax=36\,\mathrm{m^2/s^2}v2=2ax=36m2/s2. Correct it and check using elapsed time.

What to take away

Sources for this explanation

Frequently asked questions about Uniformly Accelerated Motion (Position–Time Equation)

Subject references and further reading

Related formulas

More formulas: Mechanics

Retain Uniformly Accelerated Motion (Position–Time Equation) for exams

An object starts from rest at x₀ = 3 m with a = 2 m/s². Where is it after 4 s?

v₀ = 5 m/s, a = 2 m/s², target displacement 24 m. Find the positive time.

v₀ = 10 m/s, constant a = −2 m/s². Find the distance to rest.

One-dimensional motion is $x(t)=(1\,\mathrm{m/s^2})t^2-(4\,\mathrm{m/s})t$ for 0≤t≤4 s. Find initial and final positions and the turning point. Determine displacement and distance travelled.

A particle starts from rest at x₀=5 m and accelerates at constant a=2 m/s² along +x. At x=9 m a solution sets $v^2=2ax=36\,\mathrm{m^2/s^2}$ . Correct it and check using elapsed time.

An object starts from rest at x₀ = 3 m with a = 2 m/s². Where is it after 4 s?

v₀ = 5 m/s, a = 2 m/s², target displacement 24 m. Find the positive time.

v₀ = 10 m/s, constant a = −2 m/s². Find the distance to rest.

One-dimensional motion is $x(t)=(1\,\mathrm{m/s^2})t^2-(4\,\mathrm{m/s})t$ for 0≤t≤4 s. Find initial and final positions and the turning point. Determine displacement and distance travelled.

A particle starts from rest at x₀=5 m and accelerates at constant a=2 m/s² along +x. At x=9 m a solution sets $v^2=2ax=36\,\mathrm{m^2/s^2}$ . Correct it and check using elapsed time.