Physics · Kinematics

Velocity–time law

Q: How do you calculate the final speed with $v = a\cdot{}t + v_{0}$?

With constant a, add at to signed initial velocity. For $v_{0} =\frac{ 10 m}{s}$, $a =\frac{ 1.5 m}{s^{2}}$ and $t = 8 s$ the result is $+\frac{22 m}{s}$, or $\frac{79.2 km}{h}$. Speed is the magnitude. With $v_{0} < 0$, positive a can first reduce speed and later increase it after reversal. Check $t = 0$: the result must equal v₀. Also check units, since at must be in $\frac{m}{s}$. For $a = 0$, velocity stays unchanged. A negative final value means motion opposite the positive axis. In a simple stopping problem, the modelled braking phase normally ends at the first instant of rest.

Q: What does an acceleration of $\frac{1 m}{s^{2}}$ mean intuitively?

A constant positive component of $\frac{1 m}{s^{2}}$ changes the velocity component by $+\frac{1 m}{s}$ each second. A cart starting at $-\frac{3 m}{s}$ has $-\frac{1 m}{s}$ after 2 s and has slowed down. Read the unit as $(\frac{m}{s})$ per second. Acceleration also includes changes of direction. For a second example, start at $+\frac{3 m}{s}$: after two seconds under the same positive acceleration, velocity is $+\frac{5 m}{s}$ and speed really has increased. Signs become meaningful only after choosing an axis. With varying acceleration, its time history is needed; a single instantaneous value does not describe the whole interval.

Q: How do you rearrange $v = a\cdot{}t + v_{0}$ for the time?

For $a \ne{} 0$, $t =\frac{ (v - v_{0})}{a}$. From $+\frac{26 m}{s}$ to $+\frac{20 m}{s}$ with $a = -\frac{2 m}{s^{2}}$ gives 3 s. Negative time can indicate a pre-start instant or a target unreachable after the start; it does not automatically prove a sign error. For $a = 0$ velocity remains constant and division is not allowed. Check the calculated time by substituting it into the original equation. Units reduce from $(\frac{m}{s})$ divided by $(\frac{m}{s^{2}})$ to seconds. A target velocity may occur before a reversal or afterwards. The intended motion phase and constant acceleration assumption must actually remain valid up to the calculated instant.

Q: What is the difference between velocity and acceleration?

Velocity is the rate of position change; acceleration is the rate of velocity change. Constant velocity means $a = 0$. At the top of a vertical launch $v = 0$, but $a = -g$ with upward positive. Uniform circular motion has constant speed and nonzero acceleration because direction changes. For an object with $v = -\frac{4 m}{s}$ and $a = +\frac{1 m}{s^{2}}$, speed initially decreases even though acceleration is positive. Reversing the axis changes both signs without changing the physical motion. Positive and negative alone therefore do not describe whether an object is speeding up; compare the velocity and acceleration directions.

Q: How do you read velocity and acceleration from diagrams?

Slope of the x-t graph is v; slope of the v-t graph is a. A rising straight line represents positive constant a, a falling straight line negative constant a. Signed area in the v-t plot gives displacement. For total distance, add absolute areas on both sides of the time axis. Read speed as the magnitude of the velocity component. A graph far below the time axis can therefore represent high speed. Two measured points determine only average acceleration through their connecting line. Additional measurements or a suitable modelling assumption are needed to support constant acceleration throughout the interval between those points.

The velocity–time equation describes the linear change of signed velocity under constant acceleration. Depending on signs, speed can decrease or increase.

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Formula

Calculate

v = a \cdot t + v_{0}

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

Variables & units

Variables & units
Symbol	Meaning	Unit
$v$	Velocity at time t	m/s
$a$	Constant acceleration	m/s²
$t$	Time	s
$v₀$	Initial velocity	m/s

LaTeX

v = a \cdot t + v_0

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Diagram: a v-t line meets the axis at v₀; a slope triangle shows Δv over Δt as the acceleration a. — At constant a, directed velocity changes linearly. The illustrated line has a > 0; its slope is a=Δv/Δt and its intercept is v₀.

Final velocity calculator

Find velocity, initial velocity, acceleration or time 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; if it acts against the motion, the body slows down.

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: Velocity v

v = 72 km/h

With v₀ = 36 km/h, a = 2 m/s² and t = 5 s, v = 72 km/h.

Working · rearrange for v, substitute, evaluate

1. Rearrange for v
$v=v_0+at$
2. Substitute values (in SI units)
- v₀ = 36 km/h = 10 m/s
$v=10+2\cdot 5=20$
3. Result
v = 72 km/h

Quick tips

On a v-t graph v₀ is the intercept and a the gradient: every second adds a, so 2 m/s per second at 2 m/s².
A body slows down when a and v have opposite signs. A negative a on its own does not mean braking.

Assumptions and limits

One common positive axis, constant acceleration and elapsed time t ≥ 0. Negative acceleration does not always mean slowing. Velocity is signed, not speed magnitude.

Derivation & background

Instantaneous acceleration is $a =\frac{ dv}{dt}$ ; $\frac{\Delta{}v}{\Delta{}t}$ is an interval average. Constant a gives $v=v_0+at$ . In a velocity–time plot, a is the slope and signed area is displacement. Slowing down means acceleration points against the current motion.

Exam blueprint

Validity range

The short form assumes a constant acceleration component on a fixed axis. Slowing down requires opposite signs of v and a; $a < 0$ alone does not imply slowing down. With varying a, use v(t) = v₀ + ∫₀ᵗa(τ)dτ.

Derivation steps

Integrate a constant rate of change of signed velocity.

1Choose a positive axis and set v₀ as the signed initial velocity.
2Acceleration means $a =\frac{ dv}{dt}$ ; over a finite interval ā = Δv/Δt.
3With constant a, average and instantaneous acceleration agree.
4For start time zero, $a =\frac{ (v - v_{0})}{t}$ with $t \ne{} 0$ .
5Multiply by t and add v₀: $v=v_0+at$ .
6Check $v(0) = v_{0}$ and $\frac{dv}{dt} = a$ . Speed is |v|, not the signed value.

Rearrangements

Time to reach a signed target velocity

t = \frac{v - v_0}{a}

Use signed velocities. Require $a \ne{} 0$ and check that the time lies in the modelled interval.

Average acceleration from two signed velocities

a = \frac{v - v_0}{t}

The quotient initially gives an interval average. It also equals instantaneous acceleration only if a is constant. Slowing down requires opposite velocity and acceleration directions.

Task variant

A car accelerates at $a =\frac{ 2.5 m}{s^{2}}$ from 0 to $\frac{100 km}{h}$ . How long does it take?

$\frac{100 km}{h} =\frac{ 27.8 m}{s}$ .
$t =\frac{ (27.8 - 0)}{2.5} \approx{} 11.1 s$ .

A cyclist speeds up from $\frac{20 m}{s}$ to $\frac{26 m}{s}$ in 3 s. Find a.

$a =\frac{ (26 - 20)}{3} =\frac{ 2 m}{s^{2}}$ .

Common mistakes

Substituting $\frac{km}{h}$ directly.

Always convert to $\frac{m}{s}$ by dividing by 3.6 first.

Dropping v₀ and writing $v = a\cdot{}t$ although the body is already moving.

$v = a\cdot{}t$ only holds for a start from rest.

Ignoring the sign of a when braking.

Choose an axis and put acceleration opposite the current motion: $a < 0$ for $v > 0$ , but $a > 0$ for $v < 0$ . A stopping phase normally ends at rest.

Exam context

Often a sub-step: first find v(t), then compute distance, momentum or kinetic energy from it.

Formula cluster

Laws of motion

The v-t law is the derivative of the distance-time law.

s = ½at² + v₀t

Uniformly Accelerated Motion (Position–Time Equation)

h = ½gt²

Free Fall

p = m·v

Momentum

Worked example

A train starts at $v_{0} = +\frac{10 m}{s}$ and accelerates constantly at $a = +\frac{1.5 m}{s^{2}}$ . After 8 s, $v=v_0+at=+22\,\mathrm{m/s}$ , corresponding to $\frac{79.2 km}{h}$ . Positive signs refer to the chosen direction of travel.

Applications

Acceleration measurement (0-to-100 time), braking deceleration, lift control, sports analysis (sprint start)

How do you calculate with Velocity-Time Law?

Here is how to work through a typical Velocity-Time Law (v = a·t + v₀) task step by step:

1
Task
A car accelerates at $a =\frac{ 2.5 m}{s^{2}}$ from 0 to $\frac{100 km}{h}$ . How long does it take?
Solution path
1. $\frac{100 km}{h} =\frac{ 27.8 m}{s}$ .
2. $t =\frac{ (27.8 - 0)}{2.5} \approx{} 11.1 s$ .
2
Task
A cyclist speeds up from $\frac{20 m}{s}$ to $\frac{26 m}{s}$ in 3 s. Find a.
Solution path
$a =\frac{ (26 - 20)}{3} =\frac{ 2 m}{s^{2}}$ .

Understand and apply the equation

A cart can roll left while having positive acceleration. What matters is how its signed velocity changes. First learn to read the signs, then derive the velocity–time equation and test it against a reversal and measured values.

Getting an intuitive picture

Imagine a straight rail with rightward motion chosen as positive. A reading of −6 m/s means the cart moves left. If the reading increases by +2 m/s every second, successive readings are −6, −4, −2 and 0 m/s. The cart initially slows down despite positive acceleration. Only after stopping can it speed up to the right.[1][2]

Prerequisite: a positive axis defines the meanings of plus and minus. The sign describes direction; the magnitude describes speed.
A change is final value minus initial value. Time in the equation means elapsed time since that initial instant.
Initially you need only subtraction, multiplication and division. The later integral expression explains varying acceleration.

What is the velocity–time law?

For constant acceleration, the velocity–time law is $v=v_0+at$ , where t is elapsed time from the initial state. Along a straight line, v denotes the velocity component on the chosen axis. When v = −6 m/s, speed is 6 m/s. Everyday language often blurs this distinction; calculations must separate direction and magnitude. Acceleration a describes how v changes. It is not an additional velocity but its rate of change. In free fall near Earth’s surface without air forces, acceleration is approximately constant; its sign depends on the selected vertical axis.[1][3][4]

v(t)=v_0+at

v₀ is velocity at t = 0; a must be constant throughout the interval. Speed is the magnitude |v(t)|.

Along a line, speed increases when v·a > 0 and decreases when v·a < 0. Positive a can therefore initially slow an object with v < 0. After reversal the same positive acceleration speeds it up in the positive direction. Velocity is zero at the turn, but acceleration need not be.

From acceleration to the velocity–time equation

Start with two velocities measured a time t apart. Subtract the initial reading from the final one to obtain the change. Divide by elapsed time to obtain average acceleration. Under constant acceleration, equal time intervals produce equal velocity changes. The average then also equals acceleration at every instant.[2][3]

\begin{aligned}a&=\frac{v-v_0}{t}\\at&=v-v_0\\v&=v_0+at\end{aligned}

First multiply the rate by time, then add the initial value. The quotient requires t ≠ 0; the resulting equation also holds at t = 0.[3]

The units make the step concrete: a rate of 2 m/s per second produces a change of 10 m/s in five seconds. Whether speed increases still depends on the initial sign. If the interval starts at another time, use the time difference rather than the clock reading.

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

The integral form also applies to varying acceleration.

For an interval beginning at t₀, use v(t) = v(t₀) + a(t − t₀). A later timestamp cannot be substituted without subtracting the start time. Differentiate the equation as a check: dv/dt must return the assumed constant a.

Velocity–time graph, rearrangements and acceleration units

In a velocity–time plot, v₀ is the vertical intercept and a is slope Δv/Δt. Positive a gives a rising line; negative a gives a falling line. A horizontal line represents constant velocity. Crossing the time axis at v = 0 may mark a change in motion direction. Signed area under the graph gives displacement; the displacement–time law calculates it for constant acceleration.[3][1]

\begin{aligned}[a]&=\mathrm{m\,s^{-2}}\\{}[at]&=\mathrm{m\,s^{-1}}\\a&=\frac{v-v_0}{t}\\t&=\frac{v-v_0}{a}\end{aligned}

The last rearrangement requires a ≠ 0. A calculated time must lie within the modelled motion interval.

With a = 0, v stays constant: a different target velocity is unreachable, while v = v₀ at every time. Divide km/h by 3.6 first. An acceleration of 2 m/s² means velocity changes by 2 m/s each second; it is not a fixed velocity of 2 m/s.

Velocity and speed during reversal

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Signed velocity rises uniformly; its magnitude initially falls and rises after reversal. Acceleration is +2 m/s² throughout.[3][2]

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Example: slowing down before speeding up in reverse

A cart with v₀ < 0 and a > 0

Given: v₀ = −6 m/s, a = +2 m/s², x₀ = 0 m, duration 5 s.
Find: Find final velocity, turn time, displacement and distance.

Approach: use signed velocity. Reversal occurs where it becomes zero; displacement is signed area under the velocity graph.
$\begin{aligned}v(t)&=v_0+at\\t_*&=-\frac{v_0}{a}\\s(t)&=v_0t+\frac12at^2\end{aligned}$
Substitution: use the specified initial values with their signs.
$v(5)=-6+2\cdot5=4\,\mathrm{m/s}$
The zero separates the two directions of motion.
$t_*=-\frac{v_0}{a}=3\,\mathrm s$
Evaluate the negative triangle to 3 s and positive triangle to 5 s separately.
$\begin{aligned}s_-&=-\tfrac12\cdot3\cdot6=-9\,\mathrm m\\{} s_+&=\tfrac12\cdot2\cdot4=4\,\mathrm m\end{aligned}$
Add signed values for displacement and magnitudes for distance.
$\begin{aligned}s&=-9+4=-5\,\mathrm m\\{} L&=9+4=13\,\mathrm m\end{aligned}$

Result: At 5 s: v = +4 m/s, turn at 3 s, displacement −5 m and distance 13 m.

Check: The position equation independently gives −6·5 + ½·2·5² = −5 m. Average velocity is −1 m/s; average speed is 2.6 m/s.

Example: acceleration from two measurements

Uniform deceleration

Given: Initial velocity +18 m/s, velocity after 3 s +6 m/s; constant a is assumed.
Find: Find a and predict v at 4 s assuming the same constant a continues.

Approach: the measurements initially determine an average. Using it to predict later velocity additionally requires constant acceleration.
$\begin{aligned}\Delta v&=v_f-v_i\\a&=\frac{\Delta v}{\Delta t}\\v(t)&=v_i+at\end{aligned}$
Substitution: use the specified initial values with their signs.
$\Delta v=6-18=-12\,\mathrm{m/s}$
Divide the change by elapsed time.
$a=-\frac{12}{3}=-4\,\mathrm{m/s^2}$
Use the value only while constant deceleration continues.
$v(4)=18-4\cdot4=2\,\mathrm{m/s}$

Result: a = −4 m/s²; continued unchanged acceleration gives v = +2 m/s at 4 s. Rest would occur at 4.5 s.

Check: At t = 3 s the line reproduces +6 m/s. Two measurements alone do not establish constant acceleration between them; they establish the average.

Explore further: when acceleration changes with time

For an original counterexample, let a(t) = b·t with b = 0.8 m/s³ and v₀ = 1 m/s. Integration gives v(t) = 1 m/s + ½bt². At 3 s, v = 4.6 m/s. Multiplying the instantaneous value a(3 s) = 2.4 m/s² by three seconds would give 8.2 m/s, overestimating the change.

\Delta v=\int_0^t b\tau\,\mathrm d\tau=\frac12bt^2

Linearly growing acceleration gives a triangular area in an acceleration–time plot.

An interval average ā satisfies Δv = āΔt even when a varies. It cannot automatically predict arbitrary intermediate times or justify x = x₀ + v₀t + ½āt². Position depends on when acceleration occurs.[2]

From one-dimensional motion to vectors

\vec v(t)=\vec v_0+\vec a\,t

The vector form requires constant vector acceleration, not merely a constant magnitude.

In uniform circular motion, speed is constant while velocity changes direction. The acceleration vector rotates and is not constant. Do not insert the magnitude of centripetal acceleration into the linear equation. An instantaneous velocity jump likewise cannot represent finite constant acceleration; real transitions take time. The net centripetal force points towards the circle’s centre and changes the direction of motion.[5]

For piecewise constant acceleration, calculate each interval separately. The final velocity of one interval becomes the initial velocity of the next. Continuous velocities at transitions and consistent units are simple checks on the model.

Five exercises for reliable application

Calculate · Foundation

v₀ = −3 m/s and a = −2 m/s². Find velocity and speed after 2 s. Is it speeding up?

Show hint

Give the signed value and its magnitude separately.

Learning goal: Separate signed velocity from speed change.

Show solution and calculation

Approach: constant acceleration gives v = v₀ + at.
After two seconds v = −7 m/s.
Speed increases from 3 to 7 m/s; the body speeds up to the left.

Check: each second adds −2 m/s to signed velocity. Negative acceleration therefore does not mean slowing down here.

\begin{aligned}v&=v_0+at\\&=-3\,\mathrm{m/s}+(-2\,\mathrm{m/s^2})(2\,\mathrm s)\\&=-7\,\mathrm{m/s}\\|v|&=7\,\mathrm{m/s}\end{aligned}

[2][3]

Rearrange · Foundation

A cart travels at 36 km/h and accelerates constantly at 1.5 m/s². When does it reach 54 km/h?

Show hint

Convert both velocities before taking their difference.

Learning goal: Find acceleration time after velocity conversion.

Show solution and calculation

Approach: time is required for a velocity change, so t = (v − v₀)/a.
The SI velocities are 10 and 15 m/s.
Their difference is 5 m/s, not 15 m/s.
Time is about 3.33 s.

Check: acceleration produces precisely the missing 5 m/s in this interval.

\begin{aligned}v_0&=\frac{36}{3.6}\,\mathrm{m/s}=10\,\mathrm{m/s}\\v&=\frac{54}{3.6}\,\mathrm{m/s}=15\,\mathrm{m/s}\\t&=\frac{v-v_0}{a}=\frac{15-10}{1.5}\,\mathrm s\\&=\frac{10}{3}\,\mathrm s\approx3.33\,\mathrm s\end{aligned}

[2][3]

Transfer · Application

v₀ = 0 and a(t) = (0.6 m/s³)·t. Find v after 4 s.

Show hint

Integrate a(t) rather than multiplying its final value by t.

Learning goal: Integrate time-dependent acceleration.

Show solution and calculation

Approach: velocity change is area under the acceleration graph.
Acceleration rises linearly from zero, so the area is a triangle.
At four seconds a reaches 2.4 m/s².
A triangle with base 4 s and height 2.4 m/s² gives 4.8 m/s.

Check: the rectangle using the final value would be twice as large and would overestimate acceleration at early times.

\begin{aligned}a(4\,\mathrm s)&=2.4\,\mathrm{m/s^2}\\v(4\,\mathrm s)-v_0&=\frac12(4\,\mathrm s)(2.4\,\mathrm{m/s^2})\\&=4.8\,\mathrm{m/s}\end{aligned}

[2][3]

Interpret · Application

A velocity-time plot is a straight line from v=−4 m/s at t=0 to v=+4 m/s at t=4 s. Read acceleration and reversal time. During which intervals does speed decrease or increase?

Show hint

Slope gives a; speed is the magnitude of v.

Learning goal: Identify reversal in a constant-acceleration velocity-time plot.

Show solution and calculation

Approach: slope is 2 m/s², giving v=−4+2t in SI units.
Velocity vanishes at 2 s.
Before then |v| decreases; afterward it increases, while a stays positive.

Check: at 4 s velocity is +4 m/s. Positive acceleration does not imply speeding up throughout. Speed has a cusp in its magnitude plot at reversal, while signed velocity stays smooth.

\begin{aligned}a&=\frac{4-(-4)}4\,\mathrm{m/s^2}=2\,\mathrm{m/s^2}\\t_{\mathrm{turn}}&=\frac42\,\mathrm s=2\,\mathrm s\end{aligned}

[1][2][3]

Error analysis · Application

A body moves uniformly around a circle of radius 2 m at speed 3 m/s. Acceleration magnitude is 4.5 m/s². A solution uses v=3+4.5t and claims speed 7.5 m/s after 1 s. Why is the model wrong?

Show hint

The acceleration vector points inward and rotates.

Learning goal: Avoid treating circular-motion acceleration as a constant vector.

Show solution and calculation

Approach: uniform circular motion changes velocity direction, not magnitude.
After 1 s speed remains 3 m/s.
Centripetal magnitude 4.5 m/s² is not a constant tangential component on a fixed line.

Check: v·a=0, so $\frac{d(v^2)}{dt}=0$ . The constant-a expression applies componentwise only for a genuinely constant acceleration vector.

\begin{aligned}\frac{\mathrm d}{\mathrm dt}|\mathbf v|^2&=2\mathbf v\cdot\mathbf a=0\\|\mathbf v(1\,\mathrm s)|&=3\,\mathrm{m/s}\end{aligned}

[2]

What to take away

Velocity needs an axis; speed is its magnitude. Acceleration describes change in signed velocity.
The short equation requires constant a. Two measured points alone do not establish that assumption.
Set v = 0 to find reversal. To decide whether speed increases, compare the directions of v and a rather than the sign of a alone.

Sources for this explanation

Quanta · Updated October 1, 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, 2026Answered the velocity–time law directly with constant acceleration and elapsed time; made units, rearrangements and varying acceleration easier to find. Retained reversal and v·a reasoning, and linked displacement, free fall and centripetal force.
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.

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Quanta. Velocity-Time Law. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/geschwindigkeit-beschleunigung

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Frequently asked questions about Velocity-Time Law

How do you calculate the final speed with

v = a\cdot{}t + v_{0}

With constant a, add at to signed initial velocity. For $v_{0} =\frac{ 10 m}{s}$ , $a =\frac{ 1.5 m}{s^{2}}$ and $t = 8 s$ the result is $+\frac{22 m}{s}$ , or $\frac{79.2 km}{h}$ . Speed is the magnitude. With $v_{0} < 0$ , positive a can first reduce speed and later increase it after reversal. Check $t = 0$ : the result must equal v₀. Also check units, since at must be in $\frac{m}{s}$ . For $a = 0$ , velocity stays unchanged. A negative final value means motion opposite the positive axis. In a simple stopping problem, the modelled braking phase normally ends at the first instant of rest.

What does an acceleration of

\frac{1 m}{s^{2}}

mean intuitively?+

A constant positive component of $\frac{1 m}{s^{2}}$ changes the velocity component by $+\frac{1 m}{s}$ each second. A cart starting at $-\frac{3 m}{s}$ has $-\frac{1 m}{s}$ after 2 s and has slowed down. Read the unit as $(\frac{m}{s})$ per second. Acceleration also includes changes of direction. For a second example, start at $+\frac{3 m}{s}$ : after two seconds under the same positive acceleration, velocity is $+\frac{5 m}{s}$ and speed really has increased. Signs become meaningful only after choosing an axis. With varying acceleration, its time history is needed; a single instantaneous value does not describe the whole interval.

How do you rearrange

v = a\cdot{}t + v_{0}

for the time?+

For $a \ne{} 0$ , $t =\frac{ (v - v_{0})}{a}$ . From $+\frac{26 m}{s}$ to $+\frac{20 m}{s}$ with $a = -\frac{2 m}{s^{2}}$ gives 3 s. Negative time can indicate a pre-start instant or a target unreachable after the start; it does not automatically prove a sign error. For $a = 0$ velocity remains constant and division is not allowed. Check the calculated time by substituting it into the original equation. Units reduce from $(\frac{m}{s})$ divided by $(\frac{m}{s^{2}})$ to seconds. A target velocity may occur before a reversal or afterwards. The intended motion phase and constant acceleration assumption must actually remain valid up to the calculated instant.

What is the difference between velocity and acceleration?+

Velocity is the rate of position change; acceleration is the rate of velocity change. Constant velocity means $a = 0$ . At the top of a vertical launch $v = 0$ , but $a = -g$ with upward positive. Uniform circular motion has constant speed and nonzero acceleration because direction changes. For an object with $v = -\frac{4 m}{s}$ and $a = +\frac{1 m}{s^{2}}$ , speed initially decreases even though acceleration is positive. Reversing the axis changes both signs without changing the physical motion. Positive and negative alone therefore do not describe whether an object is speeding up; compare the velocity and acceleration directions.

How do you read velocity and acceleration from diagrams?+

Slope of the x-t graph is v; slope of the v-t graph is a. A rising straight line represents positive constant a, a falling straight line negative constant a. Signed area in the v-t plot gives displacement. For total distance, add absolute areas on both sides of the time axis. Read speed as the magnitude of the velocity component. A graph far below the time axis can therefore represent high speed. Two measured points determine only average acceleration through their connecting line. Additional measurements or a suitable modelling assumption are needed to support constant acceleration throughout the interval between those points.

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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, §3.3 Average and Instantaneous Acceleration

More formulas: Mechanics

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

Newton's Second Law

F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

Newton's Law of Universal Gravitation

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

Kinetic Energy

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

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Velocity–time law

The velocity–time equation describes the linear change of signed velocity under constant acceleration. Depending on signs, speed can decrease or increase.

BasicExam-relevantLower secondaryUpper secondary

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

v

Velocity at time t

m/s

a

Constant acceleration

m/s²

t

Time

v₀

Initial velocity

m/s

Derivation & background

Exam blueprint

Validity range

Derivation steps

Integrate a constant rate of change of signed velocity.

1Choose a positive axis and set v₀ as the signed initial velocity.
2Acceleration means $a =\frac{ dv}{dt}$ ; over a finite interval ā = Δv/Δt.
3With constant a, average and instantaneous acceleration agree.
4For start time zero, $a =\frac{ (v - v_{0})}{t}$ with $t \ne{} 0$ .
5Multiply by t and add v₀: $v=v_0+at$ .
6Check $v(0) = v_{0}$ and $\frac{dv}{dt} = a$ . Speed is |v|, not the signed value.

How do you calculate with Velocity-Time Law?

Here is how to work through a typical Velocity-Time Law (v = a·t + v₀) task step by step:

Task

A car accelerates at $a =\frac{ 2.5 m}{s^{2}}$ from 0 to $\frac{100 km}{h}$ . How long does it take?

Solution path

$\frac{100 km}{h} =\frac{ 27.8 m}{s}$ .
$t =\frac{ (27.8 - 0)}{2.5} \approx{} 11.1 s$ .

Task

A cyclist speeds up from $\frac{20 m}{s}$ to $\frac{26 m}{s}$ in 3 s. Find a.

Solution path

$a =\frac{ (26 - 20)}{3} =\frac{ 2 m}{s^{2}}$ .

Getting an intuitive picture

Prerequisite: a positive axis defines the meanings of plus and minus. The sign describes direction; the magnitude describes speed.
A change is final value minus initial value. Time in the equation means elapsed time since that initial instant.
Initially you need only subtraction, multiplication and division. The later integral expression explains varying acceleration.

What is the velocity–time law?

v(t)=v_0+at

v₀ is velocity at t = 0; a must be constant throughout the interval. Speed is the magnitude |v(t)|.

From acceleration to the velocity–time equation

\begin{aligned}a&=\frac{v-v_0}{t}\\at&=v-v_0\\v&=v_0+at\end{aligned}

First multiply the rate by time, then add the initial value. The quotient requires t ≠ 0; the resulting equation also holds at t = 0.[3]

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

The integral form also applies to varying acceleration.

Velocity–time graph, rearrangements and acceleration units

\begin{aligned}[a]&=\mathrm{m\,s^{-2}}\\{}[at]&=\mathrm{m\,s^{-1}}\\a&=\frac{v-v_0}{t}\\t&=\frac{v-v_0}{a}\end{aligned}

The last rearrangement requires a ≠ 0. A calculated time must lie within the modelled motion interval.

Velocity and speed during reversal

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Signed velocity rises uniformly; its magnitude initially falls and rises after reversal. Acceleration is +2 m/s² throughout.[3][2]

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Example: slowing down before speeding up in reverse

A cart with v₀ < 0 and a > 0

Given: v₀ = −6 m/s, a = +2 m/s², x₀ = 0 m, duration 5 s.
Find: Find final velocity, turn time, displacement and distance.

Approach: use signed velocity. Reversal occurs where it becomes zero; displacement is signed area under the velocity graph.
$\begin{aligned}v(t)&=v_0+at\\t_*&=-\frac{v_0}{a}\\s(t)&=v_0t+\frac12at^2\end{aligned}$
Substitution: use the specified initial values with their signs.
$v(5)=-6+2\cdot5=4\,\mathrm{m/s}$
The zero separates the two directions of motion.
$t_*=-\frac{v_0}{a}=3\,\mathrm s$
Evaluate the negative triangle to 3 s and positive triangle to 5 s separately.
$\begin{aligned}s_-&=-\tfrac12\cdot3\cdot6=-9\,\mathrm m\\{} s_+&=\tfrac12\cdot2\cdot4=4\,\mathrm m\end{aligned}$
Add signed values for displacement and magnitudes for distance.
$\begin{aligned}s&=-9+4=-5\,\mathrm m\\{} L&=9+4=13\,\mathrm m\end{aligned}$

Result: At 5 s: v = +4 m/s, turn at 3 s, displacement −5 m and distance 13 m.

Check: The position equation independently gives −6·5 + ½·2·5² = −5 m. Average velocity is −1 m/s; average speed is 2.6 m/s.

Example: acceleration from two measurements

Uniform deceleration

Given: Initial velocity +18 m/s, velocity after 3 s +6 m/s; constant a is assumed.
Find: Find a and predict v at 4 s assuming the same constant a continues.

Approach: the measurements initially determine an average. Using it to predict later velocity additionally requires constant acceleration.
$\begin{aligned}\Delta v&=v_f-v_i\\a&=\frac{\Delta v}{\Delta t}\\v(t)&=v_i+at\end{aligned}$
Substitution: use the specified initial values with their signs.
$\Delta v=6-18=-12\,\mathrm{m/s}$
Divide the change by elapsed time.
$a=-\frac{12}{3}=-4\,\mathrm{m/s^2}$
Use the value only while constant deceleration continues.
$v(4)=18-4\cdot4=2\,\mathrm{m/s}$

Result: a = −4 m/s²; continued unchanged acceleration gives v = +2 m/s at 4 s. Rest would occur at 4.5 s.

Check: At t = 3 s the line reproduces +6 m/s. Two measurements alone do not establish constant acceleration between them; they establish the average.

Explore further: when acceleration changes with time

\Delta v=\int_0^t b\tau\,\mathrm d\tau=\frac12bt^2

Linearly growing acceleration gives a triangular area in an acceleration–time plot.

From one-dimensional motion to vectors

\vec v(t)=\vec v_0+\vec a\,t

The vector form requires constant vector acceleration, not merely a constant magnitude.

Five exercises for reliable application

Calculate · Foundation

v₀ = −3 m/s and a = −2 m/s². Find velocity and speed after 2 s. Is it speeding up?

Show hint

Give the signed value and its magnitude separately.

Learning goal: Separate signed velocity from speed change.

Show solution and calculation

Approach: constant acceleration gives v = v₀ + at.
After two seconds v = −7 m/s.
Speed increases from 3 to 7 m/s; the body speeds up to the left.

Check: each second adds −2 m/s to signed velocity. Negative acceleration therefore does not mean slowing down here.

\begin{aligned}v&=v_0+at\\&=-3\,\mathrm{m/s}+(-2\,\mathrm{m/s^2})(2\,\mathrm s)\\&=-7\,\mathrm{m/s}\\|v|&=7\,\mathrm{m/s}\end{aligned}

[2][3]

Rearrange · Foundation

A cart travels at 36 km/h and accelerates constantly at 1.5 m/s². When does it reach 54 km/h?

Show hint

Convert both velocities before taking their difference.

Learning goal: Find acceleration time after velocity conversion.

Show solution and calculation

Approach: time is required for a velocity change, so t = (v − v₀)/a.
The SI velocities are 10 and 15 m/s.
Their difference is 5 m/s, not 15 m/s.
Time is about 3.33 s.

Check: acceleration produces precisely the missing 5 m/s in this interval.

\begin{aligned}v_0&=\frac{36}{3.6}\,\mathrm{m/s}=10\,\mathrm{m/s}\\v&=\frac{54}{3.6}\,\mathrm{m/s}=15\,\mathrm{m/s}\\t&=\frac{v-v_0}{a}=\frac{15-10}{1.5}\,\mathrm s\\&=\frac{10}{3}\,\mathrm s\approx3.33\,\mathrm s\end{aligned}

[2][3]

Transfer · Application

v₀ = 0 and a(t) = (0.6 m/s³)·t. Find v after 4 s.

Show hint

Integrate a(t) rather than multiplying its final value by t.

Learning goal: Integrate time-dependent acceleration.

Show solution and calculation

Approach: velocity change is area under the acceleration graph.
Acceleration rises linearly from zero, so the area is a triangle.
At four seconds a reaches 2.4 m/s².
A triangle with base 4 s and height 2.4 m/s² gives 4.8 m/s.

Check: the rectangle using the final value would be twice as large and would overestimate acceleration at early times.

\begin{aligned}a(4\,\mathrm s)&=2.4\,\mathrm{m/s^2}\\v(4\,\mathrm s)-v_0&=\frac12(4\,\mathrm s)(2.4\,\mathrm{m/s^2})\\&=4.8\,\mathrm{m/s}\end{aligned}

[2][3]

Interpret · Application

A velocity-time plot is a straight line from v=−4 m/s at t=0 to v=+4 m/s at t=4 s. Read acceleration and reversal time. During which intervals does speed decrease or increase?

Show hint

Slope gives a; speed is the magnitude of v.

Learning goal: Identify reversal in a constant-acceleration velocity-time plot.

Show solution and calculation

Approach: slope is 2 m/s², giving v=−4+2t in SI units.
Velocity vanishes at 2 s.
Before then |v| decreases; afterward it increases, while a stays positive.

Check: at 4 s velocity is +4 m/s. Positive acceleration does not imply speeding up throughout. Speed has a cusp in its magnitude plot at reversal, while signed velocity stays smooth.

\begin{aligned}a&=\frac{4-(-4)}4\,\mathrm{m/s^2}=2\,\mathrm{m/s^2}\\t_{\mathrm{turn}}&=\frac42\,\mathrm s=2\,\mathrm s\end{aligned}

[1][2][3]

Error analysis · Application

A body moves uniformly around a circle of radius 2 m at speed 3 m/s. Acceleration magnitude is 4.5 m/s². A solution uses v=3+4.5t and claims speed 7.5 m/s after 1 s. Why is the model wrong?

Show hint

The acceleration vector points inward and rotates.

Learning goal: Avoid treating circular-motion acceleration as a constant vector.

Show solution and calculation

Approach: uniform circular motion changes velocity direction, not magnitude.
After 1 s speed remains 3 m/s.
Centripetal magnitude 4.5 m/s² is not a constant tangential component on a fixed line.

Check: v·a=0, so $\frac{d(v^2)}{dt}=0$ . The constant-a expression applies componentwise only for a genuinely constant acceleration vector.

\begin{aligned}\frac{\mathrm d}{\mathrm dt}|\mathbf v|^2&=2\mathbf v\cdot\mathbf a=0\\|\mathbf v(1\,\mathrm s)|&=3\,\mathrm{m/s}\end{aligned}

[2]

What to take away

Velocity needs an axis; speed is its magnitude. Acceleration describes change in signed velocity.
The short equation requires constant a. Two measured points alone do not establish that assumption.
Set v = 0 to find reversal. To decide whether speed increases, compare the directions of v and a rather than the sign of a alone.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

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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, 2026Answered the velocity–time law directly with constant acceleration and elapsed time; made units, rearrangements and varying acceleration easier to find. Retained reversal and v·a reasoning, and linked displacement, free fall and centripetal force.
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.

Cite this reference page

Quanta. Velocity-Time Law. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/geschwindigkeit-beschleunigung

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Frequently asked questions about Velocity-Time Law

How do you calculate the final speed with

v = a\cdot{}t + v_{0}

What does an acceleration of

\frac{1 m}{s^{2}}

mean intuitively?+

How do you rearrange

v = a\cdot{}t + v_{0}

for the time?+

What is the difference between velocity and acceleration?+

How do you read velocity and acceleration from diagrams?+

Velocity–time law

Variables & units

Final velocity calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Laws of motion

Worked example

Applications

How do you calculate with Velocity-Time Law?

Getting an intuitive picture

What is the velocity–time law?

From acceleration to the velocity–time equation

Velocity–time graph, rearrangements and acceleration units

Velocity and speed during reversal

Example: slowing down before speeding up in reverse

A cart with v0 < 0 and a > 0

Example: acceleration from two measurements

Uniform deceleration

Explore further: when acceleration changes with time

From one-dimensional motion to vectors

Five exercises for reliable application

v0 = −3 m/s and a = −2 m/s2. Find velocity and speed after 2 s. Is it speeding up?

A cart travels at 36 km/h and accelerates constantly at 1.5 m/s2. When does it reach 54 km/h?

v0 = 0 and a(t) = (0.6 m/s3)·t. Find v after 4 s.

A velocity-time plot is a straight line from v=−4 m/s at t=0 to v=+4 m/s at t=4 s. Read acceleration and reversal time. During which intervals does speed decrease or increase?

A body moves uniformly around a circle of radius 2 m at speed 3 m/s. Acceleration magnitude is 4.5 m/s2. A solution uses v=3+4.5t and claims speed 7.5 m/s after 1 s. Why is the model wrong?

What to take away

Sources for this explanation

Frequently asked questions about Velocity-Time Law

Subject references and further reading

More formulas: Mechanics

Retain Velocity-Time Law for exams

Velocity–time law

Variables & units

Final velocity calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Laws of motion

Worked example

Applications

How do you calculate with Velocity-Time Law?

Getting an intuitive picture

What is the velocity–time law?

From acceleration to the velocity–time equation

Velocity–time graph, rearrangements and acceleration units

Velocity and speed during reversal

Example: slowing down before speeding up in reverse

A cart with v0 < 0 and a > 0

Example: acceleration from two measurements

Uniform deceleration

Explore further: when acceleration changes with time

From one-dimensional motion to vectors

Five exercises for reliable application

v0 = −3 m/s and a = −2 m/s2. Find velocity and speed after 2 s. Is it speeding up?

A cart travels at 36 km/h and accelerates constantly at 1.5 m/s2. When does it reach 54 km/h?

v0 = 0 and a(t) = (0.6 m/s3)·t. Find v after 4 s.

A velocity-time plot is a straight line from v=−4 m/s at t=0 to v=+4 m/s at t=4 s. Read acceleration and reversal time. During which intervals does speed decrease or increase?

A body moves uniformly around a circle of radius 2 m at speed 3 m/s. Acceleration magnitude is 4.5 m/s2. A solution uses v=3+4.5t and claims speed 7.5 m/s after 1 s. Why is the model wrong?

What to take away

Sources for this explanation

Frequently asked questions about Velocity-Time Law

Subject references and further reading

More formulas: Mechanics

Retain Velocity-Time Law for exams

A cart with v₀ < 0 and a > 0

v₀ = −3 m/s and a = −2 m/s². Find velocity and speed after 2 s. Is it speeding up?

A cart travels at 36 km/h and accelerates constantly at 1.5 m/s². When does it reach 54 km/h?

v₀ = 0 and a(t) = (0.6 m/s³)·t. Find v after 4 s.

A body moves uniformly around a circle of radius 2 m at speed 3 m/s. Acceleration magnitude is 4.5 m/s². A solution uses v=3+4.5t and claims speed 7.5 m/s after 1 s. Why is the model wrong?

A cart with v₀ < 0 and a > 0

v₀ = −3 m/s and a = −2 m/s². Find velocity and speed after 2 s. Is it speeding up?

A cart travels at 36 km/h and accelerates constantly at 1.5 m/s². When does it reach 54 km/h?

v₀ = 0 and a(t) = (0.6 m/s³)·t. Find v after 4 s.

A body moves uniformly around a circle of radius 2 m at speed 3 m/s. Acceleration magnitude is 4.5 m/s². A solution uses v=3+4.5t and claims speed 7.5 m/s after 1 s. Why is the model wrong?