Physics · Kinematics

Velocity-Time Law

Q: How do you calculate the final speed with v = a·t + v₀?

Multiply the acceleration by the time and add the initial velocity. Example: a train travels at v₀ = 10 m/s and accelerates for 8 s at a = 1.5 m/s²: v = 1.5 × 8 + 10 = 22 m/s, which is about 79 km/h. All quantities belong in SI units; divide a speed in km/h by 3.6 before substituting. Intuitively the formula says: the acceleration states how many m/s the body gains per second. At a = 1.5 m/s² it thus gets 1.5 m/s faster every second. When braking, insert a as negative and v decreases accordingly.

Q: What does an acceleration of 1 m/s² mean intuitively?

It means the velocity increases by 1 m/s every second; after three seconds the body is 3 m/s faster. The double unit m/s² is best read as "(m/s) per s". For scale: a brisk small car manages about 3 m/s² when pulling away, emergency braking is 8 to 10 m/s², free fall is 9.81 m/s², and a Formula 1 car brakes at up to 50 m/s² (about 5g). Acceleration is not limited to speeding up: physically every change of velocity is an acceleration, including braking (negative sign) and even pure changes of direction in circular motion.

Q: How do you rearrange v = a·t + v₀ for the time?

First subtract v₀ on both sides, then divide by a: t = (v − v₀)/a. The numerator is the change in velocity, not the final velocity, which is the most common mistake. Example: a car accelerates at a = 2.5 m/s² from 0 to 100 km/h (27.8 m/s): t = (27.8 − 0)/2.5 ≈ 11.1 s. Slowing from 26 m/s to 20 m/s at a = −2 m/s²: t = (20 − 26)/(−2) = 3 s; two negative signs cancel and the time is positive as expected. If a negative time comes out, signs or start and end values are swapped. A unit check (m/s divided by m/s² gives s) secures the result.

Q: What is the difference between velocity and acceleration?

Velocity says how fast the position changes (m/s); acceleration says how fast the velocity changes (m/s²). Both can independently be large, small or zero: an aircraft in cruise is very fast but unaccelerated (a = 0). A ball at the top of a vertical throw momentarily has v = 0 yet is fully accelerated at g = 9.81 m/s², otherwise it would stay up there. The directions need not agree either: when braking, a points against the motion. This distinction is the core of many conceptual questions; anyone who equates v and a fails exactly these tasks.

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

Remember two rules: slope and area. In the s-t diagram the slope is the velocity; a steeper curve means faster, a parabola indicates acceleration. In the v-t diagram the slope is the acceleration: a line with positive slope means constant a, a horizontal line uniform motion. Additionally, the area under the v-t curve is the distance travelled; for the trapezoid made of the v₀ rectangle and the acceleration triangle this gives exactly s = v₀t + ½at². Exam problems like to combine segments: first accelerate, then drive at constant speed, then brake. Split the diagram into these phases and compute the areas separately.

The velocity-time law describes how the velocity grows linearly with time under constant acceleration.

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Formula

v = a \cdot t + v_{0}

LaTeX: v = a \cdot t + v_0

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

The velocity grows linearly with time; the slope of the line is the acceleration a, the intercept is v₀.

Variables & units – Velocity-Time Law

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

Derivation & background – Velocity-Time Law

Acceleration is defined as the change of velocity per time: a = Δv/Δt. If a is constant, v grows linearly, a straight line with slope a in the v-t diagram. The area under the line is the distance travelled, which is how the v-t law and the distance-time law are connected. Braking is the case a < 0.

Exam blueprint

Validity range

Holds for constant acceleration along a straight line. Braking is described with negative a. For varying a the relation only holds differentially: a = dv/dt.

Derivation steps

Constant acceleration means the velocity changes by the same amount every second.

1Definition: a = Δv/Δt = (v − v₀)/t.
2Solve for v: v = a·t + v₀.

Rearrangements

Time to reach a target speed

t = \frac{v - v _{0}}{a}

The numerator is the change in speed, not the final speed.

Acceleration from two speeds

a = \frac{v - v _{0}}{t}

A negative value means braking (deceleration).

Task variant

A car accelerates at a = 2.5 m/s² from 0 to 100 km/h. How long does it take?

100 km/h = 27.8 m/s. t = (27.8 − 0)/2.5 ≈ 11.1 s.

A cyclist speeds up from 20 m/s to 26 m/s in 3 s. Find a.

a = (26 − 20)/3 = 2 m/s².

Common mistakes

Substituting km/h directly.

Always convert to m/s by dividing by 3.6 first.

Dropping v₀ and writing v = a·t although the body is already moving.

v = a·t only holds for a start from rest.

Ignoring the sign of a when braking.

Insert deceleration as a < 0, otherwise the body speeds up in the calculation.

Exam context

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

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Formula cluster

Laws of motion

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

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

Uniformly Accelerated Motion (Distance-Time Law)

h = \frac{1}{2} g t^{2}

Free Fall

p = m \cdot v

Momentum

Worked example

A train moves at v₀ = 10 m/s and accelerates with a = 1.5 m/s² for t = 8 s: v = 1.5 × 8 + 10 = 22 m/s (79 km/h).

Applications

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

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Which formula describes Velocity-Time Law?

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How do you rearrange v = a·t + v₀ for Time to reach a target speed?

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Which common mistake happens with Velocity-Time Law?

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Scientific sources

[1]Tipler, P.A., & Mosca, G. (2015). Physik für Wissenschaftler und Ingenieure. Springer Spektrum.

Common notations & search queries

v=a*tv=at+v0v = a·t + v₀Geschwindigkeit-Zeit-GesetzEndgeschwindigkeit berechnenBeschleunigung Formel Zeitvelocity time formulaa=v/t umstellen

Related formulas

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

Uniformly Accelerated Motion (Distance-Time Law)

h = \frac{1}{2} g t^{2}

Free Fall

p = m \cdot v

Momentum

More Physics formulas

F = m \cdot 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_{p o t} = m \cdot g \cdot h

Potential Energy (Gravitational)

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Frequently asked questions about Velocity-Time Law

How do you calculate the final speed with v = a·t + v₀?+

Multiply the acceleration by the time and add the initial velocity. Example: a train travels at v₀ = 10 m/s and accelerates for 8 s at a = 1.5 m/s²: v = 1.5 × 8 + 10 = 22 m/s, which is about 79 km/h. All quantities belong in SI units; divide a speed in km/h by 3.6 before substituting. Intuitively the formula says: the acceleration states how many m/s the body gains per second. At a = 1.5 m/s² it thus gets 1.5 m/s faster every second. When braking, insert a as negative and v decreases accordingly.

What does an acceleration of 1 m/s² mean intuitively?+

It means the velocity increases by 1 m/s every second; after three seconds the body is 3 m/s faster. The double unit m/s² is best read as "(m/s) per s". For scale: a brisk small car manages about 3 m/s² when pulling away, emergency braking is 8 to 10 m/s², free fall is 9.81 m/s², and a Formula 1 car brakes at up to 50 m/s² (about 5g). Acceleration is not limited to speeding up: physically every change of velocity is an acceleration, including braking (negative sign) and even pure changes of direction in circular motion.

How do you rearrange v = a·t + v₀ for the time?+

First subtract v₀ on both sides, then divide by a: t = (v − v₀)/a. The numerator is the change in velocity, not the final velocity, which is the most common mistake. Example: a car accelerates at a = 2.5 m/s² from 0 to 100 km/h (27.8 m/s): t = (27.8 − 0)/2.5 ≈ 11.1 s. Slowing from 26 m/s to 20 m/s at a = −2 m/s²: t = (20 − 26)/(−2) = 3 s; two negative signs cancel and the time is positive as expected. If a negative time comes out, signs or start and end values are swapped. A unit check (m/s divided by m/s² gives s) secures the result.

What is the difference between velocity and acceleration?+

Velocity says how fast the position changes (m/s); acceleration says how fast the velocity changes (m/s²). Both can independently be large, small or zero: an aircraft in cruise is very fast but unaccelerated (a = 0). A ball at the top of a vertical throw momentarily has v = 0 yet is fully accelerated at g = 9.81 m/s², otherwise it would stay up there. The directions need not agree either: when braking, a points against the motion. This distinction is the core of many conceptual questions; anyone who equates v and a fails exactly these tasks.

How do you read velocity and acceleration from diagrams?+

Remember two rules: slope and area. In the s-t diagram the slope is the velocity; a steeper curve means faster, a parabola indicates acceleration. In the v-t diagram the slope is the acceleration: a line with positive slope means constant a, a horizontal line uniform motion. Additionally, the area under the v-t curve is the distance travelled; for the trapezoid made of the v₀ rectangle and the acceleration triangle this gives exactly s = v₀t + ½at². Exam problems like to combine segments: first accelerate, then drive at constant speed, then brake. Split the diagram into these phases and compute the areas separately.

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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 = 2.5 m/s² from 0 to 100 km/h. How long does it take?
Solution path
100 km/h = 27.8 m/s. t = (27.8 − 0)/2.5 ≈ 11.1 s.
2
Task
A cyclist speeds up from 20 m/s to 26 m/s in 3 s. Find a.
Solution path
a = (26 − 20)/3 = 2 m/s².