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Equations of Motion (Constant Acceleration)

Velocity as a function of time under constant acceleration.

PhysicsKinematicsFundamental

Equations of Motion Calculator v = u + a·t

v = u + a · t
v = final velocity  ·  u = initial velocity  ·  a = acceleration  ·  t = time
⟹ Solve v, u, a, t
m/s
m/s
m/s²
s
Please fix the errors above.
Solve for:
Presets:
Final Velocity (v)
v: u: a: t:
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Velocity Magnitude
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v = u + a·t  ·  The first equation of motion for constant acceleration. Valid for linear motion with uniform acceleration.

Interpretation

Equations of motion for constant acceleration: v = u + at, s = ut + ½at², v² = u² + 2as. These relate velocity, displacement, acceleration, and time. Example: u=0, a=2 m/s², t=5s → v=10 m/s.

v = u + at
Equations of Motion (Constant Acceleration)

Variables

SymbolQuantityUnit
vFinal velocitym/s
uInitial velocitym/s
aAccelerationm/s²
tTimes

What it means

The three equations of motion describe the relationship between displacement, initial velocity, final velocity, acceleration, and time for an object moving with constant acceleration. They are derived from the definitions of acceleration and average velocity. The first equation v = u + at gives the final velocity after time t; the second s = ut + ½at² gives the displacement; and the third v² = u² + 2as relates velocity and displacement without involving time. These equations are fundamental in kinematics and are used extensively in physics and engineering to analyse projectile motion, vehicle acceleration, and any uniformly accelerated motion. They assume constant acceleration and straight‑line motion. In practice, they are applied in designing braking systems, predicting motion of falling objects, and calculating trajectories. Understanding these equations is essential for solving a wide range of problems in mechanics.

Worked example

Equations of Motion – Two Examples

Real‑World
Scenario: A car starts from rest (u = 0) and accelerates at 2 m/s² for 5 seconds. Find its final velocity.
ParameterValue
u0 m/s
a2 m/s²
t5 s
1v = u + at = 0 + 2×5 = 10 m/s
Result v = 10 m/s ✓ 36 km/h
Scenario: A train moving at 5 m/s accelerates at 1.5 m/s² for 4 seconds. Find its final velocity.
ParameterValue
u5 m/s
a1.5 m/s²
t4 s
1v = 5 + 1.5×4 = 5 + 6 = 11 m/s
Result v = 11 m/s ✓ 39.6 km/h
Key insight: Final velocity = initial velocity + acceleration × time – the foundation of kinematics.

Common mistakes

  • Sign convention: v = u + at uses a sign convention for direction. Define positive direction consistently (e.g., upwards positive, downwards negative).
  • Constant acceleration: This equation assumes constant acceleration. Do not use it for varying acceleration unless using calculus.
  • Unit consistency: v and u in m/s, a in m/s², t in s – ensure all SI units.
  • Initial velocity u: It is the velocity at t=0; if the object starts from rest, u=0.
  • Final velocity v: The velocity after time t – not the displacement.

Applications

The equations of motion under constant acceleration – v = u + at, s = ut + ½at², and v² = u² + 2as – are the foundation of classical kinematics. They describe the relationship between displacement, velocity, acceleration, and time for objects moving with uniform acceleration. These equations are used extensively in engineering and physics to predict the motion of vehicles, projectiles, and machinery. In automotive design, they help calculate stopping distances and acceleration times. In sports science, they model the trajectory of balls and athletes. In aerospace, they are used for launch and landing trajectories. By applying these equations, engineers can design safety systems, optimise performance, and analyse dynamic systems. Their simplicity and power make them essential tools in both education and professional practice across all fields involving motion.

  • Vehicle braking distance and acceleration calculations
  • Projectile motion analysis in ballistics and sports
  • Design of amusement park rides and elevators
  • Launch and landing trajectory planning in aerospace
  • Educational foundation for introductory physics and engineering

Frequently Asked Questions

Q01What are the equations of motion for constant acceleration and when are they used?
A01

For constant acceleration, we have four kinematic equations:
1) v = u + at (velocity after time t).
2) s = ut + ½at² (displacement after time t).
3) v² = u² + 2as (velocity‑displacement relation).
4) s = ½(u+v)t (displacement from average velocity).
They apply to any motion with constant acceleration (including free‑fall, braking, and projectile motion).

Q02What do u, v, a, t, and s represent and what are their units?
A02

  • u – initial velocity (m/s).
  • v – final velocity (m/s).
  • a – constant acceleration (m/s²).
  • t – time interval (s).
  • s – displacement (m).
All are vector quantities; their signs depend on the chosen coordinate system.

Q03What is the sign convention for acceleration and how do you apply it?
A03

Choose a positive direction (e.g., upward or right). Any vector pointing in that direction is positive; opposite is negative. For example, if upward is positive, the acceleration due to gravity is a = –g (≈ –9.81 m/s²). Consistent sign use is essential to get correct answers.

Q04How do these equations apply to projectile motion?
A04

Projectile motion has constant horizontal velocity and constant vertical acceleration (g downward). Horizontally: use vx = ux and sx = ux·t. Vertically: use the full set with a = –g. The two motions are independent and are linked by time.

Q05What is the difference between average and instantaneous acceleration?
A05

Average acceleration is the change in velocity divided by the time interval: a_avg = Δv/Δt. Instantaneous acceleration is the derivative a = dv/dt at a specific instant. The equations of motion assume constant (uniform) acceleration, so average and instantaneous are equal.

Q06What happens when acceleration is not constant?
A06

For variable acceleration, you cannot use these equations directly. You must integrate: v = ∫a·dt and s = ∫v·dt. If a is given as a function of time, velocity, or position, use differential equations. Numerically, you can also integrate using step‑by‑step methods.

Q07How do you choose which equation to use for a given problem?
A07

List the known and unknown variables. Choose the equation that contains the unknown and only known quantities. For example, if you have u, v, a and want s, use v² = u² + 2as (time not needed). If you have u, a, t and want s, use s = ut + ½at².

Q08What are the common mistakes when applying these equations?
A08

  • Using the wrong sign for acceleration.
  • Forgetting that displacement, velocity, and acceleration are vectors – using magnitudes incorrectly.
  • Mixing units (e.g., using km/h and m/s together).
  • Assuming constant acceleration when it is not.
  • Applying the equations to variable acceleration without integration.

Q09How are these equations derived from calculus?
A09

They follow from the definitions: a = dv/dt → integrate to get v = u + at. Then v = ds/dt → integrate to get s = ut + ½at². The third is obtained by eliminating t: v² = u² + 2as.

Q10What are some real‑world engineering applications of these equations?
A10

  • Vehicle braking distance calculations.
  • Rocket trajectory analysis.
  • Roller coaster design.
  • Sports physics (ballistics).
  • Elevator acceleration profiles.