Kinematic Equations
Complete reference for all 5 kinematic equations — formulas, proofs, worked examples, and an interactive solver. The definitive guide to 1D kinematics equations and equations of motion.
The 4 Kinematic Equations — Complete Reference
The term "the 4 kinematic equations" is commonly used in US high school physics, but depending on your textbook, you may see 3, 4, or 5 equations. How many kinematic equations are there? The answer is: 3, 4, or 5 depending on which textbook you use. The core three are always included. A fourth (average-velocity form) is added in most courses, and a fifth (alternate displacement form) appears in British SUVAT and some university texts. This page covers all five.
These are the basic equations of kinematics for constant acceleration in one dimension (1D kinematics). They describe linear movement equations — also called equations of motion — relating displacement, initial velocity, final velocity, acceleration, and time.
Variable Definitions
| Symbol (US) | Symbol (SUVAT/UK) | Quantity | SI Unit | Notes |
|---|---|---|---|---|
| vi | u | Initial velocity | m/s | Velocity at t = 0 |
| vf | v | Final velocity | m/s | Velocity at time t |
| a | a | Acceleration | m/s² | Must be constant |
| t | t | Time | s | Duration of motion |
| d | s | Displacement | m | Change in position (vector) |
SUVAT Notation (British / A-Level)
In British A-Level physics and many university courses, the kinematic equations are written using SUVAT notation where s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time. The five SUVAT equations of motion are:
| # | SUVAT Form | US Form | Missing Variable |
|---|---|---|---|
| 1 | v = u + at | vf = vi + at | s (displacement) |
| 2 | s = ut + ½at² | d = vit + ½at² | v (final velocity) |
| 3 | v² = u² + 2as | vf² = vi² + 2ad | t (time) — time-independent |
| 4 | s = (u+v)t / 2 | d = (vi+vf)t / 2 | a (acceleration) |
| 5 | s = vt − ½at² | d = vft − ½at² | vi (initial velocity) |
Which Kinematic Equation to Use — Variable Selection Guide
The most common question students ask is: "Which kinematic equation do I use?" The answer is simple: find the variable that is missing from your problem (the unknown you want to solve for), then look up that row in the table below.
| Situation / Unknown | vi | vf | a | t | d | Use This Equation | Formula |
|---|---|---|---|---|---|---|---|
| Find vf — time known, no displacement | ✓ | ✗ | ✓ | ✓ | ✗ | 1st Equation | vf = vi + at |
| Find d — time known, no final velocity | ✓ | ✗ | ✓ | ✓ | ✗ | 2nd Equation | d = vit + ½at² |
| Find vf or d — time unknown | ✓ | ✗ | ✓ | ✗ | ✓ | 3rd Equation ⭐ | vf² = vi² + 2ad |
| Find d — both velocities known, no acceleration | ✓ | ✓ | ✗ | ✓ | ✗ | 4th Equation | d = (vi+vf)t / 2 |
| Find d — initial velocity unknown | ✗ | ✓ | ✓ | ✓ | ✗ | 5th Equation | d = vft − ½at² |
| Find t — no time given (time-independent case) | ✓ | ✓ | ✓ | ✗ | ✓ | 3rd Equation then rearrange | vf² = vi² + 2ad |
1st Kinematic Equation — vf = vi + at
vi = vf − at (solve for initial velocity)
a = (vf − vi) / t (solve for acceleration)
t = (vf − vi) / a (solve for time)
The 1st kinematic equation, vf = vi + at, is the most fundamental of all the kinematic equations. It comes directly from the definition of acceleration: acceleration is the rate of change of velocity. If an object starts with initial velocity vi and accelerates at a constant rate a for time t, its final velocity is simply vi plus the velocity gained (at).
The equation vf = vi + at appears in virtually every physics problem. In SUVAT notation this is written as v = u + at.
When to Use the 1st Kinematic Equation
Use vf = vi + at when:
- You know initial velocity, acceleration, and time — and want final velocity
- You know initial and final velocity plus time — and want acceleration
- Displacement is not needed in the calculation
Worked Example 1 — Using vf = vi + at
Problem: A car starts from rest and accelerates at 4 m/s² for 6 seconds. What is its final velocity?
Given: a = 4 m/s²
Given: t = 6 s
Find: vf = ?
vf = 0 + (4)(6)
vf = 0 + 24
✓ vf = 24 m/s
Problem: A cyclist slows from 12 m/s to 3 m/s in 3 seconds. What is the acceleration?
Given: vf = 3 m/s
Given: t = 3 s
Find: a = ?
a = (vf − vi) / t
a = (3 − 12) / 3
a = −9 / 3
✓ a = −3 m/s² (deceleration)
In plain visible HTML: vf = vi + at is the 1st kinematic equation. Remember: vf = vi + at gives you final velocity directly when you have acceleration and time. The equation vf = vi + at works in both directions — acceleration can be positive or negative. Use vf = vi + at for any problem involving velocity change over time.
2nd Kinematic Equation — d = vi×t + ½at²
vi = (d − ½at²) / t (solve for initial velocity)
a = 2(d − vit) / t² (solve for acceleration)
t = solve via quadratic (solve for time)
The 2nd kinematic equation, d = vi×t + ½at², gives displacement as a function of time. It can be understood geometrically as the area under a velocity-time graph: the rectangle vit (displacement at constant velocity) plus the triangle ½at² (extra displacement due to acceleration).
In SUVAT notation: s = ut + ½at². This is one of the most commonly used kinematic equations for distance calculations.
When to Use the 2nd Kinematic Equation
- You know vi, a, and t — and want displacement d
- You know vi, a, and d — and want time t (requires quadratic formula)
- Final velocity is not given and not needed
Worked Example — Using d = vi×t + ½at²
Problem: A ball rolls from rest with acceleration 3 m/s² for 5 seconds. How far does it travel?
Given: a = 3 m/s²
Given: t = 5 s
Find: d = ?
d = (0)(5) + ½(3)(5²)
d = 0 + ½(3)(25)
d = 0 + 37.5
✓ d = 37.5 m
Problem: A train moving at 20 m/s accelerates at 1.5 m/s² for 8 seconds. What distance does it cover?
Given: a = 1.5 m/s²
Given: t = 8 s
Find: d = ?
d = (20)(8) + ½(1.5)(64)
d = 160 + 48
✓ d = 208 m
The kinematic equation for distance d = vi×t + ½at² is the standard go-to formula in projectile problems, free-fall problems, and any 1D kinematics equation scenario where time is known. The equation d = vi×t + ½at² reduces to d = ½at² when starting from rest (vi = 0), which is the classic free-fall formula.
3rd Kinematic Equation — vf² = vi² + 2ad (Time-Independent)
vf = √(vi² + 2ad) (solve for final velocity)
vi = √(vf² − 2ad) (solve for initial velocity)
a = (vf² − vi²) / (2d) (solve for acceleration)
d = (vf² − vi²) / (2a) (solve for displacement)
The 3rd kinematic equation, vf² = vi² + 2ad, is the most powerful of all the kinematic equations. It is the only one that does not involve time — making it the time-independent kinematic equation. Whenever a problem gives you velocities and displacement (or asks for them) without mentioning time, the 3rd kinematic equation is your tool.
In SUVAT notation: v² = u² + 2as. The 3rd kinematic equation is derived by eliminating time from the first two equations. It represents a connection between kinetic energy and work done by a force — multiply both sides by ½m and you get the work-energy theorem.
The 3rd Kinematic Equation — When to Use It
Use the 3rd kinematic equation vf² = vi² + 2ad when:
- Time is not given and not needed — this is the defining situation for the time-independent equation
- You know two velocities and want to find distance
- You know initial velocity, distance, and acceleration — and want final velocity
- You want to find minimum stopping distance given deceleration
- The problem involves projectile maximum height (where vf = 0 at the peak)
The 3rd Kinematic Equation — Worked Examples
Problem: A rocket starts from rest and accelerates at 40 m/s² over a distance of 500 m. What is its final velocity? (Time not given — use the time-independent equation.)
Given: a = 40 m/s²
Given: d = 500 m
Find: vf = ?
vf² = (0)² + 2(40)(500)
vf² = 0 + 40,000
vf = √40,000
✓ vf = 200 m/s
Problem: A car travelling at 30 m/s applies its brakes and decelerates at 7.5 m/s². What is the stopping distance?
Given: vf = 0 m/s (stopped)
Given: a = −7.5 m/s²
Find: d = ?
d = (vf² − vi²) / (2a)
d = (0² − 30²) / (2 × −7.5)
d = (−900) / (−15)
✓ d = 60 m (stopping distance)
The Time-Independent Nature of the 3rd Kinematic Equation
The 3rd kinematic equation is called time-independent because it contains only four variables: vf, vi, a, and d — with no t. The time-independent equation vf² = vi² + 2ad is particularly useful in:
- Ballistics — finding bullet velocity after traveling through a medium
- Roller coasters — velocity at the bottom of a ramp
- Safety engineering — minimum braking distances
- Space launches — final velocity after a given engine burn distance
The time-independent equation vf² = vi² + 2ad is also the kinematic version of the work-energy theorem. This connection makes the 3rd kinematic equation one of the most important formulas in all of classical mechanics.
4th Kinematic Equation — d = (vi + vf)t / 2
vi = 2d/t − vf (solve for initial velocity)
vf = 2d/t − vi (solve for final velocity)
t = 2d / (vi + vf) (solve for time)
The 4th kinematic equation, d = (vi + vf)t / 2, calculates displacement using the average velocity. Under constant acceleration, the velocity changes linearly, so the average velocity is simply the arithmetic mean of the initial and final velocities. Multiplying this average velocity by time gives displacement.
In SUVAT notation: s = (u + v)t / 2. The 4th kinematic equation is particularly elegant because it doesn't require knowledge of acceleration. If you know the starting and ending velocities and the time taken, you can find displacement directly.
When to Use the 4th Kinematic Equation
Use the 4th kinematic equation when:
- Both vi and vf are known
- Acceleration is not given (and not needed)
- You want to find displacement or time
- The 4th kinematic equation is the fastest method when acceleration is missing from the problem
4th Kinematic Equation — Worked Examples
Problem: A car accelerates from 10 m/s to 30 m/s in 4 seconds. What distance does it cover? (Acceleration not given — use the 4th kinematic equation.)
Given: vf = 30 m/s
Given: t = 4 s
Find: d = ?
d = (10 + 30) × 4 / 2
d = 40 × 4 / 2
d = 160 / 2
✓ d = 80 m
Problem: An aircraft accelerates from 60 m/s to 90 m/s while traveling 600 m down a runway. How long does this take?
Given: vf = 90 m/s
Given: d = 600 m
Find: t = ?
t = 2d / (vi + vf)
t = 2(600) / (60 + 90)
t = 1200 / 150
✓ t = 8 s
The 4th kinematic equation d = (vi + vf)t / 2 is also derived from the area of a trapezoid on a velocity-time graph. The two parallel sides are vi and vf, and the height is t. The 4th kinematic equation elegantly bypasses acceleration when both velocities are known.
5th Kinematic Equation — d = vf×t − ½at²
vf = (d + ½at²) / t (solve for final velocity)
a = 2(vft − d) / t² (solve for acceleration)
t = solve via quadratic (solve for time)
The 5th kinematic equation, d = vf×t − ½at², is the mirror of the 2nd equation — it uses final velocity instead of initial velocity. It is most useful when initial velocity is unknown but final velocity is known. In SUVAT notation: s = vt − ½at².
When to Use the 5th Kinematic Equation
- Initial velocity (vi) is unknown and not needed
- You know final velocity, acceleration, and time
- The problem involves the final portion of a motion
Worked Example — Using d = vf×t − ½at²
Problem: A ball hits the ground at 25 m/s after decelerating at 2 m/s² for 5 seconds. How far did it travel in that interval? (Initial velocity unknown.)
Given: a = −2 m/s² (deceleration)
Given: t = 5 s
Find: d = ?
d = (25)(5) − ½(−2)(5²)
d = 125 − ½(−2)(25)
d = 125 − (−25)
d = 125 + 25
✓ d = 150 m
Problem: An object accelerates at 3 m/s² for 4 seconds, covering 30 m. What was its final velocity?
Given: t = 4 s
Given: d = 30 m
Find: vf = ?
vf = (d + ½at²) / t
vf = (30 + ½(3)(16)) / 4
vf = (30 + 24) / 4
vf = 54 / 4
✓ vf = 13.5 m/s
Proving the Kinematic Equations — Derivations from First Principles
The kinematic equations can be proven (derived) using only the definition of acceleration and basic algebra. Below is how to prove each kinematic equation from first principles. All derivations assume constant acceleration in 1D kinematics.
Proof of the 1st Kinematic Equation: vf = vi + at
Proof of the 2nd Kinematic Equation: d = vi×t + ½at²
Proof of the 3rd Kinematic Equation: vf² = vi² + 2ad (Time-Independent)
Proof of the 4th Kinematic Equation: d = (vi + vf)t / 2
Proof of the 5th Kinematic Equation: d = vf×t − ½at²
When Do Kinematic Equations Apply — Assumptions and Limitations
The kinematic equations are powerful tools for 1D kinematics equations and linear movement equations — but they only apply under specific conditions. Understanding these assumptions is critical for avoiding errors in physics problems.
Constant Acceleration
The most critical assumption. All five kinematic equations require that acceleration does not change during the motion interval. If acceleration varies, use calculus (integration).
Straight-Line (1D) Motion
1D kinematics equations describe motion along a single axis. For 2D or 3D motion, decompose into components and apply the equations separately to each axis.
Point Particle
The object is treated as a point — its size, shape, and rotation are ignored. For rigid body rotation, use rotational kinematics equations instead.
Classical (Non-Relativistic)
Valid only at speeds much less than the speed of light. At speeds approaching c, use special relativistic mechanics. For everyday physics problems, this is never an issue.
Single Time Interval
Apply each equation over a single phase of motion with one constant acceleration. If the motion has multiple phases (e.g., accelerate then brake), solve each phase separately.
Sign Convention
Always define a positive direction first. Velocities and displacements in the negative direction take negative values. Acceleration opposing motion is negative.
Kinematic Equations in 2D — Horizontal and Vertical Components
While the kinematic equations are written for 1D motion, they extend naturally to 2D (and 3D) by applying them independently to each component. In projectile motion — the most common 2D kinematics problem — horizontal and vertical motions are completely independent equations of motion.
⟶ Horizontal Component (x-axis)
Acceleration: ax = 0 (no air resistance)vfx = vix (constant)
x = vix × t (Eq 1 simplified)
vix = v₀ cos(θ) (from launch angle)
⟳ Vertical Component (y-axis)
Acceleration: ay = −g = −9.81 m/s²vfy = viy − gt (1st equation)
y = viyt − ½gt² (2nd equation)
vfy² = viy² − 2gy (3rd equation — time-independent)
viy = v₀ sin(θ) (from launch angle)
Projectile Motion — Key Results from 2D Kinematics
Using the kinematic equations on each component separately yields the standard results for projectile motion:
Notice that finding maximum height uses the time-independent equation vf² = vi² + 2ad (the 3rd kinematic equation) because time is not needed when we know vfy = 0 at the peak. This is a classic application of the 3rd kinematic equation in 2D projectile motion.
Common Mistakes with Kinematic Equations
These are the most frequent errors students make when using physics formulas kinematics and linear movement equations in problem-solving.
❌ Wrong Sign for Acceleration
Taking deceleration as positive when it should be negative (or vice versa), causing incorrect final answers.
❌ Confusing Displacement and Distance
The kinematic equations give displacement (vector — can be negative), not distance (scalar — always positive). If an object reverses direction, displacement ≠ distance.
❌ Using the Wrong Equation
Trying to use vf = vi + at when displacement is needed, or using the time-independent equation when time is available and needed.
❌ Forgetting to Square Root in Eq. 3
The 3rd kinematic equation gives vf² — students often forget to take the square root to find vf itself.
❌ Taking Negative Time
When solving for time using the quadratic formula in the 2nd equation, both a positive and negative root exist. Students sometimes take the negative root.
❌ Mixing Unit Systems
Mixing m/s with km/h, or metres with feet, gives incorrect answers in any kinematic equation calculation.
Worked Examples — One for Each Equation
The following comprehensive worked examples demonstrate how to approach kinematic problems using the correct equation selection process. Each example is self-contained — covering problem setup, variable identification, equation selection, and full step-by-step solution.
Problem: A sprinter starting from rest reaches a velocity of 9.5 m/s after 2.5 seconds of constant acceleration. What acceleration did the sprinter experience? How long would it take to reach 12 m/s?
vi = 0 m/s (starts from rest)
vf = 9.5 m/s
a = ? (unknown, Part A)
t = 2.5 s
d = not needed
Equation: vf = vi + at → rearranged: a = (vf − vi) / t
a = (9.5 − 0) / 2.5
a = 9.5 / 2.5
✓ a = 3.8 m/s²
Part B — Find t for vf = 12 m/s (using same acceleration):
t = (vf − vi) / a = (12 − 0) / 3.8
✓ t ≈ 3.16 s
Problem: A stone is dropped from a cliff (vi = 0). It falls for 4 seconds. How far does it fall? (g = 9.81 m/s²)
a = 9.81 m/s² (downward, taking down as positive)
t = 4 s
vf = not needed
d = ?
d = (0)(4) + ½(9.81)(4²)
d = 0 + ½(9.81)(16)
d = ½ × 156.96
✓ d = 78.48 m ≈ 78.5 m
Problem: A car enters a tunnel at 15 m/s and accelerates at 2.5 m/s². The tunnel is 200 m long. What is the car's speed when it exits? (Time not given — use the time-independent 3rd kinematic equation.)
a = 2.5 m/s²
d = 200 m
t = NOT GIVEN → use the time-independent equation
vf = ?
vf² = (15)² + 2(2.5)(200)
vf² = 225 + 1000
vf² = 1225
vf = √1225
✓ vf = 35 m/s
Problem: A lorry accelerates uniformly from 5 m/s to 25 m/s over 10 seconds. What distance does it travel? (Acceleration not given — use the 4th kinematic equation.)
vf = 25 m/s
t = 10 s
a = NOT GIVEN → use 4th kinematic equation
d = ?
d = (5 + 25) × 10 / 2
d = 30 × 10 / 2
d = 300 / 2
✓ d = 150 m
Problem: A train reaches a station at 5 m/s after decelerating at 3 m/s² for 8 seconds. What distance did it cover during braking? (Initial velocity not required — use the 5th kinematic equation.)
a = −3 m/s² (deceleration)
t = 8 s
vi = NOT NEEDED → use 5th kinematic equation
d = ?
d = (5)(8) − ½(−3)(8²)
d = 40 − ½(−3)(64)
d = 40 − (−96)
d = 40 + 96
✓ d = 136 m
Frequently Asked Questions
The answer is 3, 4, or 5 depending on which textbook you use. Most US high school physics courses teach 4 kinematic equations. Some courses and all British A-Level SUVAT courses include 5 equations. The three core equations are: vf = vi + at, d = vi×t + ½at², and the time-independent equation vf² = vi² + 2ad. The 4th kinematic equation d = (vi+vf)t/2 and 5th equation d = vf×t − ½at² are additional forms that some courses teach. All five are covered on this page.
The time-independent kinematic equation is the 3rd kinematic equation: vf² = vi² + 2ad. It is called time-independent because it contains no time variable (t). Use the time-independent equation when time is not given and not needed — for example, when finding braking distance, the speed of a projectile at a given height, or the final velocity after traveling a known distance. In SUVAT notation it is written v² = u² + 2as. The time-independent equation is the most frequently tested of all the kinematic equations.
The 3rd kinematic equation is vf² = vi² + 2ad — the time-independent kinematic equation. It connects final velocity, initial velocity, acceleration, and displacement without using time. The 3rd kinematic equation is derived by eliminating t between the 1st and 4th equations. In SUVAT notation: v² = u² + 2as. The 3rd kinematic equation is used in braking distance problems, projectile height problems, and any situation where time is unknown.
The 4th kinematic equation is d = (vi + vf) × t / 2. It calculates displacement using the average velocity (the arithmetic mean of initial and final velocity) multiplied by time. The 4th kinematic equation is ideal when acceleration is not given — only the two velocities and time are needed. In SUVAT notation: s = (u + v)t / 2. The 4th kinematic equation is valid only when acceleration is constant.
The 2nd kinematic equation is d = vi×t + ½at². It gives the displacement of an object from its initial velocity, acceleration, and time. The first term (vi×t) is the displacement at constant velocity; the second term (½at²) is the additional displacement due to acceleration. This is the kinematic equation for distance when time is known. In SUVAT notation: s = ut + ½at². For free fall from rest, it simplifies to d = ½gt².
Yes — all five kinematic equations work with negative acceleration (deceleration). Simply substitute the negative value for a. For example, a car decelerating at 5 m/s² uses a = −5 m/s². The sign convention must be consistent: if motion in the positive direction has positive velocity, then opposing acceleration is negative. The kinematic equations handle positive and negative values algebraically.
There is no single "kinematic equation for time" — time can be found by rearranging any equation that contains t. The simplest is the 1st equation rearranged: t = (vf − vi) / a. From the 4th equation: t = 2d / (vi + vf). From the 2nd equation, solving for t requires the quadratic formula. From the time-independent 3rd kinematic equation, t can be found by first finding vf, then using Eq. 1.
The primary kinematic equation for velocity is the 1st equation: vf = vi + at. This gives final velocity when you know initial velocity, acceleration, and time. Alternatively, from the time-independent 3rd kinematic equation: vf = √(vi² + 2ad) — use this when time is not known. The 4th equation rearranged also gives velocity: vf = 2d/t − vi.
The standard kinematic equations are for linear (translational) motion — they are 1D kinematics equations for straight-line motion. However, there are analogous rotational kinematic equations: ωf = ωi + αt, θ = ωit + ½αt², ωf² = ωi² + 2αθ, where ω is angular velocity, α is angular acceleration, and θ is angular displacement. The structure is identical — just with rotational variables replacing the linear ones.
Kinematics describes how objects move — using the kinematic equations to relate position, velocity, acceleration, and time — without asking why they move. Dynamics explains the causes of motion through forces and Newton's laws (F = ma). The basic equations of kinematics and linear movement equations assume acceleration is known; dynamics tells you what that acceleration is based on applied forces.
Kinematic Equations Calculators
Use our dedicated physics formulas kinematics calculators for specific problem types. Each calculator provides step-by-step solutions and instant unit conversions.