Kinematics Equations (SUVAT) Guide
Kinematics equations — often called the SUVAT equations after the five variables they connect — are the mathematical toolkit for solving motion problems under constant acceleration. The five variables are: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t).
The four equations form a complete set: given any three of the five variables, the remaining two can always be found. Each equation omits exactly one variable, making equation selection straightforward: identify which variable is absent from your problem, and use the equation that also lacks that variable.
These equations are derived from the definitions of velocity and acceleration using integration. They assume acceleration is constant throughout the motion — this is valid for free-fall, objects on frictionless slopes, and uniformly accelerating vehicles. For variable acceleration, calculus methods are needed.
Formula
v = u+at | s = ut+½at² | v² = u²+2as | s = ½(u+v)t
The Four SUVAT Equations
Equation 1: v = u + at — Final velocity equals initial velocity plus acceleration times time. This is the direct definition of constant acceleration integrated once. Missing variable: s (displacement). Use this when you know initial and final velocities and time, and want to find acceleration — or know acceleration and time and want final velocity.
Equation 2: s = ut + ½at² — Displacement equals initial velocity times time plus half of acceleration times time squared. This comes from integrating v = u + at to get position. Missing variable: v (final velocity). Use this when time is known but final velocity is not — e.g., finding how far a ball falls in 3 seconds from rest.
Equation 3: v² = u² + 2as — Eliminates time entirely. Useful when time is not given or needed — for example, finding the speed of a car after braking over a known distance. Missing variable: t (time). This is derived by eliminating t from equations 1 and 2.
Equation 4: s = ½(u + v)t — Displacement is the average velocity times time. This is a shortcut when both initial and final velocities are known. Missing variable: a (acceleration). This follows directly from the constant-acceleration assumption that velocity changes linearly with time.
| Equation | Formula | Missing Variable | When to Use |
|---|---|---|---|
| SUVAT 1 | v = u + at | s (displacement) | Find v or a when time is known |
| SUVAT 2 | s = ut + ½at² | v (final velocity) | Find s or u when time is known |
| SUVAT 3 | v² = u² + 2as | t (time) | Time not given or needed |
| SUVAT 4 | s = ½(u + v)t | a (acceleration) | Both velocities and time known |
How to Choose the Right Equation
Step 1: List what is known and what is to be found. Label each as s, u, v, a, or t. Step 2: Count — you need 3 knowns to find 1 unknown (4 variables in each equation). Step 3: Identify which variable appears in your known + unknown set. Select the equation that contains all of these variables.
Quick decision guide: No s? → use v = u + at. No v? → use s = ut + ½at². No t? → use v² = u² + 2as. No a? → use s = ½(u+v)t.
Sometimes two equations are needed in sequence. Example: find s and v for a ball dropped from rest (u=0) after t=3 s with a=9.81 m/s². First use v = 0 + 9.81×3 = 29.43 m/s (Eq.1), then s = 0×3 + ½×9.81×9 = 44.1 m (Eq.2). Alternatively, use Eq.3 and Eq.2 in any order.
Worked Examples
Example 1 — Braking car: A car at u = 30 m/s decelerates at a = −6 m/s² until it stops (v = 0). Find stopping distance s. Known: u = 30, v = 0, a = −6. Missing: t. Use v² = u² + 2as → 0 = 900 + 2(−6)s → s = 900/12 = 75 m.
Example 2 — Free fall: A stone is dropped from rest (u = 0) from a cliff. After t = 4 s, find velocity and distance fallen. Known: u = 0, a = 9.81, t = 4. v = 0 + 9.81×4 = 39.24 m/s (downward). s = 0 + ½×9.81×16 = 78.48 m.
Example 3 — Takeoff: An aircraft accelerates from rest to v = 80 m/s in s = 1200 m. Find acceleration and time. Known: u = 0, v = 80, s = 1200. Use v² = u² + 2as: 6400 = 0 + 2a×1200 → a = 2.67 m/s². Then t = (v − u)/a = 80/2.67 ≈ 30 s.
Example 4 — Ball thrown upward: Ball thrown up at u = 15 m/s. Find max height and time to reach it. At max height, v = 0, a = −9.81. v² = u² + 2as: 0 = 225 − 2×9.81×s → s = 225/19.62 ≈ 11.5 m. Time: v = u + at → 0 = 15 − 9.81t → t = 1.53 s.
Common Mistakes and How to Avoid Them
Sign errors: Always define a positive direction before starting. If upward is positive and an object is thrown upward with g = 9.81 m/s² downward, use a = −9.81 m/s². Mixing signs is the most common kinematics error. Write the sign convention at the top of your work.
Confusing distance and displacement: SUVAT equations use displacement s (which can be negative), not distance (which is always positive). A ball thrown upward that returns to its starting point has displacement s = 0 but distance = 2× max height. If the question asks for distance traveled but the object reverses, split the motion into two segments.
Using SUVAT for variable acceleration: These equations require constant acceleration. For a pendulum, a car changing gear, or any system with varying force, SUVAT will give wrong answers. Check that a is truly constant before applying SUVAT equations.