Last Updated: August 1, 2026
Quick Summary & Key Takeaways (GEO & AEO Summary)
Target Audience: High school and college physics students (AP Physics, IGCSE, A-Levels), mechanical engineers, and STEM educators.
What are SUVAT Equations?
The SUVAT equations are five foundational kinematic formulas in classical mechanics used to calculate 1D translational motion under constant (uniform) acceleration ($a$). The acronym derives from its five variables: displacement ($s$), initial velocity ($u$), final velocity ($v$), acceleration ($a$), and time ($t$).
The 5 SUVAT Formulas:
$v = u + at$ (No $s$)
$s = ut + \frac{1}{2}at^2$ (No $v$)
$v^2 = u^2 + 2as$ (No $t$)
$s = \frac{1}{2}(u + v)t$ (No $a$)
$s = vt – \frac{1}{2}at^2$ (No $u$)
Strict Physical Boundary: Applies only when acceleration is constant ($a = \text{constant}$). If acceleration varies, calculus techniques ($a = \frac{dv}{dt}$) must be applied.
Introduction to SUVAT Kinematic Equations
Understanding motion is the fundamental foundation of classical mechanics. The SUVAT equations are a set of five core mathematical formulas in kinematics that describe the movement of an object undergoing constant linear acceleration.
Whether analyzing a decelerating vehicle, an object in free fall under Earth’s gravitational force using our 5 Essential F=ma Examples Guide, or evaluating force vectors in our 3 Essential Newton’s Laws of Motion Guide, mastering these formulas is essential.
Furthermore, SUVAT dynamics connect directly to energy transformations, as discussed in our Kinetic Energy vs Potential Energy Guide, as well as wave kinematics in our 5 Essential Transverse Wave Examples Guide.
What Are the SUVAT Equations?
The acronym SUVAT derives directly from the five physical variables that define 1D translational motion under uniform acceleration. Each variable represents a specific scalar or vector quantity with standard SI units:
| Symbol | Physical Quantity | Standard SI Unit | Vector / Scalar |
| $s$ | Displacement (change in position) | Meters ($\text{m}$) | Vector |
| $u$ | Initial velocity (speed at $t = 0$) | Meters per second ($\text{m/s}$) | Vector |
| $v$ | Final velocity (speed at time $t$) | Meters per second ($\text{m/s}$) | Vector |
| $a$ | Acceleration (rate of change of velocity) | Meters per second squared ($\text{m/s}^2$) | Vector |
| $t$ | Time interval elapsed | Seconds ($\text{s}$) | Scalar |
Critical Condition: The SUVAT equations apply only when acceleration $a$ remains strictly constant throughout the entire time interval. If acceleration varies with time or position—such as in simple harmonic motion or fluid drag—calculus-based kinematics ($a = \frac{dv}{dt}$) must be used instead.
The Five SUVAT Equations
Every kinematic scenario involves four of the five SUVAT variables. Consequently, there are five distinct formulas, each omitting exactly one variable. Select the equation that excludes the variable you neither know nor need to calculate:
- Missing Displacement ($s$):$$v = u + at$$
- Missing Final Velocity ($v$):$$s = ut + \frac{1}{2}at^2$$
- Missing Time ($t$):$$v^2 = u^2 + 2as$$
- Missing Acceleration ($a$):$$s = \frac{1}{2}(u + v)t$$
- Missing Initial Velocity ($u$):$$s = vt – \frac{1}{2}at^2$$
Deriving the SUVAT Equations from First Principles
Deriving these mathematical formulas prevents rote memorization and develops analytical problem-solving skills. All five SUVAT equations stem from two foundational definitions: instantaneous velocity ($v = \frac{ds}{dt}$) and constant acceleration ($a = \frac{dv}{dt}$).
Derivation 1: First Equation ($v = u + at$)
By definition, constant acceleration is the rate of change of velocity over time:
$$a = \frac{v – u}{t}$$
Multiplying both sides by $t$ yields $at = v – u$. Rearranging gives:
$$v = u + at$$
Derivation 2: Fourth Equation ($s = \frac{1}{2}(u + v)t$)
For linear motion under uniform acceleration, average velocity ($\bar{v}$) is the arithmetic mean of initial and final velocities:
$$\bar{v} = \frac{u + v}{2}$$
Since displacement equals average velocity multiplied by time ($s = \bar{v} \cdot t$), substituting $\bar{v}$ yields:
$$s = \frac{1}{2}(u + v)t$$
Derivation 3: Second Equation ($s = ut + \frac{1}{2}at^2$)
Substitute $v = u + at$ (from Equation 1) into $s = \frac{1}{2}(u + v)t$:
$$s = \frac{1}{2} \left[ u + (u + at) \right] t$$
$$s = \frac{1}{2} (2u + at) t = ut + \frac{1}{2}at^2$$
Derivation 4: Third Equation ($v^2 = u^2 + 2as$)
Rearrange Equation 1 to solve for time: $t = \frac{v – u}{a}$. Substitute this expression into Equation 4:
$$s = \left(\frac{u + v}{2}\right) \left(\frac{v – u}{a}\right) = \frac{v^2 – u^2}{2a}$$
Multiplying by $2a$ and rearranging gives:
$$v^2 = u^2 + 2as$$
Derivation 5: Fifth Equation ($s = vt – \frac{1}{2}at^2$)
Rearrange Equation 1 to express initial velocity as $u = v – at$. Substitute this into Equation 2:
$$s = (v – at)t + \frac{1}{2}at^2 = vt – at^2 + \frac{1}{2}at^2$$
$$s = vt – \frac{1}{2}at^2$$
Systematic Step-by-Step Problem-Solving Strategy
Follow this structured method to solve any 1D kinematics problem systematically:
Step 1: Assign Positive Direction (+ Axis)
↓
Step 2: List Knowns (s, u, v, a, t) with Correct Vector Signs
↓
Step 3: Identify Target Variable & Absent Variable
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Step 4: Select the Matching SUVAT Equation
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Step 5: Rearrange Algebraically, Substitute Values, and Solve
- Establish a Coordinate System: Choose a direction (e.g., UP or RIGHT) as positive. Any quantity acting in the opposite direction takes a negative sign.
- List Your Knowns and Unknowns: Explicitly write down $s$, $u$, $v$, $a$, and $t$ with appropriate vector signs.
- Identify the Absent Variable: Determine which variable is completely omitted from the problem description.
- Select the SUVAT Equation: Pick the formula that contains your knowns and target variable, excluding the absent variable.
- Substitute and Solve: Rearrange algebraically first, then plug in numeric values with standard SI units.
Understanding Sign Conventions in Kinematics
Because displacement, initial velocity, final velocity, and acceleration are vector quantities, directional signs ($\mathbf{+}$ or $\mathbf{-}$) dictate physical behavior:
- Vertical Motion / Free Fall: If upward is defined as positive ($+y$), Earth’s gravitational acceleration points downward, so $a = -g = -9.81\text{ m/s}^2$. For precise SI physical unit standards, refer to the NIST Physical Measurement Laboratory.
- Deceleration / Braking: When an object slows down, its acceleration vector acts opposite to its velocity vector. If a car moves forward ($u > 0$) while braking, its acceleration is negative ($a < 0$).
+y (Upward Positive Direction)
↑
│ • (Object launched)
│ │ Velocity (v) > 0 (Ascending)
│ ↓ Gravitational Acceleration (a) = -9.81 m/s²
└───────────────────► +x
Velocity-Time Graphs and SUVAT Connections
Visualizing kinematic motion using velocity-time ($v$-$t$) graphs reinforces the conceptual origin of the SUVAT equations:
- Slope of $v$-$t$ Graph: The gradient represents acceleration ($a = \frac{\Delta v}{\Delta t}$).
- Area Under $v$-$t$ Curve: The area between the function line and the time axis equals total displacement ($s$).
- Y-Intercept: Represents initial velocity ($u$).
- Y-Value at Time $t$: Represents final velocity ($v$).
Interactive simulations like the PhET Interactive Motion Simulations allow students to observe these $v$-$t$ graph relationships dynamically in real-time.
Comprehensive Solved Worked Examples
Worked Example 1: Calculating Stopping Distance (Braking Car)
A vehicle traveling at $25\text{ m/s}$ applies brakes to decelerate uniformly at $5\text{ m/s}^2$ until it stops completely. Calculate its total stopping distance.
- Given: $u = +25\text{ m/s}$, $v = 0\text{ m/s}$ (at rest), $a = -5\text{ m/s}^2$ (deceleration)
- Target: $s$ | Missing Variable: $t$
- Formula: $v^2 = u^2 + 2as$
- Solution:$$0^2 = (25)^2 + 2(-5)s \implies 0 = 625 – 10s \implies 10s = 625 \implies \mathbf{s = 62.5\text{ m}}$$
Worked Example 2: Vertical Free Fall from Rest
A stone is dropped from a bridge and hits the water below after $3.5\text{ seconds}$. Calculate the height of the bridge above the water ($g = 9.81\text{ m/s}^2$).
- Given: $u = 0\text{ m/s}$ (dropped from rest), $t = 3.5\text{ s}$, $a = +9.81\text{ m/s}^2$ (downward positive)
- Target: $s$ | Missing Variable: $v$
- Formula: $s = ut + \frac{1}{2}at^2$
- Solution:$$s = (0)(3.5) + \frac{1}{2}(9.81)(3.5)^2 = 0 + 0.5 \times 9.81 \times 12.25 = \mathbf{60.09\text{ m}}$$
Worked Example 3: Upward Vertical Projection
A ball is thrown straight upward with an initial velocity of $18\text{ m/s}$. Find the maximum height reached above the release point ($g = 9.81\text{ m/s}^2$).
- Given (Upward = Positive): $u = +18\text{ m/s}$, $v = 0\text{ m/s}$ (at maximum apex height), $a = -9.81\text{ m/s}^2$
- Target: $s$ | Missing Variable: $t$
- Formula: $v^2 = u^2 + 2as$
- Solution:$$0 = (18)^2 + 2(-9.81)s \implies 0 = 324 – 19.62s \implies s = \frac{324}{19.62} = \mathbf{16.51\text{ m}}$$
Worked Example 4: Finding Acceleration and Velocity
A train accelerates uniformly from rest down a straight track, covering $200\text{ m}$ in $10\text{ seconds}$. Calculate its acceleration and final velocity.
Step 1: Calculate Acceleration ($a$)
- Given: $u = 0\text{ m/s}$, $s = 200\text{ m}$, $t = 10\text{ s}$ | Missing Variable: $v$
- Formula: $s = ut + \frac{1}{2}at^2$
- Calculation:$$200 = (0)(10) + \frac{1}{2}a(10)^2 \implies 200 = 50a \implies \mathbf{a = 4.0\text{ m/s}^2}$$
Step 2: Calculate Final Velocity ($v$)
- Formula: $v = u + at$
- Calculation:$$v = 0 + (4.0 \times 10) = \mathbf{40.0\text{ m/s}}$$
Worked Example 5: Time Taken During Deceleration
A sprinter slows down uniformly from $12\text{ m/s}$ to $4\text{ m/s}$ over a distance of $32\text{ m}$. How long did this deceleration take?
- Given: $u = 12\text{ m/s}$, $v = 4\text{ m/s}$, $s = 32\text{ m}$
- Target: $t$ | Missing Variable: $a$
- Formula: $s = \frac{1}{2}(u + v)t$
- Solution:$$32 = \frac{1}{2}(12 + 4)t \implies 32 = 8t \implies \mathbf{t = 4.0\text{ seconds}}$$
Frequently Asked Questions (FAQs)
Can SUVAT equations be used when acceleration is not constant?
No. The SUVAT equations rely entirely on the mathematical assumption that acceleration ($a$) is constant throughout the motion interval. If acceleration changes over time or distance, you must use calculus methods—integrating or differentiating acceleration functions ($a = \frac{dv}{dt}$) or numerical methods.
What is the difference between displacement and distance in SUVAT?
Displacement ($s$) is a vector quantity representing the overall change in position from start to finish in a straight line, including direction. Distance is a scalar quantity measuring total path length traveled. In 1D motion without turning back, displacement magnitude equals distance. However, if an object moves forward $10\text{ m}$ and returns $10\text{ m}$, its displacement is $0\text{ m}$, while distance is $20\text{ m}$.
Why is acceleration negative when an object slows down?
Acceleration is defined as $a = \frac{v – u}{t}$. When an object slows down, its final speed $v$ is less than initial speed $u$, resulting in a negative numerator ($v – u < 0$). In vector terms, deceleration means the acceleration vector points opposite to the direction of velocity.
What value of g should I use for free fall SUVAT problems?
Standard international physics uses $g = 9.81\text{ m/s}^2$ (or $9.8\text{ m/s}^2$). However, introductory physics curricula (like A-Level or AP Physics) frequently approximate $g = 10\text{ m/s}^2$ to simplify mental or non-calculator calculations. Always check your syllabus or exam instructions.
How do SUVAT equations apply to 2D projectile motion?
In 2D projectile motion, 1D SUVAT equations are applied independently along two perpendicular axes:
- Horizontal Axis ($x$): Zero acceleration ($a_x = 0$), so $x = u_x \cdot t$.
- Vertical Axis ($y$): Constant acceleration due to gravity ($a_y = -g$), utilizing full 1D vertical SUVAT equations.