Last updated: October 3, 2026
Quick Summary
The SUVAT equations are five equations of motion used to calculate displacement, initial velocity, final velocity, acceleration, and time when acceleration is constant. They are widely used in GCSE, IGCSE, A-Level, IB Physics, and other mechanics courses.
The five equations are:
- (v = u + at)
- (s = ut + \frac{1}{2}at^2)
- (v^2 = u^2 + 2as)
- (s = \frac{1}{2}(u+v)t)
- (s = vt – \frac{1}{2}at^2)
Quick answer: To choose the correct SUVAT equation, list the known values, identify the quantity you need, and select an equation that includes those values but omits the variable you do not know or need.
Key Takeaways
- SUVAT represents displacement ((s)), initial velocity ((u)), final velocity ((v)), acceleration ((a)), and time ((t)).
- Each of the five standard equations omits one SUVAT variable.
- SUVAT equations apply when acceleration is constant during the interval being analysed.
- Use consistent SI units: metres, metres per second, metres per second squared, and seconds.
- Choose an equation using the known values and required unknown rather than memorising formulas without understanding them.
- Displacement is not always the same as total distance travelled.
- In vertical motion, choose a positive direction and keep the signs of velocity and acceleration consistent.
What Are the SUVAT Equations?
The SUVAT equations are kinematic formulas that describe straight-line motion under constant acceleration. They connect five quantities—displacement, initial velocity, final velocity, acceleration, and time—so that an unknown quantity can be calculated from the values already known.
For example, if a car accelerates uniformly from a known initial velocity for a specified time, you can calculate its final velocity using (v=u+at). If you know its initial velocity, acceleration, and time, you can also calculate its displacement using (s=ut+\frac{1}{2}at^2).
Quick answer: SUVAT equations are used to solve motion problems when acceleration remains constant, including many questions involving cars, falling objects, trains, and vertical launches.
For a detailed explanation of motion under constant acceleration, read OpenStax’s guide to kinematic equations.
What Does SUVAT Stand for in Physics?
Each letter represents a physical quantity:
| Symbol | Quantity | SI unit | Meaning |
|---|---|---|---|
| (s) | Displacement | metre (m) | Change in position in a chosen direction |
| (u) | Initial velocity | metre per second (m/s) | Velocity at the start of the interval |
| (v) | Final velocity | metre per second (m/s) | Velocity at the end of the interval |
| (a) | Acceleration | metre per second squared (m/s²) | Rate of change of velocity |
| (t) | Time | second (s) | Duration of the motion |
Initial velocity is commonly written as (u) in SUVAT notation, while final velocity is written as (v). Acceleration describes how velocity changes with time, including changes in direction when a consistent sign convention is used.
Quick answer: SUVAT stands for displacement, initial velocity, final velocity, acceleration, and time.
When Can You Use SUVAT Equations?
You can use the standard SUVAT equations when acceleration is constant over the interval you are studying. This includes an object speeding up uniformly, slowing down uniformly, or falling under an approximately constant gravitational acceleration when air resistance is negligible.
If acceleration changes during the motion, a single SUVAT equation generally cannot describe the entire interval. You may be able to divide the motion into separate stages, each with constant acceleration, and solve each stage separately.
Quick answer: Use SUVAT when acceleration is constant for the motion interval being analysed. For variable acceleration, use an appropriate alternative method or analyse suitable separate intervals.
All 5 SUVAT Equations: Formula Table
The five standard SUVAT equations each contain four of the five variables. The variable omitted from an equation helps you decide when that equation may be useful.
| Equation | Formula | Variable omitted | When it is useful |
|---|---|---|---|
| 1 | (v=u+at) | (s) | When displacement is not needed |
| 2 | (s=ut+\frac{1}{2}at^2) | (v) | When final velocity is not needed |
| 3 | (v^2=u^2+2as) | (t) | When time is not needed |
| 4 | (s=\frac{1}{2}(u+v)t) | (a) | When acceleration is not needed |
| 5 | (s=vt-\frac{1}{2}at^2) | (u) | When initial velocity is not needed |
Quick answer: The five equations omit (s), (v), (t), (a), and (u), respectively. Choose the equation that matches the values given in the question and the quantity you need to find.
1. First SUVAT Equation: Final Velocity
[
v=u+at
]
This equation calculates final velocity from initial velocity, acceleration, and elapsed time. It is useful when an object accelerates or decelerates uniformly and the displacement is not required.
Example: An object starts at (4\text{ m/s}) and accelerates at (3\text{ m/s}^2) for (5\text{ s}).
[
v=4+(3)(5)=19\text{ m/s}
]
Quick answer: Use (v=u+at) when you know initial velocity, acceleration, and time and need final velocity.
2. Second SUVAT Equation: Displacement Without Final Velocity
[
s=ut+\frac{1}{2}at^2
]
This equation calculates displacement using initial velocity, acceleration, and time. It is particularly useful when final velocity is unknown or unnecessary.
Example: A cyclist moves at an initial velocity of (2\text{ m/s}) and accelerates at (1.5\text{ m/s}^2) for (4\text{ s}).
[
s=(2)(4)+\frac{1}{2}(1.5)(4^2)
]
[
s=8+12=20\text{ m}
]
Quick answer: Use (s=ut+\frac{1}{2}at^2) when initial velocity, acceleration, and time are known and you need displacement.
3. Third SUVAT Equation: Motion Without Time
[
v^2=u^2+2as
]
This equation relates initial velocity, final velocity, acceleration, and displacement without using time. It is often useful for braking-distance and maximum-height problems.
Example: A car moving at (20\text{ m/s}) brakes with an acceleration of (-4\text{ m/s}^2). Calculate its displacement before stopping.
Given (u=20\text{ m/s}), (v=0\text{ m/s}), and (a=-4\text{ m/s}^2):
[
0^2=20^2+2(-4)s
]
[
0=400-8s
]
[
s=50\text{ m}
]
Quick answer: Use (v^2=u^2+2as) when time is not given or is not required.
4. Fourth SUVAT Equation: Displacement From Average Velocity
[
s=\frac{1}{2}(u+v)t
]
For constant acceleration, average velocity is the arithmetic mean of the initial and final velocities. Multiplying that average velocity by the elapsed time gives displacement.
Example: An object changes velocity uniformly from (6\text{ m/s}) to (14\text{ m/s}) over (3\text{ s}).
[
s=\frac{1}{2}(6+14)(3)
]
[
s=30\text{ m}
]
Quick answer: Use (s=\frac{1}{2}(u+v)t) when initial velocity, final velocity, and time are known, and acceleration is not required.
5. Fifth SUVAT Equation: Displacement Without Initial Velocity
[
s=vt-\frac{1}{2}at^2
]
This equation calculates displacement using final velocity, acceleration, and time without requiring initial velocity as a separate input. It is useful when those three quantities are known.
Example: An object has a final velocity of (12\text{ m/s}), acceleration of (2\text{ m/s}^2), and motion time of (4\text{ s}).
[
s=(12)(4)-\frac{1}{2}(2)(4^2)
]
[
s=48-16=32\text{ m}
]
Quick answer: Use (s=vt-\frac{1}{2}at^2) when final velocity, acceleration, and time are known but initial velocity is not required.
How to Choose the Correct SUVAT Equation
Choosing an equation is easier when you identify the known and unknown quantities before substituting numbers. Write down the five symbols, fill in the values provided by the question, and mark the quantity you must calculate.
Then choose an equation that contains the known values and the required unknown. If one of the five variables is not needed, an equation that omits that variable may simplify the calculation. If the equation contains another unknown, you may need additional information or an algebraic rearrangement.
Quick answer: List the known variables, identify the required unknown, and choose a formula that connects them without introducing an unnecessary unknown.
Worked Example: Choosing an Equation
Question: A ball starts from rest and accelerates uniformly at (2.5\text{ m/s}^2) for (6\text{ s}). Calculate its displacement.
Step 1: List the known values.
- (u=0\text{ m/s})
- (a=2.5\text{ m/s}^2)
- (t=6\text{ s})
- Find (s)
Step 2: Select the equation.
The equation (s=ut+\frac{1}{2}at^2) contains all the required quantities and does not require final velocity.
Step 3: Substitute the values.
[
s=(0)(6)+\frac{1}{2}(2.5)(6^2)
]
[
s=45\text{ m}
]
Answer: The ball’s displacement is (45\text{ m}) in the chosen positive direction.
Quick answer: Choose the equation from the variables supplied in the question, not simply because it is the first formula you remember.
SUVAT Equations Rearranged for Different Variables
Rearranging a formula lets you calculate a different quantity from the same relationship. These forms are useful in exams and in questions that ask for time, acceleration, displacement, or initial velocity.
Rearrangements for Acceleration
From (v=u+at):
[
a=\frac{v-u}{t}
]
From (v^2=u^2+2as):
[
a=\frac{v^2-u^2}{2s}
]
The second form is suitable when time is not required and displacement is known. The denominator must be non-zero for the displayed rearrangement to be valid.
Quick answer: Calculate acceleration using (a=(v-u)/t) when the velocity change and time are known, or use (a=(v^2-u^2)/(2s)) when displacement is known.
Rearrangements for Time
From (v=u+at):
[
t=\frac{v-u}{a}
]
From (s=\frac{1}{2}(u+v)t):
[
t=\frac{2s}{u+v}
]
The second rearrangement requires (u+v\ne0). If the denominator is zero, use the original problem conditions to determine whether another equation is suitable.
Quick answer: Use (t=(v-u)/a) when the velocity change and acceleration are known. Use (t=2s/(u+v)) when displacement and both endpoint velocities are known and (u+v\ne0).
Rearrangements for Displacement
From (v^2=u^2+2as):
[
s=\frac{v^2-u^2}{2a}
]
From (s=\frac{1}{2}(u+v)t):
[
s=\frac{(u+v)t}{2}
]
Both forms calculate displacement, but they use different known values. Choose the one that matches the information supplied in the question.
Quick answer: Use (s=(v^2-u^2)/(2a)) when time is not needed, or (s=(u+v)t/2) when both velocities and time are known.
Rearrangement for Initial Velocity
From (v=u+at):
[
u=v-at
]
From (s=\frac{1}{2}(u+v)t):
[
u=\frac{2s}{t}-v
]
These rearrangements are useful when final velocity, acceleration, time, or displacement is known and initial velocity is the required unknown.
Quick answer: To find initial velocity, rearrange a suitable equation using the quantities given in the problem.
Rearrangement for Final Velocity
From (v=u+at):
[
v=u+at
]
From (v^2=u^2+2as):
[
v=\pm\sqrt{u^2+2as}
]
The square-root form can have two mathematical signs. Select the physically appropriate velocity using the chosen positive direction and the conditions of the motion; do not automatically choose the positive root.
Quick answer: Use (v=u+at) when time is known, or the squared-velocity equation when time is not required and the direction of motion is considered.
SUVAT Equations Without a Particular Variable
Queries such as “SUVAT equation without (u)” or “SUVAT equation without (t)” usually mean that the student wants the standard equation that does not contain a particular variable. The table below makes this choice explicit.
| Search intent | Equation to use | Variable omitted |
|---|---|---|
| SUVAT equation without (s) | (v=u+at) | Displacement |
| SUVAT equation without (v) | (s=ut+\frac{1}{2}at^2) | Final velocity |
| SUVAT equation without (t) | (v^2=u^2+2as) | Time |
| SUVAT equation without (a) | (s=\frac{1}{2}(u+v)t) | Acceleration |
| SUVAT equation without (u) | (s=vt-\frac{1}{2}at^2) | Initial velocity |
Quick answer: Match the variable you want to omit with the table, then check that the selected equation contains all the values you know and the quantity you need.
How to Derive the SUVAT Equations
The SUVAT equations follow from the definitions of acceleration, velocity, and displacement when acceleration is constant. Understanding these relationships can make the formulas easier to remember and apply.
Deriving the First Equation
Acceleration is the change in velocity divided by the elapsed time:
[
a=\frac{v-u}{t}
]
Multiply both sides by (t):
[
at=v-u
]
Rearranging gives:
[
v=u+at
]
Quick answer: The equation (v=u+at) comes directly from the definition of constant acceleration.
Deriving the Second Equation
For constant acceleration, displacement can be calculated from average velocity:
[
s=\frac{u+v}{2}t
]
Substitute (v=u+at):
[
s=\frac{u+(u+at)}{2}t
]
Simplify:
[
s=\frac{2u+at}{2}t
]
Therefore:
[
s=ut+\frac{1}{2}at^2
]
Quick answer: The second equation is obtained by substituting (v=u+at) into the average-velocity equation.
Deriving the Third Equation
Start with:
[
v=u+at
]
Rearrange for time:
[
t=\frac{v-u}{a}
]
Substitute this into (s=\frac{1}{2}(u+v)t):
[
s=\frac{1}{2}(u+v)\frac{v-u}{a}
]
Using ((u+v)(v-u)=v^2-u^2):
[
s=\frac{v^2-u^2}{2a}
]
Rearranging gives:
[
v^2=u^2+2as
]
Quick answer: The third equation is derived by eliminating time from two other constant-acceleration relationships.
Deriving the Fourth and Fifth Equations
The fourth equation is the average-velocity relationship:
[
s=\frac{1}{2}(u+v)t
]
It follows because velocity changes uniformly from (u) to (v), so the average velocity is ((u+v)/2).
For the fifth equation, rearrange (v=u+at) to obtain (u=v-at). Substitute into the fourth equation:
[
s=\frac{1}{2}(v-at+v)t
]
Simplifying gives:
[
s=vt-\frac{1}{2}at^2
]
Quick answer: The fourth equation uses average velocity, while the fifth follows by eliminating initial velocity from the average-velocity equation.
Worked SUVAT Physics Examples
The following examples demonstrate how the formulas are applied to familiar motion problems. Each solution identifies the known values, selects an appropriate equation, and includes the correct units.
Example 1: Stopping Distance of a Car
Question: A car travels at (25\text{ m/s}) and decelerates uniformly at (5\text{ m/s}^2) until it stops. Find its stopping distance.
Given:
- (u=25\text{ m/s})
- (v=0\text{ m/s})
- (a=-5\text{ m/s}^2)
Time is not given, so use:
[
v^2=u^2+2as
]
Substitute:
[
0^2=25^2+2(-5)s
]
[
0=625-10s
]
[
s=62.5\text{ m}
]
Answer: The stopping distance is (62.5\text{ m}).
Quick answer: Use (v^2=u^2+2as) for stopping-distance problems when time is unknown.
Example 2: An Object in Free Fall
Question: A stone is released from rest and falls for (3.5\text{ s}). Calculate its vertical displacement, taking downward as positive and (g=9.81\text{ m/s}^2).
Given:
- (u=0\text{ m/s})
- (a=9.81\text{ m/s}^2)
- (t=3.5\text{ s})
Use:
[
s=ut+\frac{1}{2}at^2
]
Substitute:
[
s=(0)(3.5)+\frac{1}{2}(9.81)(3.5^2)
]
[
s=60.09\text{ m}
]
Answer: The stone’s displacement is approximately (60.1\text{ m}) downward.
Quick answer: For an object released from rest, use (s=\frac{1}{2}gt^2) when downward is chosen as the positive direction and air resistance is negligible.
Example 3: Maximum Height of a Vertically Thrown Ball
Question: A ball is thrown vertically upward at (18\text{ m/s}). Find its maximum height above the launch point. Use (g=9.81\text{ m/s}^2).
At maximum height, the instantaneous vertical velocity is zero. Taking upward as positive:
- (u=18\text{ m/s})
- (v=0\text{ m/s})
- (a=-9.81\text{ m/s}^2)
Use:
[
v^2=u^2+2as
]
Substitute:
[
0=18^2+2(-9.81)s
]
[
19.62s=324
]
[
s\approx16.51\text{ m}
]
Answer: The ball reaches a maximum height of approximately (16.5\text{ m}) above its launch point, neglecting air resistance.
Quick answer: At the maximum height of a vertical launch, set vertical final velocity to zero and use a consistent sign convention.
Example 4: Two Trains Meeting
Question: Train A passes a station at (10\text{ m/s}) and accelerates at (0.5\text{ m/s}^2). Five seconds later, Train B leaves the same station at a constant (25\text{ m/s}). When and where do they meet?
Let (t) be the time in seconds measured from when Train A passes the station.
For Train A:
[
s_A=10t+\frac{1}{2}(0.5)t^2
]
[
s_A=10t+0.25t^2
]
For Train B, which starts (5\text{ s}) later:
[
s_B=25(t-5),\quad t\ge5
]
At a meeting point, their displacements from the station are equal:
[
10t+0.25t^2=25(t-5)
]
Rearrange:
[
0.25t^2-15t+125=0
]
Multiply by four and solve:
[
t^2-60t+500=0
]
[
(t-10)(t-50)=0
]
Therefore, the trains have equal positions at (t=10\text{ s}) and (t=50\text{ s}). The first meeting occurs at (t=10\text{ s}), when Train B has travelled for (5\text{ s}).
[
s_B=25(10-5)=125\text{ m}
]
Answer: They first meet (125\text{ m}) from the station, (10\text{ s}) after Train A passes it. Under the stated motion model, they meet again at (t=50\text{ s}), at (1125\text{ m}).
Quick answer: When solving a multi-stage journey, define the time variable carefully and account for any delayed start before equating displacements.
SUVAT Equations in Vertical Motion and Gravity
For vertical motion near Earth’s surface, gravitational acceleration is approximately (9.81\text{ m/s}^2) downward if air resistance is neglected. The sign of acceleration depends on the direction you choose as positive.
Upward Motion
If upward is positive, gravitational acceleration is:
[
a=-9.81\text{ m/s}^2
]
A ball thrown upward has positive initial velocity if it starts moving upward. At its highest point, its instantaneous vertical velocity is zero, although gravity continues to act downward.
Quick answer: With upward positive, use negative acceleration for gravity and set (v=0) at the highest point of a vertical launch.
Downward Motion
If downward is positive, gravitational acceleration is:
[
a=+9.81\text{ m/s}^2
]
An object released from rest has (u=0). If the object is already moving downward, its initial velocity is positive under this convention.
Quick answer: With downward positive, gravitational acceleration is positive; keep the same sign convention throughout the calculation.
Displacement Versus Distance
Displacement describes the change in position from start to finish, including direction. Distance describes the total length of the path travelled and is not negative.
For example, if a ball travels (15\text{ m}) upward and returns to its launch point, its total distance travelled is (30\text{ m}), but its net displacement is zero.
Quick answer: In SUVAT, (s) represents displacement, not total distance. For motion that reverses direction, separate intervals may be needed to calculate the full journey.
SUVAT Equations and Projectile Motion
Projectile motion can often be analysed by separating horizontal and vertical motion. Under the usual ideal model, air resistance is neglected and gravitational acceleration acts vertically downward.
For horizontal motion, acceleration is zero:
[
a_x=0
]
Therefore:
[
s_x=u_xt
]
For vertical motion, taking upward as positive:
[
a_y=-g
]
The vertical velocity is:
[
v_y=u_y-gt
]
If the projectile is launched at speed (u) and angle (\theta) above the horizontal, the initial components are:
[
u_x=u\cos\theta
]
[
u_y=u\sin\theta
]
The horizontal and vertical calculations share the same elapsed time, but their displacement and velocity components must be treated separately.
Quick answer: Apply SUVAT separately to the horizontal and vertical components of projectile motion, using the appropriate acceleration for each direction.
For more detail, see Projectile Motion Formulas: Range, Height & Time.
SUVAT Equations on Inclined Planes
An object moving along an inclined plane may also be analysed using SUVAT if its acceleration along the slope remains constant. On a frictionless incline at angle (\theta), the component of gravitational acceleration parallel to the slope has magnitude:
[
a=g\sin\theta
]
If the object slides down the slope and friction acts against its motion, a simple model gives:
[
a=g(\sin\theta-\mu\cos\theta)
]
Here, (\mu) is the coefficient of kinetic friction, and the expression assumes the object slides down the incline with the positive direction chosen down the slope. Different motion directions or forces require an appropriate sign convention and force model.
Quick answer: Resolve the forces along the slope first. If the resulting acceleration is constant, use SUVAT with the acceleration along the incline.
When Do SUVAT Equations Not Apply?
The standard SUVAT equations assume constant acceleration. They should not be applied to an entire interval in which acceleration changes continuously unless a suitable approximation or separate-interval analysis is justified.
Examples include:
- Strong air resistance: Drag changes with speed, so acceleration may change during the fall.
- Variable thrust: A rocket’s mass and net force can change during flight.
- Simple harmonic motion: Acceleration depends on displacement from equilibrium, so it is not generally constant over the motion.
- Multi-stage journeys: A vehicle may accelerate, travel at constant speed, and then brake. Each stage may require a separate calculation.
When acceleration changes between stages, calculate the motion over each constant-acceleration interval and use the ending conditions of one stage as the starting conditions for the next.
Quick answer: SUVAT is not a universal formula set for every motion problem; its standard equations require constant acceleration during the interval being solved.
Common SUVAT Exam Mistakes
Avoiding a few common errors can make your solutions more accurate and easier to follow.
- Mixing up initial and final velocity: (u) is the initial velocity, while (v) is the final velocity for the interval.
- Using the wrong sign for acceleration: Deceleration is not automatically negative; its sign depends on the chosen positive direction.
- Confusing distance and displacement: The symbol (s) represents displacement in SUVAT.
- Using inconsistent units: Convert values to compatible units before substituting into a formula.
- Using SUVAT with changing acceleration: Check the constant-acceleration assumption before choosing an equation.
- Choosing an equation before listing known values: Identify what is given and what must be calculated first.
- Ignoring the physical meaning of an answer: Check whether the sign, magnitude, units, and direction make sense for the problem.
Quick answer: Write down the variables, choose a consistent direction, check units, and verify that constant acceleration applies before calculating.
How to Remember the SUVAT Equations
Rather than memorising the equations as unrelated expressions, remember what each one omits. The first omits displacement, the second omits final velocity, the third omits time, the fourth omits acceleration, and the fifth omits initial velocity.
Practise choosing equations from a list of known quantities before doing the arithmetic. This builds the skill needed for unfamiliar exam questions and reduces dependence on memorisation alone.
Quick answer: Learn the five omitted-variable patterns and practise identifying the correct equation from the values given in a question.
Frequently Asked Questions About SUVAT Equations
Are there four or five SUVAT equations?
There are five commonly taught SUVAT equations in the standard set. Some courses or textbooks emphasise four core equations because the fifth can be derived from the others, but the fifth form is still useful when initial velocity is omitted.
Are SUVAT equations the same as kinematic equations?
SUVAT equations are a standard set of kinematic equations for one-dimensional motion with constant acceleration. In other courses, similar relationships may be called the equations of motion or constant-acceleration kinematic equations, with different symbols such as (v_0) for initial velocity.
Can SUVAT equations be used when an object starts from rest?
Yes. Starting from rest means the initial velocity is (u=0). Substitute zero for (u) into a suitable equation, provided acceleration is constant during the interval.
What value of gravity should I use in a SUVAT question?
Use the value specified in the question or by your course. If no value is given, (9.81\text{ m/s}^2) is a common approximation near Earth’s surface; some questions use (9.8\text{ m/s}^2) or (10\text{ m/s}^2).
How do you know whether acceleration is positive or negative?
Choose a positive direction first. Acceleration in that direction is positive, while acceleration in the opposite direction is negative. Keep that convention consistent for displacement and velocity as well.
Can SUVAT equations calculate stopping time?
Yes, if the initial velocity, final velocity, and constant acceleration are known, use (t=(v-u)/a). For a vehicle that stops, set final velocity to zero and use a consistent sign for its deceleration.
What should I do if a SUVAT question gives too few values?
Check whether another relationship, such as a force equation or a condition at the highest point of a vertical launch, supplies additional information. If there is not enough independent information to determine the unknown, the problem cannot be solved uniquely from the stated data alone.
Do SUVAT equations work for negative velocity?
Yes. Negative velocity can represent motion opposite to the chosen positive direction. The equations still apply when acceleration is constant and all signs are used consistently.
Final Summary
The SUVAT equations connect displacement, initial velocity, final velocity, acceleration, and time for motion with constant acceleration. The key to solving problems is to identify the known values, select an equation that contains the required unknown, use consistent units and signs, and check whether the result is physically reasonable.
With a clear equation table, correct rearrangements, and practice on worked examples, SUVAT becomes a systematic method for solving many straight-line motion problems in physics.