SUVAT Equations: The Complete Kinematic Equations Guide

A car brakes from 30 m/s to rest in 50 m. How long does it take? A ball is thrown upward at 15 m/s — how high does it go? A stone is dropped from a cliff — how fast is it moving after 3 seconds? These questions all involve motion with constant acceleration, and they are all solved with the same four equations: the SUVAT equations. Named after the five variables they connect — s, u, v, a, t — these are among the most used equations in all of physics and engineering, forming the foundation of kinematics.

The SUVAT Variables

s = displacement (m) — the vector distance from start to end position (not total path length)
u = initial velocity (m/s) — speed at the start of the time interval
v = final velocity (m/s) — speed at the end of the time interval
a = acceleration (m/s²) — constant throughout (SUVAT only applies for constant a)
t = time (s) — duration of the motion

Know any three → find the remaining two.

The Four SUVAT Equations

v = u + at [1]

s = ut + ½at² [2]

v² = u² + 2as [3]

s = ½(u + v)t [4]

Each equation uses four of the five variables and omits one — making them easy to select: identify which variable you don't know and don't need, then pick the equation that omits it.

Equation	Variables used	Variable omitted	Use when
v = u + at	v, u, a, t	s	No displacement given/needed
s = ut + ½at²	s, u, a, t	v	Final velocity not given/needed
v² = u² + 2as	v, u, a, s	t	Time not given/needed
s = ½(u+v)t	s, u, v, t	a	Acceleration not given/needed

Deriving the SUVAT Equations

Equation 1 (v = u + at) is the definition of constant acceleration: a = (v−u)/t, rearranged.

Equation 4 (s = ½(u+v)t) comes from the area under a v-t graph. For constant acceleration, the v-t graph is a straight line from u to v — a trapezium with area = ½(u+v) × t = displacement.

Equation 2 (s = ut + ½at²) is found by substituting v = u + at into equation 4: s = ½(u + u + at)t = ½(2u + at)t = ut + ½at².

Equation 3 (v² = u² + 2as) is found by eliminating t: from eq. 1, t = (v−u)/a. Substituting into eq. 4: s = ½(u+v)(v−u)/a = (v²−u²)/(2a), giving v² = u² + 2as.

Sign Conventions — The Most Common Source of Errors

SUVAT variables are signed — positive or negative depending on direction. Before solving any problem:

1. Choose a positive direction (e.g. upward, or the direction of initial motion). Stick to it throughout.

2. Assign signs consistently: if upward is positive, then for a thrown ball u = +15 m/s, a = −9.8 m/s² (gravity acts downward).

3. A negative s means displacement is opposite to your chosen positive direction. A negative v means moving in the negative direction. Neither is "wrong" — it is just directional information.

Common Mistake: Confusing Displacement and Distance

s in SUVAT is displacement — the vector from start to end. A ball thrown upward and returning to its starting point has s = 0, not s = 2 × maximum height. If you want the total distance travelled (not displacement), you must split the motion at any turning points and calculate each phase separately.

Worked Examples

Example 1: Braking car

A car decelerates from 30 m/s to rest in a distance of 45 m. Find the deceleration and the time taken.

Known: u = 30 m/s, v = 0, s = 45 m. Find: a and t.

v² = u² + 2as → 0 = 900 + 2a(45) → a = −900/90 = −10 m/s²

v = u + at → 0 = 30 + (−10)t → t = 3.0 s

Example 2: Ball thrown upward

A ball is thrown vertically upward at 20 m/s. Taking upward as positive, a = −9.8 m/s². Find: (a) maximum height; (b) time to return to the starting point.

(a) At maximum height, v = 0:

v² = u² + 2as → 0 = 400 − 19.6s → s = 400/19.6 = 20.4 m

(b) When ball returns, s = 0:

s = ut + ½at² → 0 = 20t − 4.9t² → t(20 − 4.9t) = 0 → t = 0 or t = 4.08 s

Example 3: Freely falling object

A stone is dropped from rest off a 80 m cliff. Find the time to hit the ground and the speed of impact.

Taking downward as positive: u = 0, a = 9.8 m/s², s = 80 m.

s = ut + ½at² → 80 = 0 + ½(9.8)t² → t² = 16.33 → t = 4.04 s

v² = u² + 2as = 0 + 2(9.8)(80) = 1568 → v = 39.6 m/s

Example 4: Rocket launch

A rocket accelerates from rest at 15 m/s² for 8 seconds. Find displacement and final speed.

v = u + at = 0 + 15 × 8 = 120 m/s

s = ut + ½at² = 0 + ½(15)(64) = 480 m

SUVAT and Projectile Motion

Projectile motion applies SUVAT independently to horizontal and vertical components (using velocity as a vector):

Horizontal: a = 0, so s_x = u_x t (constant velocity — no SUVAT needed beyond this)

Vertical: a = −g = −9.8 m/s², so apply all four SUVAT equations with u_y and a = −9.8 m/s²

The two components share the same time t — this is the coupling that links the horizontal and vertical motion.

When SUVAT Does NOT Apply

SUVAT requires constant acceleration throughout the time interval. It does not apply when:

• Acceleration varies with time (e.g. a rocket burning fuel as mass decreases)

• Acceleration varies with position (e.g. a falling object with significant air resistance, where drag ∝ v²)

• The motion changes direction mid-interval without you accounting for it (split the interval at the turning point)

For variable acceleration, calculus is required: v = ∫a dt, s = ∫v dt.

Frequently Asked Questions

What are the SUVAT equations?

The four SUVAT equations are: (1) v = u + at; (2) s = ut + ½at²; (3) v² = u² + 2as; (4) s = ½(u+v)t. They relate displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t) for motion with constant acceleration. Know any three variables to find the remaining two.

When can SUVAT equations be used?

SUVAT applies only when acceleration is constant throughout the motion. This includes free fall (a = g, ignoring air resistance), car braking at constant deceleration, and constant-thrust rocket motion. It does not apply when acceleration varies with time or position.

Which SUVAT equation should I use?

Identify which variable you neither have nor need — then choose the equation that omits that variable. 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.

What does s represent in SUVAT?

s is displacement — the vector change in position from start to end. It is not the total distance travelled. If an object goes up then comes back down, s may be zero even though it has travelled a large distance. For objects that change direction, split the motion at the turning point and apply SUVAT to each phase separately.

What is the acceleration due to gravity in SUVAT problems?

Near Earth's surface, g = 9.8 m/s² directed downward. In SUVAT, assign a = −9.8 m/s² when upward is positive, or a = +9.8 m/s² when downward is positive. Always define your positive direction before starting and stay consistent. Objects in free fall have a = ±g throughout (if air resistance is neglected).