SUVAT Equations — The Complete Physics Guide
The SUVAT equations are five kinematic relationships that describe motion under constant acceleration. They connect five variables — displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t) — and each equation relates four of the five, omitting one. Given any three variables, you can always determine the remaining two. SUVAT is the essential toolkit for A-level mechanics and first-year university physics.
The name is an acronym: s (displacement), u (initial velocity), v (final velocity), a (acceleration), t (time). These equations apply whenever acceleration is constant — a condition that covers free fall, motion on frictionless surfaces, and many standard exam scenarios. When acceleration varies, calculus-based methods or numerical integration are required.
The Five Equations and Their Derivation
All five equations derive from two definitions: acceleration a = (v − u)/t and displacement = average velocity × time = ½(u + v)t. Everything else follows by algebraic substitution.
The geometric derivation is elegant and worth understanding: on a velocity-time graph with constant acceleration, the graph is a straight line from u to v over time t. Displacement equals the area under this graph — a trapezoid with parallel sides u and v and height t. Area = ½(u + v)t, giving the fourth equation directly. The triangle area ½at² and rectangle area ut combine to give the second equation.
How to Choose the Right Equation
The systematic approach: identify the three known variables, identify the one unknown you want to find, and pick the equation that omits the fifth (irrelevant) variable. This works every time without confusion.
Example: you know initial velocity u, acceleration a, and time t. You want displacement s, and you don't know (and don't need) final velocity v. Pick the equation that omits v: s = ut + ½at². Simple, systematic, no guesswork.
Common mistake: students sometimes pick v² = u² + 2as when they need t, then wonder why they can't extract time. This happens because they don't first identify which variable they want. Always state s, u, v, a, t with known/unknown labels before picking an equation.
Worked Example 1 — Braking Car
Problem: A car moving at 25 m/s brakes with a constant deceleration of 5 m/s². How long does it take to stop, and what distance is covered during braking?
Known: u = 25 m/s, v = 0 m/s (stops), a = −5 m/s² (deceleration, so negative)
Time (omits s): v = u + at → 0 = 25 − 5t → t = 25/5 = 5 s
Distance (omits t, or use t found above): v² = u² + 2as → 0 = 625 − 10s → s = 62.5 m
Check: s = ½(u+v)t = ½ × 25 × 5 = 62.5 m ✓. Note: at 50 m/s (double the speed), deceleration would need 4× the distance (250 m) to stop — the v² relationship means stopping distance quadruples with doubled speed.
Worked Example 2 — Projectile (Vertical Component)
Problem: A ball is thrown vertically upward with initial speed 20 m/s. Using g = 9.81 m/s², find: (a) the maximum height, (b) the time to reach maximum height, (c) the total time in the air.
Take upward as positive. a = −9.81 m/s² (gravity acts downward).
(a) Max height (v = 0 at peak): v² = u² + 2as → 0 = 400 + 2(−9.81)s → s = 400/19.62 = 20.4 m
(b) Time to peak (v = 0): v = u + at → 0 = 20 − 9.81t → t = 20/9.81 = 2.04 s
(c) Total time (s = 0 when returns to launch height): 0 = 20t − ½(9.81)t² → 0 = t(20 − 4.905t) → t = 0 (launch) or t = 20/4.905 = 4.08 s — exactly double the time to peak, confirming symmetry.
Worked Example 3 — Finding Initial Velocity
Problem: A ball bearing is released from rest and falls 4.9 m under gravity. Find the time of fall and the speed on impact.
Known: u = 0 (dropped from rest), s = 4.9 m, a = 9.81 m/s² (taking downward positive).
Time (omits v): s = ut + ½at² → 4.9 = 0 + ½(9.81)t² → t² = 4.9/4.905 = 0.999 → t = 1.0 s
Speed (omits t): v² = u² + 2as = 0 + 2(9.81)(4.9) = 96.1 → v = 9.80 m/s
SUVAT in Two Dimensions — Projectile Motion
Projectile motion applies SUVAT independently in the horizontal and vertical directions. This works because horizontal and vertical motions are completely independent — horizontal acceleration is zero (no horizontal force in ideal projectile motion), while vertical acceleration is g downward.
Horizontal: s_x = u_x · t (constant velocity, a_x = 0). Vertical: all five SUVAT equations apply with a_y = −g = −9.81 m/s². Both components share the same time variable — this is the link that ties them together. The standard approach: find the time of flight from the vertical equations, then use that time in the horizontal equation to find range.
At launch angle θ and speed v₀: u_x = v₀cosθ, u_y = v₀sinθ. Range R = u_x × T where T = 2u_y/g = 2v₀sinθ/g. So R = v₀cosθ × 2v₀sinθ/g = v₀²sin(2θ)/g. Maximum range at θ = 45° where sin(2θ) = sin(90°) = 1.
When SUVAT Doesn't Apply
SUVAT requires truly constant acceleration. The equations break down when: air resistance makes drag force (and therefore acceleration) dependent on velocity; a spring provides a restoring force proportional to position; a rocket's thrust-to-mass ratio changes as fuel burns; any time-varying or position-dependent force acts.
Exam tip: if a problem mentions air resistance, drag, or a spring and asks you to solve for motion, SUVAT cannot be applied directly. In these cases, energy methods (work-energy theorem) or differential equations are required. Most A-level problems explicitly state "ignore air resistance" — that is your signal that SUVAT is safe to use.
Historical Background
The kinematic equations were developed during the Scientific Revolution. Galileo Galilei derived the key result s ∝ t² for free fall in the early 17th century, using inclined planes to slow the motion to a measurable rate. By timing balls rolling down ramps with a water clock, he established that displacement increases as the square of time — a hallmark of constant acceleration.
Newton's Second Law (F = ma) provided the theoretical foundation that Galileo lacked: constant net force on constant mass gives constant acceleration, and SUVAT is the mathematical consequence. The equations themselves are a direct result of integrating constant acceleration with respect to time — a fact that becomes transparent when you study calculus-based mechanics.