The Equations of Motion (suvat)

A car brakes hard from 30 m/s and comes to a stop; a ball is thrown straight up and falls back; a stone is dropped down a well. In every one of these, the acceleration stays constant for the whole journey — the velocity changes at a steady rate. Whenever that is true, you don't need calculus to predict the motion exactly. Four short algebraic equations — the equations of motion, known to every physics student as suvat — tie the whole trip together.

The name is just the five letters they use. Learn what the letters mean, learn to pick the right equation, and you can answer almost any "how fast / how far / how long" question about constant-acceleration motion.

The five variables: s,u,v,a,t

Every constant-acceleration problem is described by these five quantities, in consistent SI units:

s — the displacement (how far it has moved from the start), in metres (m).
u — the initial velocity (the velocity at the moment the clock starts), in m/s.
v — the final velocity (at time t), in m/s.
a — the acceleration, in m/s^2 — and it must be constant.
t — the time for which the motion lasts, in seconds (s).

A typical question hands you three of these and asks for a fourth. That is exactly what the four equations are for — each one links four of the five variables, leaving one out.

The four equations — and where they come from

Everything grows from the two definitions you already met. Acceleration is the rate of change of velocity, so over a time t at constant a:

a = \frac{v - u}{t} \quad\Longrightarrow\quad \boxed{v = u + at}.

Because the acceleration is constant, the velocity rises in a straight line, so the average velocity over the trip is just the mean of the start and end values, \tfrac12(u+v). Displacement is average velocity times time:

s = \tfrac12 (u+v)\,t.

Substitute v = u + at into that and simplify, and you get displacement directly from u, a and t:

s = ut + \tfrac12 a t^2.

Finally, eliminate t between the first two (rearrange v = u + at for t and substitute) to get the equation with no time in it at all:

v^2 = u^2 + 2as.

For motion in a straight line with constant acceleration a, the five quantities are linked by:

v = u + at — no s.
s = ut + \tfrac12 a t^2 — no v.
v^2 = u^2 + 2as — no t.
s = \tfrac12 (u + v)\,t — no a.

Each equation omits exactly one variable — that is the key to choosing between them.

Choosing the right equation

There is a reliable three-step recipe:

List the five letters s,u,v,a,t and write next to each the number you know.
Mark the one you want to find, and the one you neither know nor care about.
Pick the equation that leaves out that unwanted variable — it uses only the three you have plus the one you want.

For example: you know u, a and s, and want v, with no interest in t. The equation without t is v^2 = u^2 + 2as — done. Get comfortable with this and the whole topic becomes bookkeeping.

suvat is one-dimensional but directional: s, u, v and a are all vectors along a line, so each may be positive or negative. Before you start, choose a positive direction and stick with it for the whole problem. If you call "up" positive, then gravity's acceleration is a = -g \approx -9.8\ \text{m/s}^2, an upward throw has u > 0, and a downward displacement comes out negative. Mixing signs part-way through is the single most common way to get a suvat answer wrong.

Worked example 1 — a braking car

A car travels at u = 30\ \text{m/s} when the driver brakes with a constant deceleration of 5\ \text{m/s}^2. Taking the direction of travel as positive, a deceleration is a negative acceleration: a = -5\ \text{m/s}^2. How far does the car travel before stopping, and how long does it take?

Knowns: u = 30, v = 0 (it stops), a = -5. First find the distance s — that leaves out t, so use v^2 = u^2 + 2as:

0 = 30^2 + 2(-5)s \;\Rightarrow\; 0 = 900 - 10s \;\Rightarrow\; s = 90\ \text{m}.

For the time, we now want t and can use the simplest equation with it, v = u + at:

0 = 30 + (-5)t \;\Rightarrow\; t = \frac{30}{5} = 6\ \text{s}.

The car covers 90 m and takes 6 s to stop.

Worked example 2 — a ball thrown straight up

A ball is thrown vertically upward at u = 20\ \text{m/s}. Taking up as positive, gravity acts downward, so a = -g = -9.8\ \text{m/s}^2 throughout the flight. How high does it rise, and how long is it in the air before it returns to the thrower's hand?

Highest point: at the very top the ball is instantaneously at rest, so v = 0 there. Want s, don't want t — use v^2 = u^2 + 2as:

0 = 20^2 + 2(-9.8)s \;\Rightarrow\; s = \frac{400}{19.6} \approx 20.4\ \text{m}.

Time to return: back at the start its displacement is s = 0 (it has come back to where it left). Use s = ut + \tfrac12 a t^2:

0 = 20t + \tfrac12(-9.8)t^2 = t\,(20 - 4.9t) \;\Rightarrow\; t = 0 \text{ or } t = \frac{20}{4.9} \approx 4.1\ \text{s}.

The root t = 0 is the launch; the physical answer is t \approx 4.1\ \text{s} in the air, rising about 20.4 m. Notice the time up equals the time down.

Worked example 3 — free fall from rest

A stone is dropped (not thrown) from the top of a cliff and falls freely for t = 3\ \text{s} before hitting the sea. "Dropped" means it starts from rest, u = 0. Take down as positive this time, so a = +g = 9.8\ \text{m/s}^2. How high is the cliff, and how fast is the stone travelling on impact?

s = ut + \tfrac12 a t^2 = 0 + \tfrac12(9.8)(3^2) = 44.1\ \text{m}.

v = u + at = 0 + 9.8 \times 3 = 29.4\ \text{m/s}.

The cliff is about 44 m high and the stone hits the water at roughly 29 m/s (over 100 km/h). Because we chose "down" as positive here, both a and s came out positive — the signs stay tidy as long as you pick a direction and commit to it.

See it move: the velocity–time graph

Because the acceleration is constant, a graph of velocity against time is a straight line: it starts at u and climbs (or falls) at a steady rate. That line hides both of the first two equations inside its geometry:

its gradient is the acceleration a — this is v = u + at;
the area beneath it (a trapezium) is the displacement s = \tfrac12(u+v)t.

Drag the sliders for the initial velocity u and acceleration a, and slide the time t to watch the shaded area (the displacement so far) and the final velocity grow.

The equations only work for CONSTANT acceleration. If the acceleration changes during the motion — a car whose braking eases off, a rocket burning fuel, anything with air resistance that grows with speed — suvat is simply invalid. Split the journey into constant-acceleration stages, or reach for calculus.
"At rest" means v = 0, not a = 0. A ball at the very top of its flight is momentarily stationary, but gravity is still pulling it down: its acceleration is still -g. That is exactly why it doesn't hang there — it immediately speeds up downward.
Watch the signs. Choose one positive direction at the start. A downward displacement, a deceleration, or a velocity pointing "the wrong way" must go in as a negative number. Feed a deceleration in as a positive a and your car will speed up forever.

In an emergency stop, your stopping distance has two parts. During your reaction time — typically about 0.7\ \text{s} — the car hasn't slowed at all, so a = 0 and you travel at a steady speed: "thinking distance" = u \times t_{\text{react}}. At a motorway speed of 30\ \text{m/s} that is over 20 metres covered before the brakes bite. Then the constant-deceleration suvat phase begins and adds the "braking distance". This is the physics behind the numbers in the Highway Code — and Galileo worked out the constant-acceleration part four centuries ago by rolling balls down gentle inclines and timing them with a water clock, precisely because free fall was too fast for him to measure directly.