Kinematics defines four quantities that fully describe any 1D motion. Everything else is an equation relating them.
sdisplacementm
uinitial velocitym s⁻¹
vfinal velocitym s⁻¹
aaccelerationm s⁻²
ttime elapseds
Displacement ≠ distance. Displacement is a vector — it carries direction. If you walk 5 m forward then 5 m back, distance = 10 m but displacement = 0 m.
Velocity and speed carry the same distinction: velocity is displacement per unit time (vector), speed is distance per unit time (scalar).
Average acceleration. Slope of the v–t graph between two points.
Instantaneous velocity is the limit as $\Delta t \to 0$, which is the derivative $\dfrac{ds}{dt}$. You don't need calculus to use SUVAT, but understanding this link deepens everything.
§ 2
Uniform motion
When no net force acts, velocity is constant. The displacement grows linearly with time.
UNIFORM MOTION
$$s = vt$$
On an $s$–$t$ graph this is a straight line. The slope of that line is the velocity. A steeper line = faster object. A horizontal line = stationary object. A negative slope = moving backwards.
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Slope = velocity
$v = \Delta s / \Delta t$ — rise over run on any $s$–$t$ graph gives velocity at that segment.
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Zero acceleration
Constant velocity means $a = 0$. The $v$–$t$ graph is a flat horizontal line.
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Area = displacement
The area under any $v$–$t$ graph equals displacement. Here it's a rectangle: $s = v \cdot t$.
§ 3
Uniform acceleration
A constant net force produces constant acceleration. Velocity now changes at a fixed rate, so the $v$–$t$ graph becomes a straight line with non-zero slope.
The two primary equations follow from the definitions of velocity and acceleration by simple rearrangement — no calculus needed.
Average velocity under constant acceleration is the mean of start and end velocities
$s = \bar{v} \cdot t = \left(u + \tfrac{1}{2}at\right) t$
Displacement = average velocity × time (area under $v$–$t$ trapezoid)
$$\boxed{s = ut + \tfrac{1}{2}at^2}$$
Expand. Two terms: the uniform-motion part $ut$, plus an extra quadratic contribution from acceleration.
Notice that the $s$–$t$ graph is now a parabola. The curvature of that parabola encodes $a$ — more curved means larger acceleration.
THIRD EQUATION (eliminates t)
$$v^2 = u^2 + 2as$$
Obtained by substituting $t = (v-u)/a$ into $s = ut + \frac{1}{2}at^2$ and simplifying. Useful when time is unknown.
§ 4
The SUVAT equations
There are five kinematic equations. Each one omits exactly one of the five variables $\{s, u, v, a, t\}$. To solve any problem: identify which variable is missing, pick the equation that doesn't contain it.
Equation
Missing variable
Form
$v = u + at$
$s$
Velocity as function of time
$s = ut + \tfrac{1}{2}at^2$
$v$
Displacement as function of time
$v^2 = u^2 + 2as$
$t$
Velocity as function of displacement
$s = \tfrac{1}{2}(u+v)t$
$a$
Displacement as mean velocity × time
$s = vt - \tfrac{1}{2}at^2$
$u$
Displacement from final velocity
Strategy: write down the 3 variables you know and the 1 you want. The missing variable (the 5th) tells you which equation to use.
§ 5
Projectile motion
A projectile is launched with speed $v_0$ at angle $\theta$ above the horizontal. The key insight — and the entire reason the maths works out cleanly — is that horizontal and vertical motion are completely independent.
RESOLVING INTO COMPONENTS
$$u_x = v_0 \cos\theta \qquad a_x = 0$$
No horizontal force → constant horizontal velocity.
$$u_y = v_0 \sin\theta \qquad a_y = -g$$
Gravity acts downward at $g \approx 9.8\ \text{m s}^{-2}$.
Position at time $t$
$$x(t) = v_0 \cos\theta \cdot t \qquad y(t) = v_0 \sin\theta \cdot t - \tfrac{1}{2}g t^2$$
Key results
TIME OF FLIGHT
$$T = \frac{2\, v_0 \sin\theta}{g}$$
Set $y = 0$ and solve for $t$. The trajectory is symmetric — time to peak equals time to fall back down.
MAXIMUM HEIGHT
$$H = \frac{v_0^2 \sin^2\!\theta}{2g}$$
At peak, $v_y = 0$. Use $v^2 = u^2 + 2as$ in the vertical direction.
HORIZONTAL RANGE
$$R = \frac{v_0^2 \sin 2\theta}{g}$$
Substitute $T$ into $x(T)$. Since $\sin 2\theta$ is maximised at $2\theta = 90°$, maximum range occurs at $\theta = 45°$.
The parabola emerges naturally. Eliminate $t$ from $x(t)$ and $y(t)$: substitute $t = x / (v_0 \cos\theta)$ into the $y$ equation. The result is $y = x\tan\theta - \dfrac{g}{2v_0^2\cos^2\!\theta}\,x^2$ — a parabola in $x$ and $y$.
§ 6
Live simulation
Adjust the parameters and watch the equations act. Every readout is computed from the SUVAT equations above — the simulation is the maths.