What Are the SUVAT Equations of Motion (Kinematics Equations)?

SL and HL: The SUVAT equations relate displacement, initial velocity, final velocity, acceleration and time when motion occurs with uniform acceleration. The IB data booklet gives four core forms; a fifth commonly taught form follows directly from them.

The Equations and Variables

The letters represent:

The equations are:

Select the equation that contains the known quantities and the required unknown. These equations apply only when acceleration is constant; they should not be used directly for changing acceleration.

These equations follow from the definitions of acceleration and average velocity. With constant acceleration, $a=(v-u)/t$, which rearranges to $v=u+at$. The velocity-time graph is a straight line, so displacement equals the area under it: $s=\frac{u}{}$. Substituting gives ; eliminating time gives . This mechanism explains why every equation depends on constant acceleration.

Before substituting, list all five quantities and mark the unknown. Choose the row omitting the quantity you neither know nor need. If acceleration changes, use a graph, calculus or a suitable numerical model instead. On a velocity-time graph, the gradient is acceleration and the signed area is displacement.

Velocity, displacement and acceleration are vector quantities, so signs must follow a declared positive direction. For vertical motion, if upward is positive, acceleration due to gravity is $a=-9.8\,\text{m s}^{-2}$.

For example, a ball is projected upward at $20\,\text{m s}^{-1}$. At maximum height, $v=0$, so

$$0^2=20^2+2(-9.8)s$$

and therefore $s=20.4\,\text{m}$.

A common misconception is that $s$ means distance. It means displacement, which can be negative. For projectile motion, apply the equations separately to horizontal and vertical components; without fluid resistance, horizontal acceleration is zero.

Exam Technique

In a Calculate question, state the positive direction, identify the known variables, select an equation, substitute values and give the result. Do not assume that acceleration is positive, and remember that $v=0$ at maximum height but $a\neq0$.