Motion with constant acceleration

Concept & Definition

Calculus handles any motion, but a special case covers an enormous slice of everyday physics: motion in a straight line where the acceleration never changes. Cars pulling away from lights, falling stones, balls thrown straight up — all of these are (to good approximation) constant-acceleration problems.

When $a$ is constant, the average and instantaneous accelerations coincide, and the definitions of acceleration and average velocity can be rearranged into a small set of equations that connect position, velocity, acceleration, and time. These are often called the equations of motion.

The Equations of Motion

With initial position $x_0$, initial velocity $v_0$, constant acceleration $a$, and elapsed time $t$, three equations answer nearly every question:

Equation 1

Velocity from Time

$$v = v_0 + at$$

where $v$ is the velocity at time $t$, $v_0$ is the initial velocity, and $a$ is the constant acceleration.

Equation 2

Position from Time

$$x = x_0 + v_0 t + \tfrac{1}{2} a t^2$$

where $x$ is the position at time $t$ and $x_0$ is the initial position of the particle.

Equation 3

Velocity from Position

$$v^2 = v_0^2 + 2a\,(x - x_0)$$

where $(x - x_0)$ is the displacement from the starting point — the equation to reach for when time is neither given nor asked.

Choosing the Right Equation

List what is given and what is asked. Each equation links four of the five quantities $x$, $v_0$, $v$, $a$, $t$ — pick the one that contains your knowns and your unknown, and leaves out the quantity you neither have nor want.

Common Pitfall

These equations are only valid while $a$ is truly constant. The moment acceleration varies — air resistance at high speed, a rocket burning fuel — they stop applying, and the problem needs the full calculus treatment.

Quick Check

A car starts from rest and accelerates uniformly at 2 m/s². How fast is it moving, and how far has it traveled, after 5 s?

Reveal Explanation

Velocity from Equation 1 with $v_0 = 0$: $v = 0 + (2)(5) = 10\ \text{m/s}$.

Position from Equation 2 with $x_0 = 0$: $x = 0 + 0 + \tfrac{1}{2}(2)(5^2) = 25\ \text{m}$.

Check with Equation 3: $v^2 = 0 + 2(2)(25) = 100$, so $v = 10\ \text{m/s}$ — consistent.