We're actually going to talk about what is normally called the fourth kinematic equation, before we discuss the third, due to it being a bit simpler in nature.

When an object has constant acceleration, its velocity changes at a steady rate. This creates a straight line on a velocity-time graph, which leads to a surprisingly simple and powerful result.

The Average Velocity Formula

For constant acceleration, the average velocity equals the arithmetic mean of the initial and final velocities:

Two Equivalent Forms

Since average velocity is also defined as displacement divided by time, we have:

$$\bar{v} = \dfrac{\Delta x}{t} = \dfrac{x - x_0}{t}$$

Combining these two expressions for average velocity gives us:

$$\dfrac{x - x_0}{t} = \dfrac{v_i + v_f}{2}$$

Both sides equal $\bar{v}$, which means we can use either form depending on what information we have:

• If we know initial and final velocities, use $\bar{v} = \dfrac{v_i + v_f}{2}$
• If we know displacement and time, use $\bar{v} = \dfrac{x - x_0}{t}$
• Or, if we know $\bar{v}$ and $v_{i}$ for example, we can get $v_{f}$.

Finding Displacement

Rearranging the equation above, we get a useful formula for displacement:

$$x - x_0 = \dfrac{v_i + v_f}{2} \cdot t$$

or simply:

$$\Delta x = \dfrac{v_i + v_f}{2} \cdot t$$

Why This Works

For constant acceleration, velocity changes uniformly from $v_i$ to $v_f$. The average is simply the midpoint between them—exactly like finding the average of any two numbers.

On a velocity-time graph:

• The velocity trace forms a straight line (constant slope = constant acceleration)
• The displacement equals the area under this line
• This area forms a trapezoid with parallel sides of heights $v_i$ and $v_f$
• The area of a trapezoid is $\frac{v_i + v_f}{2} \cdot t$, which matches our displacement formula