Average Velocity

What this formula says

Over a time interval from $t_1$ to $t_2$, a body moves from position $x_1$ to position $x_2$. The average velocity over this interval is:

$\bar{v} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\Delta x}{\Delta t}$

It is the total displacement divided by the total time taken. It does not tell you anything about how the body moved during that interval — only the net result.

Average velocity vs average speed

These are two different things and are often confused.

Average velocity uses displacement — the straight-line distance from start to end, with direction:

$\bar{v} = \frac{\text{displacement}}{\text{time}} = \frac{\Delta x}{\Delta t}$

Average speed uses total path length — the actual distance travelled, regardless of direction:

$\text{average speed} = \frac{\text{total distance}}{\text{time}}$

If a body goes 10 m forward and then 4 m back in 7 seconds:

• Displacement = $+6$ m
• Total distance = $14$ m
• Average velocity = $\frac{6}{7} \approx 0.86$ m/s
• Average speed = $\frac{14}{7} = 2$ m/s

They are equal only when the body never reverses direction.

Derivation

This is a definition, not a derived result. We define average velocity as the ratio of displacement to time elapsed:

$\bar{v} = \frac{\Delta x}{\Delta t}$

where $\Delta x = x_2 - x_1$ is the displacement and $\Delta t = t_2 - t_1$ is the time interval.

The symbol $\Delta$ (delta) always means "final minus initial" — change in a quantity.

When average velocity equals instantaneous velocity

For uniform motion (constant velocity), the velocity is the same at every instant. So the average velocity over any interval equals the instantaneous velocity at every point:

$\bar{v} = v \quad \text{(uniform motion only)}$

For uniformly accelerated motion, the average velocity over an interval equals the instantaneous velocity at the midpoint in time of that interval:

$\bar{v} = v\left(\frac{t_1 + t_2}{2}\right) = \frac{u + v}{2}$

This is the result used to derive the second equation of motion.

Sign of average velocity

Since $\Delta x = x_2 - x_1$:

• If the body ends up to the right of where it started: $\Delta x > 0$, $\bar{v} > 0$
• If the body ends up to the left: $\Delta x < 0$, $\bar{v} < 0$
• If the body returns to its starting point: $\Delta x = 0$, $\bar{v} = 0$ — even if it travelled a large distance