Velocity: Speed with a Direction is speed with a direction. With positions on an axis, we can make that exact. Over a time interval $\Delta t$, divide the displacement by the time.
Key idea
Average velocity: $v_{avg} = \dfrac{\Delta x}{\Delta t}$, with a sign. Average speed: $\dfrac{\text{total distance}}{\Delta t}$, never negative.
Why it works
The average velocity is the one steady velocity that would carry you from the same start to the same end in the same time. That is why only the displacement matters: a steady motion can't turn around, so it only "sees" where you began and where you finished. Average speed answers a different question: how fast, on average, you covered ground.
Worked examples
Worked example: One leg
A toy car moves from $x = 2\ \text{m}$ to $x = -10\ \text{m}$ in $4\ \text{s}$. Find its average velocity.
- $\Delta x = -10 - 2 = -12\ \text{m}$.
- $v_{avg} = \dfrac{-12\ \text{m}}{4\ \text{s}} = -3\ \text{m/s}$, so it moved in the $-x$ direction at 3 m/s on average.
Answer: $-3\ \text{m/s}$
Worked example: Velocity and speed differ
You walk $40\ \text{m}$ east in $30\ \text{s}$, then $10\ \text{m}$ west in $10\ \text{s}$. Take east as $+x$.
- Displacement: $+40 + (-10) = 30\ \text{m}$. Time: $40\ \text{s}$.
- $v_{avg} = \dfrac{30}{40} = 0.75\ \text{m/s}$ (east).
- Distance: $40 + 10 = 50\ \text{m}$, so the average speed is $\dfrac{50}{40} = 1.25\ \text{m/s}$.
Answer: $v_{avg} = 0.75\ \text{m/s}$; average speed $1.25\ \text{m/s}$.
Worked example: The averaging trap
You drive $60\ \text{km}$ at $30\ \text{km/h}$, then $60\ \text{km}$ at $60\ \text{km/h}$. What is your average speed?
- Times: $\dfrac{60}{30} = 2\ \text{h}$ and $\dfrac{60}{60} = 1\ \text{h}$.
- Average speed $= \dfrac{120\ \text{km}}{3\ \text{h}} = 40\ \text{km/h}$.
- Not $45\ \text{km/h}$: you spent twice as long at the slow speed, so it counts twice as much.
Answer: $40\ \text{km/h}$
Common mistake
Averaging the starting and ending speeds, or the speeds of the legs, as if each counted equally. That only works when each speed lasts the same amount of time (or when the velocity changes at a steady rate). Safe method: always go back to the definition, total displacement (or distance) divided by total time.