$\def \u#1{\,\mathrm{#1}}$ $\def \abs#1{\left|#1\right|}$ $\def \ast{*}$ $\def \deg{^{\circ}}$ $\def \ten#1{\times 10^{#1}}$ $\def \redcancel#1{{\color{red}\cancel{#1}}}$ $\def \BLUE#1{{\color{blue} #1}}$ $\def \RED#1{{\color{red} #1}}$ $\def \PURPLE#1{{\color{purple} #1}}$ $\def \th#1,#2{#1,\!#2}$ $\def \lshift#1#2{\underset{\Leftarrow\atop{#2}}#1}}$ $\def \rshift#1#2{\underset{\Rightarrow\atop{#2}}#1}}$ $\def \dotspot{{\color{lightgray}{\circ}}}$

Average Velocity

The displacement vector is a lot like the velocity vector. Both point in the general direction of the car's motion, and both vectors become larger when the car speeds up. But the two are not exactly the same. Consider the following two motion diagrams showing a car speeding up to the right: both diagrams represent the exact same motion.

Two horizontal motion diagrams describing identical motion, one with dots every 1s, one with dots every 2s

The lower diagram is different from the upper diagram in two ways: its displacement vectors are twice as long, and its time interval $\Delta t$ (the time between successive dots) is also twice as large. If for every displacement vector we defined the ratio $\Delta x/\Delta t$, then both “twices” would cancel out and the ratio would be the same for both diagrams.

Thus we define the average velocity of an object at a given time by the equation

$$\vec v_{avg}={\Delta\vec x\over\Delta t}$$

Because this is a distance divided by a time, velocity (and speed) has units of meters per second (m/s). The average velocity at any moment points in the same direction as the displacement vector, which is the direction of the object's motion.