P01-050 – Average Velocity Vector

Average Velocity Vector

Average velocity vector for one-dimensional motion:

While the displacement vector describes how the position vector changes as the particle moves along the $x$-axis, the average velocity vector describes how fast the position vector changes over time.

Consider a particle with a position vector ${\overrightarrow{r}}_i$ at time $t_i$ and a position vector ${\overrightarrow{r}}_f$ at time $t_f$.

The average velocity ${\overrightarrow{v}}_{avg\ x}$ of an object during the time interval $\mathrm{\Delta }t=t_f-t_i$ over which the object undergoes a displacement $\mathrm{\Delta }\overrightarrow{r}=\overrightarrow{r_f}-\overrightarrow{r_i}$, is the vector defined by

\boxed{{\overrightarrow{v}}_{avg\ x}=\frac{\mathrm{\Delta }\overrightarrow{r}}{\mathrm{\Delta }t}=\frac{{\overrightarrow{r}}_f-{\overrightarrow{r}}_i}{t_f-t_i}=\frac{x_f-x_i}{t_f-t_i}\ \hat{x}\ \ \ \ \ \ \ \ (m/s)}

The average velocity represents how fast the particle moves in a given direction and its sign indicates the direction of motion.

If the particle moves in the positive $x$-direction from $x_i$ to $x_f$ in time $\mathrm{\Delta }t=t_f-t_i$ then its average velocity is positive and equal to

v_{avg\ x}=\frac{x_f-x_i}{t_f-t_i}\ > 0

For example, in the figure below, the particle moves in the positive direction from $x_i=-4\ m$ to $x_f=2\ m$ in $\mathrm{\Delta }t=2\ s$ and therefore its velocity is $v_{avg\ x}=3\ m/s$.

If the particle moves in the negative $x$-direction from $x_i$ to $x_f$ in time $\mathrm{\Delta }t=t_f-t_i$ then its average velocity is negative and equal to

v_{avg\ x}=\frac{x_f-x_i}{t_f-t_i} < 0

For example, in the figure below, the particle moves in the negative direction from $x_i=1\ m$ to $x_f=-3\ m$ in $\mathrm{\Delta }t=2\ s$ and therefore its velocity is $v_{avg\ x}=-2\ m/s$.

In both instances, the average velocity is written $\displaystyle{v_{avg\ x}=\frac{x_f-x_i}{t_f-t_i}}$. However, the sign of $v_{avg\ x}$ indicates the direction in which the particle moves. Indeed, in the case where $v_{avg\ x}>0$ the particle moves in the positive $x$ direction while in the case where $v_{avg\ x}<0$ the particle moves in the negative $x$ direction.

Average speed for one-dimensional motion:

The magnitude of velocity is called ”speed” and is a positive number. Therefore, speed does not have a direction, only a value, and the speed of a particle with velocity ${\overrightarrow{v}}_{avg\ x}$ is equal to

\left|{\overrightarrow{v}}_{avg\ x}\right|=\left|\frac{\mathrm{\Delta }\overrightarrow{r}}{\mathrm{\Delta }t}\right|=\frac{\left|x_f-x_i\right|}{\mathrm{\Delta }t}\ \ \ \ \ (m/s)

as long as $x_i \ne x_f$.

Important distinction:

Speed is equal to the total distance divided by the total time i.e.

\boxed{v_{avg\ x}=\frac{total\ distance}{\mathrm{\Delta }t}}

and therefore is never zero if there is any amount of motion. This differs from the average velocity vector which can be zero if the net displacement is zero. For instance, if a runner runs once around a circular track with circumference $400\ m$ in $80\ s$, then her average velocity is zero because she comes back to her starting point. However, her average speed is equal to

v_{avg\ x}=\frac{total\ distance}{\mathrm{\Delta }t}=\frac{400}{80}=5\ \ m/s