Speed and Velocity

Definition: Average Speed

Average speed is the distance ($D$) travelled divided by the time ($\Delta t$) taken for the journey.

Quantity: average speed (${v}_{av}$) Unit name: metre per second Unit symbol: $\text{m·s$^{-1}$}$

Definition: Average Velocity

Average velocity is the change in position of a body divided by the time it took for the displacement to occur.

Quantity: average velocity ($\vec{v}_{av}$) Unit name: metre per second Unit symbol: $\text{m·s$^{-1}$}$

Before moving on review the difference between distance and displacement. Sometimes the average speed can be a very big number while the average velocity is zero.

Average velocity is the rate of change of position. It tells us how much an object's position changes per unit of time. Velocity is a vector. We use the symbol $\vec{v}_{av}$ for average velocity. If we have a displacement of $\Delta \vec{x}$ and a time taken of $\Delta t$, $\vec{v}_{av}$ is then defined as:

\begin{align*} \text{average velocity (in } \text{m·s$^{-1}$} \text{) }& = \frac{\text{change in position (in } \text{m} \text{)}}{\text{change in time (in } \text{s} \text{)}} \\ \vec{v}_{av} & = \frac{\Delta \vec{x}}{\Delta t} \end{align*}

Velocity can be positive or negative. A positive velocity points in the direction you chose as positive in your coordinate system. A negative velocity points in the direction opposite to the positive direction.

Average speed (symbol ${v}_{av}$) is the distance travelled ($D$) divided by the time taken ($\Delta t$) for the journey. Distance and time are scalars and therefore speed will also be a scalar. Speed is calculated as follows:

\begin{align*} \text{average speed (in } \text{m·s$^{-1}$} \text{) }& = \frac{\text{change in position (in } \text{m} \text{)}}{\text{change in time (in } \text{s} \text{)}} \\ v_{av} & = \frac{D}{\Delta t} \end{align*}

Example: Average Speed and Average Velocity

Question

James walks $\text{2}$ $\text{km}$ away from home in $\text{30}$ minutes. He then turns around and walks back home along the same path, also in $\text{30}$ minutes. Calculate James' average speed and average velocity.

Step 1: Identify what information is given and what is asked for

The question explicitly gives

the distance and time out ($\text{2}$ $\text{km}$ in $\text{30}$ minutes)
the distance and time back ($\text{2}$ $\text{km}$ in $\text{30}$ minutes)

Step 2: Check that all units are SI units.

The information is not in SI units and must therefore be converted.

To convert $\text{km}$ to $\text{m}$, we know that:

\begin{align*} \text{1}\text{ km} & = \text{1 000}\text{ m} \\ \therefore \text{2}\text{ km} & = \text{2 000}\text{ m} \text{ (multiply both sides by } 2 \end{align*}

Similarly, to convert $\text{30}$ minutes to seconds,

\begin{align*} \text{1}\text{ min} & = \text{60}\text{ s} \\ \therefore \text{30}\text{ min} & = \text{1 800}\text{ s} \text{ (multiply both sides by } 20 \end{align*}

Step 3: Determine James' displacement and distance.

James started at home and returned home, so his displacement is $\text{0}$ $\text{m}$.

$\Delta \vec{x} = \text{0}\text{ m}$

James walked a total distance of $\text{4 000}$ $\text{m}$ ($\text{2 000}$ $\text{m}$ out and $\text{2 000}$ $\text{m}$ back).

$D = \text{4 000}\text{ km}$

Step 4: Determine his total time.

James took $\text{1 800}$ $\text{s}$ to walk out and $\text{1 800}$ $\text{s}$ to walk back.

$\Delta t = \text{3 600}\text{ s}$

Step 5: Determine his average speed

\begin{align*} v_{av} & = \frac{D}{\Delta t} \\ & = \frac{\text{4 000}\text{ m}}{\text{3 600}\text{ km}} \\ & = \text{1,11}\text{ m·s$^{-1}$} \end{align*}

Step 6: Determine his average velocity

\begin{align*} \vec{v}_{av} & = \frac{\Delta \vec{x}}{\Delta t} \\ & = \frac{\text{0}\text{ m}}{\text{3 600}\text{ km}} \\ & = \text{0}\text{ m·s$^{-1}$} \end{align*}

Speed and Velocity