# FHSST Physics/Rectilinear Motion/Speed and Velocity

Rectilinear Motion The Free High School Science Texts: A Textbook for High School Students Studying Physics Main Page - << Previous Chapter (Forces) - Next Chapter (Momentum) >> Definition - Speed and Velocity - Graphs - Equations of Motion - Important Equations and Quantities

# Speed and Velocity

Let's take a moment to review our definitions of velocity and speed by looking at the worked example below:

## Worked Example 23 Speed and Velocity

Question: A cyclist moves from A through B to C in 10 seconds. Calculate both his speed and his velocity.

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

• the distance between A and B
• the distance between B and C
• the total time for the cyclist to go from A through B to C

all in the correct units!

Step 2 :

What is being asked? We are asked to calculate the average speed and the average velocity of the cyclist.

His speed - a scalar - will be

$\begin{matrix}v&=&\frac{s}{t}\\&=&\frac{30m+40m}{10s}\\&=&7\frac{m}{s}\end{matrix}$

Since velocity is a vector we will first need to find the resultant displacement of the cyclist. His velocity will be

$\begin{matrix}\overrightarrow{v}=\frac{\overrightarrow{s}}{t}\end{matrix}$

The total displacement is the vector from A to C, and this is just the resultant of the two displacement vectors, i.e.

$\begin{matrix}\overrightarrow{s} = \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}\end{matrix}$

Using the rule of Pythagoras:

$\begin{matrix}\overrightarrow{s} &=& \sqrt{{(30m)}^2+{(40m)}^2}\\&=& 50m \ in \ the \ direction \ from \ A \ to \ C\end{matrix}$

$\begin{matrix}\overrightarrow{v} &=&\frac{50\ \mbox{m}}{10\ \mbox{s}}\\&=& 5\ \frac{\mbox{m}}{\mbox{s}} \ in\ the \ direction \ from \ A \ to \ C\end{matrix}$

For this cyclist, his velocity is not the same as his speed because there has been a change in the direction of his motion. If the cyclist traveled directly from A to C without passing through B his speed would be

$\begin{matrix}v &=& \frac{50m}{10s}\\&=& 5 \frac{m}{s}\end{matrix}$

and his velocity would be

$\begin{matrix}\overrightarrow{v} &=& \frac{50m}{10s}\\&=& 5 \frac{m}{s} \ in \ the\ direction \ from \ A \ to \ C\end{matrix}$

In this case where the cyclist is not undergoing any change of direction (i.e. he is traveling in a straight line) the magnitudes of the speed and the velocity are the same. This is the defining principle of rectilinear motion.

 Important: For motion along a straight line the magnitudes of speed and velocity are the same, and the magnitudes of the distance and displacement are the same.