# FHSST Physics/Rectilinear Motion/Speed and Velocity

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# 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

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

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

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

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

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

Using the rule of Pythagoras:

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

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

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

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

and his velocity would be

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

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.