It is often important to know both the speed of an object and the direction in which it is moving.
Speed and direction are combined in another quantity, called velocity. The velocity of an object can be thought of as its speed in a particular direction. So, like displacement, velocity is a vector quantity. Speed is the corresponding scalar quantity, because it does not have a direction.
So, to give the velocity of something, we have to state the direction in which it is moving. For example, ‘an aircraft flies with a velocity of $300m\,{s^{ - 1}}$ due north’.
Since velocity is a vector quantity, it is defined in terms of displacement:

$Velocity = \frac{{change\,in\,displacement}}{{time\,taken}}$
We can write the equation for velocity in symbols:
$v = \frac{s}{t}$

Alternatively, we can say that velocity is the rate of change of an object’s displacement:

$v = \frac{{\Delta s}}{{\Delta t}}$

where the symbol $\Delta $ (the Greek letter delta) means ‘change in’. It does not represent a quantity (in the way that s and t do). Another way to write $\Delta s$ would be ${s_2} - {s_1}$, but this is more time-consuming and less clear.
From now on, you need to be clear about the distinction between velocity and speed, and between displacement and distance. Table 1.2 shows the standard symbols and units for these quantities.

Table 1.2: Standard symbols and units. (Take care not to confuse italic s for displacement with s for seconds. Notice also that v is used for both speed and velocity.)

Symbol for unit	Symbol for quantity	Quantity
m	d	distance
m	s, x	displacement
s	t	time
$m\,{s^{ - 1}}$	v	speed, velocity

Speed and velocity calculations

The equation for velocity, $v = \frac{{\Delta s}}{{\Delta t}}$ , can be rearranged as follows, depending on which quantity we want to determine:
change in displacement $\Delta s = v \times \Delta t$
change in time $\Delta t = \frac{{\Delta s}}{v}$
Note that each of these equations is balanced in terms of units. For example, consider the equation for displacement. The units on the right-hand side are $m\,{s^{ - 1}} \times s$, which simplifies to m, the correct unit for displacement.
We can also rearrange the equation to find distance s and time t:

$\begin{array}{l}
\Delta s = v \times t\\
t = \frac{{\Delta s}}{v}
\end{array}$

Making the most of units

In Worked example 1 and Worked example 2, units have been omitted in intermediate steps in the calculations. However, at times it can be helpful to include units as this can be a way of checking that you have used the correct equation; for example, that you have not divided one quantity by another when you should have multiplied them. The units of an equation must be balanced, just as the numerical values on each side of the equation must be equal.
If you take care with units, you should be able to carry out calculations in non-SI units, such as kilometres per hour, without having to convert to metres and seconds.
For example, how far does a spacecraft travelling at $40000km\,{h^{ - 1}}$ travel in one day? Since there are 24 hours in one day, we have:

$\begin{array}{l}
dis\tan ce = 40000km\,{h^{ - 1}} \times 24h\\
 = 960000km
\end{array}$