9. In 2018 , a racing driver was allowed to use the Drag Reduction System provided that the car was within 1 second of the car ahead. Suppose that two cars were 1 second apart, each travelling at $180 \mathrm{~km} / \mathrm{h}$ (in the same direction!).
How many metres apart were they?
A 100
B 50
C 10
D 5
E 1
Solution
B
The distance apart of the cars was the distance that a car travelling at $180 \mathrm{~km} / \mathrm{h}$ travels in 1 second. There are 1000 metres in one kilometre. Hence there are $180 \times 1000$ metres in $180 \mathrm{~km}$.

There are $60 \times 60$ seconds in each hour.
It follows that at $180 \mathrm{~km} / \mathrm{h}$ a car travels $180 \times 1000$ metres in $60 \times 60$ seconds.
Therefore the number of metres that it travels in 1 second is
$$
\frac{180 \times 1000}{60 \times 60}=\frac{3 \times 1000}{60}=\frac{1000}{20}=50 \text {. }
$$

Therefore when the cars are 1 second apart, they are 50 metres apart.
FOR INVESTIGATION
9.1 The Highway Code gives the following table of typical stop ping distances in metres for motor vehicles travelling at different velocities.
\begin{tabular}{rc}
\hline velocity & stopping distance \\
\hline $32 \mathrm{~km} / \mathrm{h}$ & $12 \mathrm{~m}$ \\
$48 \mathrm{~km} / \mathrm{h}$ & $23 \mathrm{~m}$ \\
$64 \mathrm{~km} / \mathrm{h}$ & $36 \mathrm{~m}$ \\
$80 \mathrm{~km} / \mathrm{h}$ & $53 \mathrm{~m}$ \\
$96 \mathrm{~km} / \mathrm{h}$ & $73 \mathrm{~m}$ \\
$112 \mathrm{~km} / \mathrm{h}$ & $96 \mathrm{~m}$ \\
\hline
\end{tabular}

For each velocity, how far apart in seconds should two cars travelling in the same direction at that velocity be so that their distance apart is the same as the corresponding stopping distance given in the above table?
9.2 Note: The Highway Code adds that "The distances shown are a general guide. The distance will depend on your attention (thinking distance), the road surface, the weather conditions and the condition of your vehicle at the time. "
The Highway Code gives a thinking distance of $12 \mathrm{~m}$ for a car travelling at $64 \mathrm{~km} / \mathrm{h}$. How much thinking time does that correspond to?