Answer:
The problem is
equivalent to the motion of molecules.
Mean
free path, \[\lambda =\,\frac{1}{\sqrt{2}\,\pi \,n\,{{d}^{2}}}\]
Where,
\[n=\]number of planes volume
\[=\,\frac{10}{20\,\times
20\,\times 1.5}\,\,=\,0.0167\,\,k{{m}^{-3}}\]
\[=0.0167\times
{{10}^{9}}{{m}^{3}}\]
\[d=20\,m\]
\[\therefore
\] \[\lambda \,=\,\frac{1}{1.414\,\times \,3.14\,\times \,400\,\times
\,0.0167\,\times \,{{10}^{-9}}}\]
\[=3.37\times
{{10}^{7}}m.\]
\[\upsilon
=\,150\,km/h=150\,\times \,\frac{5}{18}=\,41.67\,m\,{{s}^{-1}}\]
Time
that will elapse between near collision,
\[t=\,\frac{\lambda
}{\upsilon }=\,\frac{3.37\,\times
{{10}^{7}}m}{41.67\,m{{s}^{-1}}}\,=\,8.09\,\,\times \,{{10}^{5}}s\]
=
224.7 hours.
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