Solve the following :

$\vec{V}_{1}=60^{\times \frac{5}{18}}=16.6 \mathrm{~m} / \mathrm{s}$

$\vec{V}_{2}=42^{\times \frac{2}{18}}=11.6 \mathrm{~m} / \mathrm{s}$

Relative velocity $=16.6-11.6$

$V_{\text {rel }}=5 \mathrm{~m} / \mathrm{s}$

$\mathrm{d}_{\mathrm{res}}=10 \mathrm{~m}$

Time to cross $=\frac{\mathrm{d}_{\mathrm{rel}}}{\mathrm{v}_{\mathrm{rel}}}=\frac{10}{5}=2 \mathrm{sec}$

In $2 \mathrm{sec}, 1^{\text {st }}$ car moves $=16.6 \times 2$ $=33.2$

Length of road $=33.2+$ length of car $=33.2+5$

$\approx 38 \mathrm{~m}$