A car travelling at

Question:

A car travelling at $60 \mathrm{~km} / \mathrm{h}$ overtakes another car travelling at $42 \mathrm{~km} / \mathrm{h}$. Assuming each car to be $5.0 \mathrm{~m}$ long, find the time taken during the overtake and the total road distance used for the overtake.

Solution:

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

$\overrightarrow{\mathrm{V}}_{2}=42 \times \frac{5}{18}=11.6 \mathrm{~m} / \mathrm{s}$

Relative velocity $=16.6-11.6$

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

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

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

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

Length of road $=33.2+$ length of car

$=33.2+5$

$\approx 38 \mathrm{~m}$

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