On a two lane road, car \(A\) is travelling with a speed of \(36 \mathrm{kmh}^{-1}\). Two cars \(B\) and…

1. 5 \(\mathrm{ms}^{-2}\)
2. 10 \(\mathrm{ms}^{-2}\)
3. 6 \(\mathrm{ms}^{-2}\)
4. 1 \(\mathrm{ms}^{-2}\)

Solution

At the instant when car \(B\) decides to overtake \(\operatorname{car} A\), the velocities of cars are:
\(\begin{array}{l}
v_{A}=36 \times \frac{5}{18}=10 \mathrm{~m} \mathrm{~s}^{-1} \\
v_{B}=54 \times \frac{5}{18}=15 \mathrm{~m} \mathrm{~s}^{-1}
\end{array}\)
and \(\quad v_{C}=-54 \times \frac{5}{18}=-15 \mathrm{~m} \mathrm{~s}^{-1}\)
Velocity of car \(B\) relative to \(A\),
\(\vec{v}_{B A}=\vec{v}_{B}-\vec{v}_{A}=15-10=5 \mathrm{~m} \mathrm{~s}^{-1}\)
Velocity of car C relative to A,
\(\vec{v}_{C A}=\vec{v}_{C}-\vec{v}_{A}=-15-10=-25 \mathrm{~m} \mathrm{~s}^{-1}\)
Time required by car \(C\) to just cross \(A\)
\(=\frac{1000}{v_{C A}}=\frac{1000}{25}=40 \mathrm{~s}\)
In order to avoid accident, car \(B\) must overtake \(A\) in this time. So
\(1000=v_{B A} t+\frac{1}{2} a_{B A} t^{2}\)
\(1000=5 \times 40+\frac{1}{2} a_{B A} \times 40^{2}\)
\(a_{BA}=1 \mathrm{~m} \mathrm{~s}^{-2}\). acceleration that car \(B\) requires to avoid accident.
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Asked in: JEE Mains - Motion In One Dimension - Test 1

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