The driver of a train moving at a speed \(v_{1}\) sights another train at a distance \(d\) ahead of him,…
The driver of a train moving at a speed \(v_{1}\) sights another train at a distance \(d\) ahead of him, moving in the same direction with a slower speed \(v_{2}\). He immediately applies brakes to achieve a constant retardation \(a\). There will be no collision if \(d\) is greater than
\(\frac{\left(v_{1}-v_{2}\right)^{2}}{a}\)
\(\frac{\left(v_{1}^{2}-v_{2}^{2}\right)}{a}\)
\(\frac{\left(v_{1}-v_{2}\right)^{2}}{2 a}\)
\(\frac{\left(v_{1}^{2}-v_{2}^{2}\right)}{2 a}\)
Solution
The two trains will not collide if the initial relative velocity \(u=\left(v_{1}-v_{2}\right)\) reduces to zero relative velocity \((v=0)\) in a minimum distance \(S=d_{\min }\) under a retardation \((-a)\). Using \(v^{2}-u^{2}=2 a S\), we have
$\begin{aligned}
& 0-\left(v_{1}-v_{2}\right)^{2}=2(-a) d_{\min } \\
\Rightarrow \quad d_{\min }=\frac{\left(v_{1}-v_{2}\right)^{2}}{2 a}
\end{aligned}$
So the correct choice is \((\mathrm{c})\).
Asked in: JEE Mains - Motion In One Dimension - Test 4