Two cars \(A\) and \(B\) are travelling in the same direction with velocities \(V_{A}\) and…

Two cars \(A\) and \(B\) are travelling in the same direction with velocities \(V_{A}\) and \(V_{B}\left(V_{A} > V_{B}\right)\). When car A is at a distance s ahead of car B, the driver of car \(A\) applies the brakes producing a uniform retardation \(a\); there will be no collision when
  1. \(s < \frac{\left(V_{A}-V_{B}\right)^{2}}{2 a}\)
  2. \(s=\frac{\left(V_{A}-V_{B}\right)^{2}}{2 a}\)
  3. \(s \geq \frac{\left(V_{A}-V_{B}\right)^{2}}{2 a}\)
  4. \(s \leq \frac{\left(V_{A}-V_{B}\right)^{2}}{2 a}\)

Solution

For no collision, the speed of car \(A\) should be reduced to \(\mathrm{v}_{B}\) before the cars meet, i.e. final relative velocity of car \(\mathrm{A}\) with respect to car \(B\) is zero, i.e., \(V_{r}=0\) Here \(u_{r}=\) initial relative velocity \(=V_{A}-V_{B}\) Relative acceleration \(=a_{r}=-a-0=-a\) Let relative displacement \(=s_{r}\) Then using the equation, \(v_{r}^{2}=u_{r}^{2}+2 a_{r} s_{r}\) \(0^{2}=\left(V_{A}-V_{B}\right)^{2}-2 a s_{r} \cdot \operatorname{or} s_{r}=\frac{\left(V_{A}-V_{B}\right)^{2}}{2 a}\) For no collision, \(s_{r} \leq s\), i.e., \(\frac{\left(V_{A}-V_{B}\right)^{2}}{2 a} \leq s\)

Asked in: JEE Mains - Motion In One Dimension - Chapter Test

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