The driver of train \(A\) moving at a speed of \(30 \mathrm{~ms}^{-1}\) sights another train \(B\) moving on…

The driver of train \(A\) moving at a speed of \(30 \mathrm{~ms}^{-1}\) sights another train \(B\) moving on the same track at a speed of \(10 \mathrm{~ms}^{-1}\) in the same direction. He immediately applies brakes and achieves a uniform retardation of \(2 \mathrm{~ms}^{-2}\). To avoid collision, what must be the minimum distance between trains \(A\) and \(B\) when the driver of \(A\) sights \(B\) ?
  1. 120 m
  2. 100 m
  3. 180 m
  4. 130 m

Solution

Relative initial velocity \(A\) w.r.t. \(B\) is,
\(u_{A B}=u_{A}-u_{B}=30-10=20 \mathrm{~ms}^{-1}\)
Relative retardation of \(A\) w.r.t \(B\) is
\(a_{A B}=a_{A}-a_{B}=-2-0=-2 \mathrm{~ms}^{-2}\)
To avoid collision, the relative final velocity of \(A\) w.r.t. \(B\) \(=v_{A B}\) must be zero. Minimum relative displacement of \(A\)
w.r.t. \(B\) is \(s_{A B}\) which is found by using the relation \(v^{2}-u^{2}\) \(=2\) as which gives
\(0-(20)^{2}=2 \times(-2) s_{A B} \Rightarrow s_{A B}=100 \mathrm{~m}\) ^

Asked in: JEE Mains - Motion In One Dimension - Test 4

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