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Two trains are moving with velocities $𝑣_{1} = 10 𝑚 𝑠^{- 1} a n d 𝑣_{2} = 20 𝑚 𝑠^{- 1}$ on the same track in opposite directions. After the application of brakes if their retarding rates are $a_{1} = 2 m s^{- 2} a n d a_{2} = 1 m s^{- 2}$ respectively, then the minimum distance of separation between the trains to avoid collision is

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150 m

225 m

450 m

300 m

answer is B.

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Detailed Solution

$d = d_{1} + d_{2} = \frac{v_{1}^{2}}{2 a_{1}} + \frac{v_{2}^{2}}{2 a_{2}} = \frac{(10)^{2}}{2 \times 2} + \frac{(20)^{2}}{(2 \times 1)} = 225 m$

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