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Train A:
Moves with velocity 10 m/s and retards at 2m/s^2,
Hence, substituting values in v^2 = u^2 –2as
0 = 100 – 2 x2x s
Or s= 25 m
Train B:
Moves with velocity 20 m/s and retards at 1m/s^2,
Hence, substituting values in v’^2 = u’^2 –2as’
0 = 400 – 2 x1x s’
Or s= 200 m
So, the minimum distance of separation to avoid collision = 200+25 =225 m
Thanks & Regards,
Vasantha Sivaraj,
askIITians faculty
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