A) 100 m
B) 1150 m
C) 1300 m
D) 1250 m
Correct Answer: D
Solution :
Given that, \[{{u}_{A}}={{u}_{B}}=72\,km{{h}^{-1}}=72\times \frac{5}{18}=20\,\,m{{s}^{-1}}\] Using the relations, \[s=ut+\frac{1}{2}a{{t}^{2}}\], we get \[{{s}_{B}}={{u}_{B}}t+\frac{1}{2}a{{t}^{2}}=20\times 50+\frac{1}{2}\times 1\times {{(50)}^{2}}\] \[{{s}_{B}}=100+1250=2250\,\,m\] Also, let \[{{s}_{A}}\] be the distance covered by the train A, then \[{{s}_{A}}={{u}_{A}}\times t\] \[=20\times 50=1000\,m\] Original distance between the two trains \[={{s}_{B}}-{{s}_{A}}\] \[=2250-1000=1250\,\,m\]
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