A) \[\text{1}00\text{m}\]
B) \[\text{115}0\text{m}\]
C) \[\text{13}00\text{m}\]
D) \[\text{125}0\text{m}\]
Correct Answer: D
Solution :
Given that,\[{{u}_{A}}={{u}_{B}}=72km{{h}^{-1}}=72\times \frac{5}{18}=\text{20 m}{{\text{s}}^{\text{-1}}}\] Using the relations, \[s=ut+\frac{1}{2}a{{t}^{2}},\] we get \[{{s}_{B}}={{u}_{B}}t\frac{1}{2}a{{t}^{2}}=20\times 50+\frac{1}{2}\times 1\times {{(50)}^{2}}\] \[{{s}_{B}}=1000+1250=2250m\] Also, let \[{{s}_{A}}\] be the distance covered by the train A, then \[{{s}_{A}}={{u}_{A}}\times t\] = 20 x 50 = 1000 m Original distance between the two trains \[={{s}_{B}}-{{s}_{A}}\] = 2250 -1000 = 1250 m
You need to login to perform this action.
You will be redirected in
3 sec
