A) \[6\,\,s\]
B) \[2\,\,s\]
C) \[4\,\,s\]
D) \[5\sqrt{\frac{2}{3}}\,\,s\]
Correct Answer: C
Solution :
Key idea: Trains are moving towards each other, hence resultant speed is sum of two. The trains are moving in \[\xrightarrow[I=50\,m]{{{\text{v}}_{1}}{{=}_{10\,\text{m}/\text{s}}}}\] opposite directions. Relative speed \[{{v}_{1}}-\left( -{{v}_{2}} \right)\] \[=10+15=25\,\,m\xleftarrow[_{l=50m}^{{{v}_{2}}{{=}_{15m/s}}}]{}\] Relative distance \[=50\,m+50\,m=100\,m\] \[\xrightarrow[+ve]{}\] \[\therefore \text{Time}=\frac{\text{distance}}{\text{speed}}=\frac{100}{25}=4\,s\]
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