Two trains measuring 100, m and 80, m respect | vedclass

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(D) Let the speeds of the two trains be $S_1$ and $S_2$ in $m/s$,where $S_1 > S_2$.When moving in opposite directions,the relative speed is $(S_1 + S_

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$54, 18$

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(D) Let the speeds of the two trains be $S_1$ and $S_2$ in $m/s$,where $S_1 > S_2$.When moving in opposite directions,the relative speed is $(S_1 + S_2)$. The total distance covered to pass each other is the sum of their lengths: $100 + 80 = 180\, m$.Given time $= 9\, s$,so $S_1 + S_2 = 180 / 9 = 20\, m/s$ ... $(i)$.When moving in the same direction,the relative speed is $(S_1 - S_2)$. The total distance covered is still $180\, m$.Given time $= 18\, s$,so $S_1 - S_2 = 180 / 18 = 10\, m/s$ ... $(ii)$.Adding equations $(i)$ and $(ii)$:$2S_1 = 30 \implies S_1 = 15\, m/s$.Subtracting $(ii)$ from $(i)$:$2S_2 = 10 \implies S_2 = 5\, m/s$.Converting to $km/h$ by multiplying by $18/5$:$S_1 = 15 	imes (18/5) = 54\, km/h$.$S_2 = 5 	imes (18/5) = 18\, km/h$.

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