Two trains of equal lengths take 10 ; s and 1 | vedclass

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(B) Speed of the first train $= \frac{120 \; m}{10 \; s} = 12 \; m/s$.Speed of the second train $= \frac{120 \; m}{15 \; s} = 8 \; m/s$.When two train

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$12$

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(B) Speed of the first train $= \frac{120 \; m}{10 \; s} = 12 \; m/s$.Speed of the second train $= \frac{120 \; m}{15 \; s} = 8 \; m/s$.When two trains travel in opposite directions,their relative speed is the sum of their individual speeds.Relative speed $= 12 \; m/s + 8 \; m/s = 20 \; m/s$.To cross each other,the total distance covered must be the sum of the lengths of both trains.Total distance $= 120 \; m + 120 \; m = 240 \; m$.Time taken to cross each other $= \frac{	ext{Total distance}}{	ext{Relative speed}} = \frac{240 \; m}{20 \; m/s} = 12 \; s$.

Two trains of equal lengths take $10 \; s$ and $15 \; s$ respectively to cross a telegraph post. If the length of each train is $120 \; m$,in what time (in $s$) will they cross each other when travelling in opposite directions?

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