A train travels at a speed of 10, m/s and lea | vedclass

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(B) Speed of the first train $(S_1)$ = $10\, m/s = 10 	imes \frac{18}{5} = 36\, km/h$.Speed of the second train $(S_2)$ = $36 	imes (1 + \frac{1}{3}

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$32$

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(B) Speed of the first train $(S_1)$ = $10\, m/s = 10 	imes \frac{18}{5} = 36\, km/h$.Speed of the second train $(S_2)$ = $36 	imes (1 + \frac{1}{3}) = 36 	imes \frac{4}{3} = 48\, km/h$.Total distance between Trivandrum and Nagercoil = $68\, km$.The first train starts at $8:00\, AM$ and the second at $8:20\, AM$. The first train travels alone for $20\, 	ext{minutes}$ $(1/3\, 	ext{hour})$.Distance covered by the first train in $1/3\, 	ext{hour} = 36 	imes \frac{1}{3} = 12\, km$.Remaining distance to be covered by both trains = $68 - 12 = 56\, km$.Relative speed of the two trains = $S_1 + S_2 = 36 + 48 = 84\, km/h$.Time taken to meet after $8:20\, AM$ = $\frac{	ext{Remaining Distance}}{	ext{Relative Speed}} = \frac{56}{84} = \frac{2}{3}\, 	ext{hours}$.The second train travels from Nagercoil for $2/3\, 	ext{hours}$ at $48\, km/h$.Distance from Nagercoil = $48 	imes \frac{2}{3} = 32\, km$.

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