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(D) Speed of the train $= 120\, km/h$.Converting to $m/s$: $120 	imes \frac{5}{18} = \frac{100}{3}\, m/s$.Let the length of the platform be $x\, m$.W

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(D) Speed of the train $= 120\, km/h$.Converting to $m/s$: $120 	imes \frac{5}{18} = \frac{100}{3}\, m/s$.Let the length of the platform be $x\, m$.When the train crosses the platform,the total distance covered is $(x + 320)\, m$.Using the formula $	ext{Distance} = 	ext{Speed} 	imes 	ext{Time}$:$x + 320 = \frac{100}{3} 	imes 24$.$x + 320 = 100 	imes 8 = 800$.$x = 800 - 320 = 480\, m$.The man crosses the same platform $(480\, m)$ in $4\, minutes$.Time in seconds $= 4 	imes 60 = 240\, s$.Speed of the man $= \frac{	ext{Distance}}{	ext{Time}} = \frac{480}{240} = 2\, m/s$.

$A$ $320\, m$ long train moving at an average speed of $120\, km/h$ crosses a platform in $24\, seconds$. $A$ man crosses the same platform in $4\, minutes$. What is the speed of the man in $m/s$?

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