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Question

(B) First,convert the speed of the train from $km/h$ to $m/s$:$108 \, km/h = 108 	imes \frac{5}{18} = 30 \, m/s$.Let the length of the platform be $L

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) First,convert the speed of the train from $km/h$ to $m/s$:$108 \, km/h = 108 	imes \frac{5}{18} = 30 \, m/s$.Let the length of the platform be $L \, m$.When the train crosses the platform,the total distance covered is the sum of the length of the train and the length of the platform:$	ext{Speed} = \frac{	ext{Total Distance}}{	ext{Time}}$$30 = \frac{280 + L}{12}$$360 = 280 + L$$L = 360 - 280 = 80 \, m$.The man crosses the same platform of length $80 \, m$ in $10 \, s$. Assuming the man's length is negligible:$	ext{Speed of the man} = \frac{	ext{Length of platform}}{	ext{Time}} = \frac{80}{10} = 8 \, m/s$.

$A$ $280 \, m$ long train moving with an average speed of $108 \, km/h$ crosses a platform in $12 \, s$. $A$ man crosses the same platform in $10 \, s$. What is the speed of the man in $m/s$?

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