A train passes a station platform in 36 ; tex | vedclass

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(B) First, convert the speed of the train from $	ext{km/hr}$ to $	ext{m/s}$:Speed $= 54 	imes \frac{5}{18} = 15 \; 	ext{m/s}$.Let the length of th

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$240$

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(B) First, convert the speed of the train from $	ext{km/hr}$ to $	ext{m/s}$:Speed $= 54 	imes \frac{5}{18} = 15 \; 	ext{m/s}$.Let the length of the train be $L_t$ and the length of the platform be $L_p$.When the train passes a man standing on the platform, it covers its own length in $20 \; 	ext{seconds}$:$L_t = 	ext{Speed} 	imes 	ext{Time} = 15 	imes 20 = 300 \; 	ext{m}$.When the train passes the platform, it covers the sum of its own length and the platform's length in $36 \; 	ext{seconds}$:$L_t + L_p = 	ext{Speed} 	imes 	ext{Time} = 15 	imes 36 = 540 \; 	ext{m}$.Substituting the value of $L_t$:$300 + L_p = 540$$L_p = 540 - 300 = 240 \; 	ext{m}$.Thus, the length of the platform is $240 \; 	ext{m}$.

$A$ train passes a station platform in $36 \; \text{seconds}$ and a man standing on the platform in $20 \; \text{seconds}$. If the speed of the train is $54 \; \text{km/hr}$, what is the length (in $m$) of the platform?

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