A train running at 25 , km/hr takes 18 , seco | vedclass

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(A) Let the length of the train be $L_t$ and the length of the platform be $L_p$.Convert the speed of the train from $km/hr$ to $m/s$: $25 	imes \fra

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) Let the length of the train be $L_t$ and the length of the platform be $L_p$.Convert the speed of the train from $km/hr$ to $m/s$: $25 	imes \frac{5}{18} = \frac{125}{18} \, m/s$.When the train passes a man walking in the opposite direction at $5 \, km/hr$ $(5 	imes \frac{5}{18} = \frac{25}{18} \, m/s)$,the relative speed is $\frac{125}{18} + \frac{25}{18} = \frac{150}{18} = \frac{25}{3} \, m/s$.The length of the train $L_t = 	ext{Relative Speed} 	imes 	ext{Time} = \frac{25}{3} 	imes 12 = 100 \, m$.When the train passes the platform,the total distance covered is $L_t + L_p$.$L_t + L_p = 	ext{Speed of train} 	imes 	ext{Time} = \frac{125}{18} 	imes 18 = 125 \, m$.Substituting $L_t = 100 \, m$,we get $100 + L_p = 125$,so $L_p = 25 \, m$.The sum of the length of the train and the platform is $L_t + L_p = 100 + 25 = 125 \, m$.

$A$ train running at $25 \, km/hr$ takes $18 \, seconds$ to pass a platform. Next,it takes $12 \, seconds$ to pass a man walking at $5 \, km/hr$ in the opposite direction. Find the sum of the length of the train and that of the platform (in $m$).

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