A train 120 ; m long moving at a speed of 60 | vedclass

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Question

(D) Speed of the first train $= 60 \; km/hr = 60 	imes \frac{5}{18} \; m/s = \frac{50}{3} \; m/s$.Let the speed of the second train be $x \; m/s$.Sin

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(D) Speed of the first train $= 60 \; km/hr = 60 	imes \frac{5}{18} \; m/s = \frac{50}{3} \; m/s$.Let the speed of the second train be $x \; m/s$.Since the trains are moving in opposite directions,their relative speed is the sum of their individual speeds: $	ext{Relative speed} = (\frac{50}{3} + x) \; m/s$.The total distance covered to cross each other is the sum of their lengths: $	ext{Total distance} = 120 \; m + 130 \; m = 250 \; m$.Using the formula $	ext{Speed} = \frac{	ext{Distance}}{	ext{Time}}$:$\frac{50}{3} + x = \frac{250}{6}$.$\frac{50}{3} + x = \frac{125}{3}$.$x = \frac{125}{3} - \frac{50}{3} = \frac{75}{3} = 25 \; m/s$.Converting the speed back to $km/hr$: $25 	imes \frac{18}{5} = 90 \; km/hr$.

$A$ train $120 \; m$ long moving at a speed of $60 \; km/hr$ crosses a train $130 \; m$ long coming from the opposite direction in $6 \; seconds$. The speed of the second train is (in $km/hr$):

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