A train travelling with a speed of 60 , km/h | vedclass

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(C) Speed of the first train $(v_1)$ $= 60 \, km/h$.Converting to $m/s$: $v_1 = 60 	imes \frac{5}{18} = \frac{50}{3} \, m/s$.Let the speed of the sec

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$36$

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(C) Speed of the first train $(v_1)$ $= 60 \, km/h$.Converting to $m/s$: $v_1 = 60 	imes \frac{5}{18} = \frac{50}{3} \, m/s$.Let the speed of the second train be $v_2 \, m/s$.Since both trains are moving in the same direction,the relative speed is $(v_1 - v_2) = (\frac{50}{3} - v_2) \, m/s$.The first train covers a relative distance of $120 \, m$ in $18 \, seconds$.Using the formula: $	ext{Relative Speed} = \frac{	ext{Distance}}{	ext{Time}}$.$\frac{50}{3} - v_2 = \frac{120}{18} = \frac{20}{3}$.$v_2 = \frac{50}{3} - \frac{20}{3} = \frac{30}{3} = 10 \, m/s$.Converting back to $km/h$: $10 	imes \frac{18}{5} = 36 \, km/h$.

$A$ train travelling with a speed of $60 \, km/h$ catches another train travelling in the same direction and then leaves it $120 \, m$ behind in $18 \, seconds$. The speed of the second train is (in $km/h$):

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