A train 75 ; m long overtook a person who was | vedclass

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(B) The length of the train is $75 \; m$.For the first person:Relative speed $= \frac{	ext{Distance}}{	ext{Time}} = \frac{75}{7.5} = 10 \; m/s$.Conv

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(B) The length of the train is $75 \; m$.For the first person:Relative speed $= \frac{	ext{Distance}}{	ext{Time}} = \frac{75}{7.5} = 10 \; m/s$.Converting to $km/h$: $10 	imes \frac{18}{5} = 36 \; km/h$.Since the train is overtaking the person,the relative speed is (Speed of train - Speed of person).$36 = 	ext{Speed of train} - 6 \implies 	ext{Speed of train} = 42 \; km/h$.For the second person:Time taken $= 6 \frac{3}{4} = 6.75 \; s = \frac{27}{4} \; s$.Relative speed $= \frac{75}{27/4} = \frac{300}{27} = \frac{100}{9} \; m/s$.Converting to $km/h$: $\frac{100}{9} 	imes \frac{18}{5} = 40 \; km/h$.Let the speed of the second person be $v_2$.$40 = 42 - v_2 \implies v_2 = 2 \; km/h$.

$A$ train $75 \; m$ long overtook a person who was walking at the rate of $6 \; km/h$ and passed him in $7 \frac{1}{2}$ seconds. Subsequently,it overtook a second person,and passed him in $6 \frac{3}{4}$ seconds. At what rate (in $km/h$) was the second person travelling?

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