Heena and Renu are competing in a 100, m race | vedclass

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(C) Let the initial speed of Renu be $v$ and the initial speed of Heena be $2v$.For the first $50\, m$:Time taken by Heena $t_{H1} = \frac{50}{2v} = \

Vedclass · Accepted Answer

$25$

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(C) Let the initial speed of Renu be $v$ and the initial speed of Heena be $2v$.For the first $50\, m$:Time taken by Heena $t_{H1} = \frac{50}{2v} = \frac{25}{v}$.Time taken by Renu $t_{R1} = \frac{50}{v}$.After $50\, m$,Heena's speed becomes $\frac{1}{4} 	imes 2v = 0.5v$.Renu continues at speed $v$.Let $d$ be the additional distance covered by both until Renu catches Heena.Time taken by Heena for distance $d$ is $t_{H2} = \frac{d}{0.5v} = \frac{2d}{v}$.Time taken by Renu for distance $d$ is $t_{R2} = \frac{d}{v}$.Since Renu catches Heena,the total time taken by both must be equal:$t_{R1} + t_{R2} = t_{H1} + t_{H2}$$\frac{50}{v} + \frac{d}{v} = \frac{25}{v} + \frac{2d}{v}$$50 + d = 25 + 2d$$d = 25\, m$.The total distance covered from the start is $50 + 25 = 75\, m$.The distance from the finish line $N = 100 - 75 = 25\, m$.

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