A thief is spotted by a policeman from a dist | vedclass

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Question

(C) Initial distance between the policeman and the thief $= 400\, m$.Relative speed of the policeman with respect to the thief $= 9\, km/h - 5\, km/h

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(C) Initial distance between the policeman and the thief $= 400\, m$.Relative speed of the policeman with respect to the thief $= 9\, km/h - 5\, km/h = 4\, km/h$.Converting relative speed into $m/s$: $4 	imes \frac{5}{18} = \frac{10}{9}\, m/s$.Time taken by the policeman to overtake the thief $= \frac{	ext{Distance}}{	ext{Relative Speed}} = \frac{400}{10/9} = 400 	imes \frac{9}{10} = 360\, s$.Speed of the thief in $m/s = 5 	imes \frac{5}{18} = \frac{25}{18}\, m/s$.Distance covered by the thief in $360\, s = 	ext{Speed} 	imes 	ext{Time} = \frac{25}{18} 	imes 360 = 25 	imes 20 = 500\, m$.

$A$ thief is spotted by a policeman from a distance of $400\, m$. When the policeman starts the chase,the thief also starts running. Assuming the speed of the thief is $5\, km/h$ and that of the policeman is $9\, km/h$,how far will the thief have run before he is overtaken by the policeman? (in $m$)

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