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

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(A) The relative speed of the policeman with respect to the thief is $(11 - 10)\, km/h = 1\, km/h$.To convert the relative speed into $m/s$,we multipl

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) The relative speed of the policeman with respect to the thief is $(11 - 10)\, km/h = 1\, km/h$.To convert the relative speed into $m/s$,we multiply by $\frac{5}{18}$:Relative speed $= 1 	imes \frac{5}{18} = \frac{5}{18}\, m/s$.The time given is $6\, minutes$,which is $6 	imes 60 = 360\, seconds$.The distance covered by the policeman relative to the thief in $360\, seconds$ is:Distance $= 	ext{Relative speed} 	imes 	ext{Time} = \frac{5}{18} 	imes 360 = 5 	imes 20 = 100\, m$.The initial distance between them was $200\, m$. After $6\, minutes$,the policeman has closed the gap by $100\, m$.Therefore,the remaining distance between them is $(200 - 100)\, m = 100\, m$.

$A$ thief is noticed by a policeman from a distance of $200\, m$. The thief starts running and the policeman chases him. The thief and the policeman run at the speed of $10\, km/h$ and $11\, km/h$ respectively. What is the distance between them after $6\, minutes$? (in $m$)

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