A hare sees a dog 100, m away from her and ru | vedclass

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(D) First,convert the speeds from $km/h$ to $m/s$ by multiplying by $\frac{5}{18}$.Speed of hare $= 12 	imes \frac{5}{18} = \frac{10}{3} \, m/s$.Spee

Vedclass · Accepted Answer

$900$

Vedclass · Answer

(D) First,convert the speeds from $km/h$ to $m/s$ by multiplying by $\frac{5}{18}$.Speed of hare $= 12 	imes \frac{5}{18} = \frac{10}{3} \, m/s$.Speed of dog $= 16 	imes \frac{5}{18} = \frac{40}{9} \, m/s$.In $1$ minute ($60$ seconds),the hare runs a distance of $d_1 = \frac{10}{3} 	imes 60 = 200 \, m$.At the moment the dog starts chasing,the total distance between them is $D = 100 + 200 = 300 \, m$.The relative speed of the dog with respect to the hare is $v_{rel} = (16 - 12) \, km/h = 4 	imes \frac{5}{18} = \frac{10}{9} \, m/s$.The time taken by the dog to catch the hare is $t = \frac{D}{v_{rel}} = \frac{300}{10/9} = 300 	imes \frac{9}{10} = 270 \, s$.The distance covered by the hare after being spotted (during the chase) is $d_2 = 	ext{speed of hare} 	imes t = \frac{10}{3} 	imes 270 = 900 \, m$.

$A$ hare sees a dog $100\, m$ away from her and runs off in the opposite direction at a speed of $12\, km/h$. $A$ minute later,the dog perceives her and gives chase at a speed of $16\, km/h$. How much distance is covered by the hare after being spotted? (in $m$)

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