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(B) Time taken to reach the market: $t_1 = \frac{	ext{distance}}{	ext{speed}} = \frac{2.5 \,km}{5 \,km/h} = 0.5 \,h = 30 \,min$. Time taken to retur

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$\frac{5}{3}$

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(B) Time taken to reach the market: $t_1 = \frac{	ext{distance}}{	ext{speed}} = \frac{2.5 \,km}{5 \,km/h} = 0.5 \,h = 30 \,min$. Time taken to return to the house: $t_2 = \frac{	ext{distance}}{	ext{speed}} = \frac{2.5 \,km}{7.5 \,km/h} = \frac{1}{3} \,h = 20 \,min$. Total time taken for the round trip is $30 \,min + 20 \,min = 50 \,min$. Total distance covered = $2.5 \,km + 2.5 \,km = 5 \,km = 5000 \,m$. Total time in seconds = $50 \,min 	imes 60 \,s/min = 3000 \,s$. Average speed = $\frac{	ext{Total distance}}{	ext{Total time}} = \frac{5000 \,m}{3000 \,s} = \frac{5}{3} \,m/s$.

$A$ person walks along a straight road from his house to a market $2.5 \,km$ away with a speed of $5 \,km/h$ and instantly turns back and reaches his house with a speed of $7.5 \,km/h$. The average speed of the person during the time interval $0$ to $50 \,min$ is (in $m/s$)

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