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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 home: $t_2 = \frac{	ext{Distance}}{	ext{Speed}} = \frac{2.5 \ km}{7.5 \ km/h} = \frac{1}{3} \ h = 20 \ min$. Total time taken for the round trip is $30 \ min + 20 \ min = 50 \ min$. Since the total time interval given is $50 \ min$,the person completes the entire journey. 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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