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

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$4 \frac{2}{3}$

$\frac{5}{3}$

$\frac{5}{6}$

$\frac{1}{3}$

answer is B.

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Detailed Solution

Time required to reach market = 30 min = 1/2 hr
remaining time = 50 – 30 = 20 min
distance in next 20 min = $7.5 \times \frac{20}{60} = 2.5 km$
$\begin{array}{l} ∴ v_{avg} = \frac{2 v_{1} v_{2}}{v_{1} + v_{2}} = \frac{2 \times 5 \times 7.5}{5 + 7.5} = 6 km / hr \\ v_{avg} = 6 \times \frac{5}{18} m / s = \frac{5}{3} m / s \end{array}$

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