Question

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

• A. $4\frac{2}{3}$
• B. $\frac{5}{3}$
• C. $\frac{5}{6}$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

$\frac{5}{3}$

Distance between house and market $d=2.5$ km

From House to market :

Speed of the person $v_1=5km/hr$

$\therefore$ Time taken $t_1=\frac{2.5}{5}$ hr $=0.5$ hr $=30$ min

From market to house :

Speed of the person $v_2=7.5km/hr$

$\therefore$ Time taken $t_2=\frac{2.5}{7.5}$ hr $=\frac{1}{3}$ hr $=20$ min

Total distance covered $D=2.5+2.5=5$ km $=5000$ m

Total time taken $T=30+20=50$ min $=3000$ s

$\therefore$ Average speed of the person $v_{avg}=\frac{5000}{3000}=\frac{5}{3}$ $m/s$