Q158 / 160 · TS EAMCET 2014 · Physics
A person walks along a straight road from his house to a market $2.5 \mathrm{~km}$ away with a speed of $5 \mathrm{~km} / \mathrm{h}$ and instantly turns back and reaches his house with a speed of $7.5 \mathrm{~km} / \mathrm{h}$. The average speed of the person during the time interval 0 to $50 \mathrm{~min}$ is $(\mathrm{in} \mathrm{m} / \mathrm{s})$
- A$4 \frac{2}{3}$
- Correct answerB$\frac{5}{3}$
- C$\frac{5}{6}$
- D$\frac{1}{3}$
Solution
A according to question, $t_1=\frac{2.5}{5}=\frac{1}{2} \mathrm{~h}=30 \mathrm{~min}$
$$
\text { and } t_2=\frac{2.5}{7.5}=\frac{1}{3} \mathrm{~h}=20 \mathrm{~min}
$$
So, the average speed
$$
\begin{aligned}
& =\frac{\text { Total distance }(\text { in metre })}{\text { Total time }(\text { in second })} \\
& =\frac{(2.5+2.5) \times 1000}{(30+20) \times 60} \\
& =\frac{5 \times 1000}{50 \times 60}=\frac{50}{30}=\frac{5}{3} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.