AP EAMCET · PHYSICS · Motion In One Dimension
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 \(4 \frac{2}{3}\)
- B \(\frac{5}{3}\)
- C \(\frac{5}{6}\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{3}\)
Step-by-step Solution
Detailed explanation
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…
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