A man walks to a market $2.5 ext{ km}$ away at $5 ext{ km/h}$ and instantly returns home at $7.5

Question

A man walks to a market $2.5	ext{ km}$ away at $5	ext{ km/h}$ and instantly returns home at $7.5	ext{ km/h}$. What is his average speed over the time interval $0$ to $40	ext{ min}$?

Professor PC Thomas Classes · Answer

For the first $30	ext{ min}$, he walks $2.5	ext{ km}$ to the market. In the remaining $10	ext{ min}$ ($\frac{1}{6}	ext{ h}$), he walks back at $7.5	ext{ km/h}$, covering a distance of $7.5 	imes \frac{1}{6} = 1.25	ext{ km}$. Total distance covered in $40	ext{ min} = 2.5 + 1.25 = 3.75	ext{ km}$. Average speed = $\frac{3.75}{40/60} = \frac{3.75 	imes 6}{4} = 5.625	ext{ km/h}$.