A man walks to a market $2.5 ext{ km}$ away with a speed of $5 ext{ km/h}$ and immediately returns

Question

A man walks to a market $2.5	ext{ km}$ away with a speed of $5	ext{ km/h}$ and immediately returns home with a speed of $7.5	ext{ km/h}$. What is his average speed over the time interval $0$ to $50	ext{ min}$?

Professor PC Thomas Classes · Answer

He travels $2.5	ext{ km}$ to the market and $2.5	ext{ km}$ back. Total distance $= 5.0	ext{ km}$. Total time $= 50	ext{ min} = \frac{5}{6}	ext{ h}$. Average speed = $\frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{5}{5/6} = 6	ext{ km/h}$.