A man walks on a straight road from his home to a market $2.5 ext{ km}$ away with a speed of $5 ex

Question

A man walks on a straight road from his home to a market $2.5	ext{ km}$ away with a speed of $5	ext{ km/h}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5	ext{ km/h}$. What is the magnitude of his average velocity over the time interval $0$ to $50	ext{ min}$?

Professor PC Thomas Classes · Answer

Time to reach the market $t_1 = \frac{2.5}{5} = 0.5	ext{ h} = 30	ext{ min}$. Time to return home $t_2 = \frac{2.5}{7.5} = \frac{1}{3}	ext{ h} = 20	ext{ min}$. Total time $= 30 + 20 = 50	ext{ min}$. Since he returns to his starting point at exactly $50	ext{ min}$, his total displacement is $0$. Hence, average velocity is $0$.