Question

A boy walks on a straight road from his home to a market 2.5 km with a speed of 5 km $h^{-1}$. Finding the market closed he instantly turns and walks back with a speed of 7.5 km $h^{-1}$. What is the average speed and average velocity of the boy between t = 0 to t = 50 min?

• A. 0,0
• B. 6 km $h^{-1}$, 0
• C. 0, 6 km $h^{-1}$
• D. 6 km $h^{-1}$, 6 km $h^{-1}$

Answer 1

The correct option is B 6 km $h^{-1}$, 0

Initial and final position of body is same (home)

Displacement $=0$

time taken from home to market and market to home is

$t=\frac{2.5}{5}+2.57.5=\frac{1}{2}+\frac{1}{3}$

$=\frac{5}{6}$

Average speed $=\frac{2x}{t}=\frac{2\times 2.5}{\frac{5}{6}}=6m/s$

average velocity $=\frac{displacement}{time}=\frac{0}{t}=0$