Question

A boy walks on a straight road from his home to a market 2.5 km with a speed of 5 dm 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

Answer: (B)

6 km h^-1, 0

Answer 2

Time taken by the boy to go from his home to the market, $t_{1} = \frac{2.5km}{5kmh^{- 1}} = \frac{1}{2}h$

Times taken by the boy to return back from the market to his home, $t_{2} = \frac{2.5km}{7.5kmh^{- 1}} = \frac{1}{3}h$

$\therefore$Total times taken

$= t_{1} + t_{2} = \frac{1}{2}h + \frac{1}{3}h = \frac{5}{6}h = 50\min$

In t = 0 to 50 min

Total distances travelled =$2.5km + 2.5km$= 5 km

Displacement = 0 (As the boy returns back home)

$\therefore$Average speed

$= \frac{Dis\tan cetravelled}{Timetaken} = \frac{5km}{\frac{5}{6}h} = 6kmh^{- 1}$

Average velocity $= \frac{Displacement}{Timetaken} = 0$