Q.
A boy begins to walk eastward along a street infront of his house and the graph of his displacement from home is shown in the following figure. His average speed for in the whole time interval is equal to
A) $8\ m\min^{- 1}$
B) $6\ m\min^{- 1}$
C) $\frac{8}{3}m\min^{- 1}$
D) $2\ m\min^{- 1}$
Explanation
(d) $Displacement\ form\ 0\ to\ 5\ s\ = \ 40\ m$
$Displacement\ from\ 5\ to\ 10\ s\ = \ 40\ m$
$Displacement\ from\ 0\ to\ 15\ s\ = \ - \ 20\ m$
$And\ displacement\ from\ 15\ to\ 20\ s = 0\ m$
$\therefore Net\ displacement = 40 + 40 - 20 + 0 = 60\ m$
$\ \ \ \ \ \ \ Total\ time\ taken = 5 + 5 + 15 + 5 = 30\ min.$
$Hence,\ average\ speed = \frac{displacement\ (m)}{time\ \left( \min \right)} = \frac{60}{30} = 2\ m\ \min^{- 1}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.