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

Answer: (B)

$6\,km \,h^{-1}$, $0$

Answer 2

Time taken by the boy to go from his home to Time taken by the boy to go from his home to the market, $t_1 =\frac{2.5\,km}{5\,km\,h^{-1}}=\frac{1}{2}h$ Time taken by the boy to return back from the market to his home, $t_{2}=\frac{2.5\,km}{7.5\,km\,h^{-1}}=\frac{1}{3}h$ $\therefore$ Total time 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 distance travelled $= 2.5 \,km + 2.5 \,km = 5\, km$ Displacement $= 0$ (As the boy returns back home) $\therefore$ Average speed $=\frac{\text{Distance travelled}}{\text{Time taken}}=\frac{5\,km}{\frac{5}{6}h}=6\,km\,h^{-1}$ Average velocit $=\frac{\text{Displacement}}{\text{Time taken}}=0$