Question

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

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

Answer 1

Answer: (B)

Time taken by the boy to go from his home to Time taken by the boy to go from his home to the market, $t_1=\frac{2.5km}{5km{h}^{-1}}=\frac{1}{2}h$
Time taken by the boy to return back from the market to his home, $t_2=\frac{2.5km}{7.5km{h}^{-1}}=\frac{1}{3}h$
$\therefore$ Total time taken $=t_1+t_2=\frac{1}{2}h+\frac{1}{3}h=\frac{5}{6}h=50min$
In $t=0$ to $50min,$
Total distance travelled $=2.5km+2.5km=5km$
Displacement $=0$ (As the boy returns back home)
$\therefore$ Average speed $=\frac{\text{Distance travelled}}{\text{Time taken}}=\frac{5km}{\frac{5}{6}h}=6km{h}^{-1}$
Average velocit $=\frac{\text{Displacement}}{\text{Time taken}}=0$