Q. A man walks on a straight road from his home to a market $2.5\ km$ away with a speed of $5\ km/h$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5\ km/h$. The average speed of the man over the interval of time $0$ to $40\ min$. Is equal to
A) $5\ km/h$
B) $\frac{25}{4}\ km/h$
C) $\frac{30}{4}\ km/h$
D) $\frac{45}{8}\ km/h$
Explanation
(d) Man walks from his home to market with a speed of $5\ km/h$. Distance$= 2.5\ km$ and time $= \frac{d}{v} = \frac{2.5}{5} = \frac{1}{2}\ hr$ and he returns back with speed of $7.5\ km/h$ in rest of time of $10\ minutes$
Distance $= 7.5 \times \frac{10}{60} = 1.25\ km$
So, Average speed $= \frac{Total\ distance}{TOtal\ time}$
$= \frac{(2.5 + 1.25)km}{(40/60)hr} = \frac{45}{8}km/hr$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.