A man walks on a straight road from his home to a market $2.5km$ away with a speed of $5km/h$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5km/h$. The average speed of the man over the interval of time $0$ to $40$ minutes, is equal to

The correct option is D $\frac{45}{8}km/h$
Time taken in going to the market=$\frac{distancecovered}{timetaken}$ $=\frac{2.5}{5}=\frac{1}{2}hr=30min$

As we are told to find the average speed for the interval $0$ to $40$ minutes, so remaining time for consideration of motion is $10$ min

This will be included in return trip with speed of 7.5 km/h

So, distane travelled in remaining $10$ minutes :
Distance travelled = speed * time $=7.5\times \frac{10}{60}=1.25km$

Hence, $\text{average speed}$ = $\frac{Totaldistance}{Totaltime}$ = $\frac{2.5+1.25}{\frac{40}{60}}=\frac{45}{8}km/hr$