A man walks on a straight road from hishome to a market 2.5 𝑘𝑚 away with aspeed of 5 𝑘𝑚/ℎ. Finding the marketclosed, he instantly turns and walksback home with a speed of 7.5 𝑘𝑚/ℎ. Theaverage speed of the man over theinterval time of 0 to 40 𝑚𝑖𝑛 is equal to

First, let's calculate the time it takes for the man to walk to the market and back home.

The distance to the market is 2.5 km and he walks at a speed of 5 km/h. Therefore, the time to the market is distance/speed = 2.5 km / 5 km/h = 0.5 hours.

The distance back home is also 2.5 km and he walks at a speed of 7.5 km/h. Therefore, the time back home is distance/speed = 2.5 km / 7.5 km/h = 0.33 hours.

So, the total time for the man to walk to the market and back home is 0.5 hours + 0.33 hours = 0.83 hours.

The total distance he walked is 2.5 km to the market + 2.5 km back home = 5 km.

Therefore, the average speed is total distance / total time = 5 km / 0.83 hours = 6.02 km/h.

However, the question asks for the average speed over the interval from 0 to 40 min, not the total time it takes for the man to walk to the market and back home. 40 min is approximately 0.67 hours.

So, the average speed over the interval from 0 to 40 min is total distance / time interval = 5 km / 0.67 hours = 7.46 km/h.