A person walks along a straight road fro...

A person walks along a straight road from his house to a market 2.5kms away with a speed of 5 km/hr and instantly turns back and reaches his house with a speed of 7.5 kms/hr. The average speed of the person during the time interval 0 to 50 minutes is (in m/sec)

`4 2/3`

`5/3`

`5/6`

`1/3`

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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 man walks on a straight road from his home to a market 2.5 km away with a speed of 5km h^-1 . Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1 . What is the (a) magnitude of average velocity and (n) average speed of the man over the time interval 0 to 50 minutes?

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^(-1) . Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^(-1) . What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

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