An woman walks on a straight road from her 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 with a speed of 7.5 km/hr. What is the
(a) magnitude of an average velocity ,
(b) average speed of the man,over the interval of time .
0 to 50 min and

Explanation

Time taken by woman to go from her home to market
`t_(1) =("Distance") /("Speed") = (2.5)/(5) = (1)/(2)` h
Time taken by woman to go from market to her home,
`t_(2) = (2.5)/(7 . 5) = (1)/(3) h`
`:.` Total time `= t_(1) + t_(2) = (1)/(2) +(1)/(3) = (5)/(6)` h
=50 min
(a) Average velocity
`vec(v) _(av) = ("Displacement")/("Time")`
(b) Average sped
`v_(av) = ("Distance")/("Time")`
Between 0 to 50 min
` {:("Total distance"),("travelled"):}}= 2 . 5 + 2 . 5 = 5 km `
`{:("Total"),("displacement"):}}= zero`
`vec(v)_(av) = 0`
`v_(av) = (5)/((5)/(6))= 6 ` km /hr