A man walks on a straight road from his home to a market 2.5 km away with speed of 5 `\frac { k m } { h \r }`. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 `\frac { k m } { h \r }`. The average speed of the man over the interval of time 0 to 40 min is equal to

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 40 min
Distance moved in 30 min (from home to market)
2 . 5 km
Distance moved in 10 min (from market to home) with
Speed ` 7 . 5 km / h = 7 . 5 xx (10)/(60) = 1.25 km`
So displacement = 2 . 5 - 1.25
= 1 . 25 kms (towards market)
Distance travelled = 2 . 5 + 1 . 25 = 3 . 75 m
`vec(v)_(av) = (1.25)/((40)/(60))`
= 1 . 875 km /h (towards market)
`v_(av) = (3 . 75)/((40)/(60)) = 5 . 625` km /h