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

A man walks on a straight road from his home to a market 2.5 km away w

1.	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
Answer» Solution :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 30 min `vec(v) _(av) = (2.5)/((1)/(2))` = 5 KM /h TOWARDS market