A man walks on a straight road from his home to a market 2.5 km away with a speed of `"kmh"^(-1)` Finding the market closed he instantly turns and walks back home with a speed of.7.5 `"kmh"^(-1)` What is the (a) magnitude of average velocity and (b) average speed of the man over the time interval o to 50 minutes ?

Explanation

Time taken by man to go from his home to market,
`t_(1)=("distance")/("speed")=(2.5)/(5)=(1)/(2)h` Time take by man to go from market to his home , `t_(2)=(2.5)/(7.5)=(1)/(3)` h
`therefore` Total time taken `=t_(1)+t_(2)=(1)/(2)+(1)/(3)=(5)/(6)h=50` min
In time interval 0 to 50 min ,
Total distance travelled =2.5 +2.5 =5 km.
Total displacement =zero
(a) Average velocity `=("displacement")/("time")=0`
(b) Average speed `=("distance")/("time")=(5)/(5//6)=6` km/h