A man walks on a straight road from his home to a market 2,5 km away with a speed of `5kmh^(-1)` . Finding the market closed, he in: stantly 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 intervalo 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 take `t_(r) + 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`