A man walks on a straight road from his home to a market2.5 km away with a speed of5 km //h. Finding the market closed, he instantly turns and walks back with a speed of 7.5 km//h. What is the (a) magnitude of average velocity and (b) average speed of the man, over the interval of time (i)0 to30 min . (ii) 0 to 50 min (iii) 0 to 40 min ?

A man walks on a straight road from his home to a market2.5 km away wi

1.	A man walks on a straight road from his home to a market2.5 km away with a speed of5 km //h. Finding the market closed, he instantly turns and walks back with a speed of 7.5 km//h. What is the (a) magnitude of average velocity and (b) average speed of the man, over the interval of time (i)0 to30 min . (ii) 0 to 50 min (iii) 0 to 40 min ?
Answer» 0,0 `6 km h^(-1), 0` 0, `6 km^(-1)` `6 km h^(-1), 6 km h^(-1)` Solution :Time taken by the boy to go from his home to the market, `t_1 = (2.5 ("km"))/(5 "km" h^(-1)) = 1/2` h Time taken by the boy to return back from the market to his home, `t_2= (2.5 km)/(7.5 km h^(-1))= 1/3` h `therefore` Total time taken = `t_1 + t_2 = 1/2 h + 1/3 h = 5/6 h = 50` min In t = 0 to 50 min. Total DISTANCE TRAVELLED = 2.5 km + 2.5 km = 5 km) Displacement = 0 (As the boy returns back home) `therefore` Average speed `= ("Distance travelled") /("Time taken")`=`(5 "km")/(5/6h) = 6 km h^(-1)` Average velocity `= ("Displacement")/("Time taken") = 0`