A man walks on a straight road from his home to market 2.5 km away with speed of 5 km/h Finding the market closed, he instantly turns and walks back with a speed of 7.5 km/h. What is the magnitude of average velocity and average speed of the man over the interval of time (i) 0 to 30 min (ii) 0 to 50 min (iii) 0 to 40 min ?

Explanation

Time taken by man to go from his home to market, 1 `t=("distance")/("speed") =(2.5)/(5) =(1)/(2) hr =30` min
Time taken by man to go from market to his home, `t_(2) =(2.5)/(7.5) =(1)/(3) hr =20` min
? Total time taken `t_(1)+t_(2)=(1)/(2) +(1)/(3)=(5)/(6) hr =50` min
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/hr "=(7.5 x)/(60) =10=1.25 km`
So, displacement = 2.5 –1.25 = 1.25 km
Distance traveled = 2.5 +1.25 = 3.75 km
(a) Average velocity `=(1.25)/(40//60) =1.875 km//hr`
(b) Average speed `=(3.75)/(40//60) =5.625 km//hr`