Question

a man walks on a straight road from his home to a market 3 km away with a speed of 6 km/h. finding the market closed, he instantly turns and walks back with a speed of 5 km/h. what is the
(a) magnitude of average velocity. (b) average speed of man in the time interval 0 to 40 min.

Answer 1

Average speed from home to market :  6 kmph
time taken from home to market : 3 km/ 6 kmph = 1/2 hr = 30 min
So this is the average speed for 0 to 30 min. : 6 kmph

Average speed from market to home = 9 kmph
Time taken from market to home on the way back : 3 km/9kmph = 1/3 hour
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Average speed for the total journey : from to market and market to home :
         = total distance / total time
         =  ( 3 km + 3 km ) / [ 1/2 + 1/3 ]hr
         = 6 / (5/6)  = 36/5 = 7.2  kmph
     This is average speed for 0 to 50 min
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    distance traveled from 30 to 40 minutes, that is 10 min: 1/6 hr
             = 9 kmph * 1/6 hour = 1.5 km

     Average speed for time = 0 to 40 min is
               = distance traveled / time taken = (3 + 1.5) / (40/60)
                = 4.5 * 60 / 40  = 6.75 kmph
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Velocity = change in displacement vector =
           vector joining initial position of man and  final position of man

Velocity is dependent on the current position of man and not the distance traveled by man.

Velocity initially at t=0 is 0, as the man is at home.
Velocity for t = 0 to 30 minutes is : 6 kmph towards the market.
Velocity at t = 50 minutes is zero as the man is at home. THat is, displacement of the man from initial position is zero. So displacement/time = 0.

Distance of the man from home at t = 40 min is :  3 - 1.5 km = 1.5 km
Velocity at time 40 minutes is :  1.5 km / (40/60) hr     =  2.25 kmph
The direction of velocity is in the direction towards market from home.