# A man walks on a straight road from his home to a market 3kms away with a speed of 6km/h. Finding the market closed, he instantly turns and walks back with a speed of 9km/h. What is the (i)magnitude of average velocity (Give answers:- in first case(A)6km/h and second case (B)6km/h)(ii)average speed of the man over the interval of time (I) 0 to 30 mins (II) 0 to 50 mins (III) 0 to 40 mins. (Given answers :- (I)(A)6km/h (B)6km/h (II)(A)0 (B)7.2 km/h (III)(A) 2.25 km/h (B)6.75 km/h.Can you explain how did we get these answers?

1
by megha9

2014-08-19T02:20:03+05:30

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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.

thanks
my pleasure sir