Q. 143.7( 65 Votes )

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

[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]

Answer :

Given,


Speed from home to market, s1 = 5 km/h


Speed from market to home, s2 = 7.5 km/h


Distance between market and home, d = 2.5 km


Average velocity, v =  m/s


Average speed, s = m/s


Magnitude of total length of the path = 5 km


Time taken for home to market, t1 = h = 30 min


Time taken for market to home, t2 = h= 20 min


Total journey time, T = t1 + t2 = 50 min


For,


(i)0 to 30 min


(a) Average velocity = = 5 km/h


(b) Average speed = = 5 km/h


(ii)0 to 50 min


Net displacement = 0 m


Net distance = 5 km


(a) Average velocity = 0 m/s


(b) Average speed = = 6 km/h


(iii)0 to 40 min


Distance travelled in return journey in 10 min = 7.5× =1.25 km


Net displacement = 2.5 – 1.25 km = 1.25 km


Net distance = 2.5+1.25 km = 3.75 km


(a) Average velocity =  = 1.875 km/h


(b) Average speed =  = 5.625 km/h

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