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Question -

A man walks on a straight road from his home to a market 2.5 km away witha
speed of 5 km/h. Finding the marketclosed, he instantly turns and walks back
home with a speed of 7.5 km h–1.What is the
(a) Magnitude of average velocity, and
(b) Average speed of the man over theinterval of time (i) 0 to 30 min, (ii) 0 to
50 min, (iii) 0 to 40 min? [Note: Youwill appreciate from this exercise why it
is better to define average speed astotal path length divided by time, and not
as the magnitude of average velocity.You would not like to tell the tired man on
his return home that his average speedwas zero !]



Answer -

Distance to the market= 2.5 km = 2500 m

Speed of the man walking to the market= 5 km/h = 5 x (5/18) = 1.388 m/s

Speed of the manwalking when he returns back home = 7.5 km/h = 7.5  x (5/18) = 2.08m/s

(a) Magnitude of theaverage speed is zero since the displacement is zero

(b)

(i) Time taken toreach the market = Distance/Speed = 2500/1.388 = 1800 seconds = 30 minutes

So, the average speedover 0 to 30 minutes is 5 km/h or  1.388 m/s

(ii) Time taken toreach back home = Distance/Speed = 2500/2.08 = 1200 seconds = 20 minutes

So, the average speedis

Average Speed over ainterval of 50 minutes= distance covered/time taken = (2500 + 2500)/3000 =5000/3000 = 5/3 = 1.66 m/s

= 6 km/h

(ii) Average speedover an interval of 0 – 40 minutes = distance covered/ time taken = (2500+1250)/2400 = 1.5625 seconds = 5.6 km/h

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