Q 3.14  A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h–1. Finding the market closed, 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 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 !]

 

Answers (1)
S Sayak

(a).(i) Time taken by the man to reach the market is t1

\\t_{1}=\frac{2.5}{5}\\ t_{1}=0.5\ h\\ t_{1}=30\ min

Displacement in 30 minutes is d1=2.5 km

Magnitude of average velocity=d1/t1= 5 km h-1

(b).(i) Distance travelled in 30 minutes is s1=2.5 km

Average speed=s1/t1= 5 km h-1

The first 30 minutes the man travels from his home to the market.

During the next 10 minutes, he travels with a speed of 7.5 km h-1 towards his home covering a distance s3

\\s_{3}=7.5\times \frac{10}{60}\\ =1.25\ km

t3=40 min

Magnitude of average velocity =(2.5-1.25)/t3=1.875 km h-1

Average speed=(2.5+1.25)/t3=5.625 km h-1

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