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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. (b) What is the average speed of the man

Q 3.13     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

                  (ii)      0 to 50 min,

Answers (1)
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S Sayak

(a).(ii) Time taken by the man to reach the market is t1= 30 minutes.

Time taken by the man to reach back home is t2

\\t_{2}=\frac{2.5}{7.5}\\ t_{2}=\frac{1}{3}\ hr\\ t_{2}=20\ min

t1+t2=50 min

Displacement in 50 minutes is d2=0 km

Magnitude of average velocity=d2/(t1+t2)= 0 km h-1

(b).(ii) Distance travelled in 50 minutes is s2=5 km

Average speed=s2/(t1+t2)= 0.6 km h-1

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