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Question

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 $$40\ min$$? 


Solution

Distance to market $$s=2.5 km=2.5 \times 10^3=2500 m$$
Speed with which he goes to market $$=5 km/h=5 \dfrac{10^3}{3600}=\dfrac{25}{18} m/s$$
Speed with which he comes back $$=7.5 km/h=7.5 \times \dfrac{10^3}{3600}=\dfrac{75}{36} m/s$$
(a)Average velocity is zero since his displacement is zero.
(b) 
(i)Since the initial speed is 5km/s and the market is 2.5 km away,time taken to reach market:$$\dfrac{2.5}{5}=1/2 h=30$$ minutes.
Average speed over this interval $$=5 km/h$$
(ii)After 30 minutes,the man is travelling wuth 7.5 km/h speed for 50-30=20 minutes.The distance he covers in 20 minutes :$$7.5 \times \dfrac{1}{3}=2.5 km$$
His average speed in 0 to 50 minutes: $$V_{avg}=\dfrac{distance \quad traveled}{time}$$
$$=\dfrac{2.5+2.5}{(50/60)}=6 km/h$$
(iii)In 40-30=10 minutes he travels a distance of :$$7.5 \times \dfrac{1}{6}=1.25 km$$
$$V_{avg}=\dfrac{2.5+1.25}{(40/60)}=5.625 km/h$$

Physics

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