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Grade 9General Physics

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. Find the given time intervals (1) 0 to 30 min. (2) 0 to 59 min , (3) 0 to 49 min , what is the (a) magnitude of average velocity and (b) average speed of the man.

Shinjini Ghosh

8 Years agoGrade 9

1 Answer

Eshan

8 Years ago

Dear student,

(1) Time taken to reach the market= $\dfrac{s}{v}=\dfrac{2.5km}{5km/hr}=0.5hr=30min$
Sin ce it moves with constant speed for the time interval 0 min to 30 min, the average speed and the magnitude of velocity both is 5km/hr.

Time taken to reach from market to home= $\dfrac{s}{v}=\dfrac{2.5km}{7.5km/hr}=\dfrac{1}{3}hr=20min$

(2) Total distance covered from 0 min to 59 min=2.5km+2.5km=5km

Hence average speed= $\dfrac{5km}{1hr}=5km/hr$
Average velocity=0km/hr

(3)Total distance covered from 0 min to 49 min=2.5km+2.5km=5km

Hence average speed= $\dfrac{5km}{(\dfrac{50}{60})hr}=6km/hr$

Average velocity=0km/hr

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