A boy runs from his home to the market 1 km away. He reaches the market in 30 min, seeing the market closed he once runs back to his home and stops after 15 min at his friends house which is 1/2 km away from the market. Calculate his average speed and m magnitude of av. velocity during the whole journey.

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Solution

$A v e r a g e s p e e d = t o t a l d i s \tan c e / t o t a l t i m e t a k e n = 1000 + 500 / 45 * 60 m / s = 1500 / 2700 m / s e c = 0.55 m / s e c A v e r a g e v e l o c i t y = t o t a l d i s p l a c e m e n t / t o t a l t i m e = 500 / 2700 m / s e c = 0.185 m / s e c$

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Average speed= total distance /total time taken= 1000+ 500/ 45*60 m/s =1500/2700 m/sec = 0.55 m/secAverage velocity = total displacement /total time = 500/2700 m/sec = 0.185 m/sec

$A v e r a g e s p e e d = t o t a l d i s \tan c e / t o t a l t i m e t a k e n = 1000 + 500 / 45 * 60 m / s = 1500 / 2700 m / s e c = 0.55 m / s e c A v e r a g e v e l o c i t y = t o t a l d i s p l a c e m e n t / t o t a l t i m e = 500 / 2700 m / s e c = 0.185 m / s e c$