A man walks on a straight road from his home to market 3 km away with a speed of 6 km/hr.The market was closed so he turns back and walks with a speed of 9 km/hr. What is the magnitude of the average velocity of the man between time interval 0 to 40 min. ?
I want the solution

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Answers

The Brainliest Answer!
2015-11-24T16:43:07+05:30
Hi there,
distance b/w home and market=3km
time taken to reach market=3/6=1/2 hrs=.5 hrs
time taken to reach home=3/9=1/3 hrs=20min
average velocity=T.displacement/T.time

average  velocity=36/5=7.2km/h

i hope this helps u...


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2015-11-24T16:43:42+05:30
velocity = 6 km/h
velocity
= 9 km/h
distance₁ = 3 km
distance₁ = 3 km
therefore time₁ = d₁ / s₁ = 3/6 = 1/2 h
time₂ = d₂ / s₂ = 3/9 = 1/3 h
now,
av. velocity = total displacement / total time 
av. velocity = ( 3 + 3 ) / 1/2 + 1/3 
= 6 / 5/6
= 6 * 6/5
= 36 / 5
= 7.2 
therefore 7.2 kmp .
         i think it will helps u ............................^_^

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