A man walks on a straight road from his home to market 2.5km

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SUMMARY

A man walks 2.5 km to a market at a speed of 5 km/h, taking 30 minutes to reach. Upon finding the market closed, he returns home at a speed of 7.5 km/h, which takes an additional 20 minutes, resulting in a total travel time of 50 minutes. The average speed is calculated based on the total distance traveled and the total time taken, while the average velocity considers the net displacement over time intervals of 0 to 30 minutes, 0 to 50 minutes, and 0 to 40 minutes.

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Homework Statement


A man walks on a straight road from his home to market 2.5km away with a speed of 5km/h. Finding the market closed, he instantly turns and walk back home with a speed of 7.5km/h.
For the given time intervals (i) 0 to 30minutes. (ii) 0 to 50minutes., (iii) 0 to40 min. What is the
a) magnitude of average velocity and
b) average speed of the man.


Homework Equations





The Attempt at a Solution


distance=displacement
d=s*t
2.5=5*x
x=5/2.5=2
 
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welcome to pf!

hi reshmi! welcome to pf! :smile:

you're obviously not understanding the question, so let's rewrite it …

A man walks on a straight road from his home to market 2.5km away with a speed of 5km/h. So it takes him 30 minutes.

Finding the market closed, he instantly turns and walk back home with a speed of 7.5km/h. So that takes him 20 minutes, making a total of 50 minutes.

For the given time intervals (i) 0 to 30minutes. (ii) 0 to 50minutes., (iii) 0 to40 min. What is the
a) magnitude of total displacement over total time
b) average speed of the man.​

:wink:
 

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