Average speed of person walking

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SUMMARY

The discussion focuses on calculating the average speed and average velocity of a person walking between two points at different speeds: 5.00 m/s from A to B and 3.00 m/s from B to A. The average speed over the entire trip is determined by the total distance divided by the total time, while the average velocity is 0 m/s since the person returns to the starting point. Participants clarify that using 0 m/s is appropriate as it includes units, and they discuss why simply averaging the two speeds does not yield the correct average speed due to differing travel times.

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  • Understanding of average speed and average velocity concepts
  • Familiarity with basic physics equations, particularly distance = speed × time
  • Knowledge of unit dimensions in physical quantities
  • Ability to set up and solve equations with variables
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  • Research the formula for average speed in multi-segment journeys
  • Learn about the implications of average velocity in physics
  • Study the concept of dimensional analysis in physical quantities
  • Explore examples of variable speed scenarios in physics problems
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babysnatcher
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A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s. What are (a) her average speed over the entire trip and (b) her average velocity over the entire trip?

I already solved this problem but I think I had to think way outside the box for it so I want to see an easier way to figure this out. For part "(b)", I just wrote as the answer 0 m/s because she ends up in the same spot. I'm not going to talk about "(a)" because I do not want to disrupt peoples problem solving process. I also want to know if it is correct to use 0 m/s rather than 0; I think it is because 0 pressure is not the same as 0 m/s.

Another thing I want to know is why does dividing the sum of the speeds by 2 not work since it is average speed? I think it is because the it took longer travel at 3.0 m/s. And this affects the results because their are two different times to add to each other.
 
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babysnatcher said:
A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s. What are (a) her average speed over the entire trip and (b) her average velocity over the entire trip?

I already solved this problem but I think I had to think way outside the box for it so I want to see an easier way to figure this out. For part "(b)", I just wrote as the answer 0 m/s because she ends up in the same spot. I'm not going to talk about "(a)" because I do not want to disrupt peoples problem solving process.

For part a, average speed is total distance over total time. The key to the problem is just setting the unknown distance between point A and point B as a variable "d". Use the relation distance = speed*time to find the time required to travel each leg of the journey (in terms of d). You will find that this variable d cancels out in the final answer for v.


babysnatcher said:
I also want to know if it is correct to use 0 m/s rather than 0; I think it is because 0 pressure is not the same as 0 m/s.

Yes, physical quantities have dimension, and you must include units when expressing them.

babysnatcher said:
Another thing I want to know is why does dividing the sum of the speeds by 2 not work since it is average speed? I think it is because the it took longer travel at 3.0 m/s. And this affects the results because their are two different times to add to each other.

Yes, your reasoning is correct.
 

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