How is Average Velocity Calculated for a Two-Part Journey?

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Homework Help Overview

The discussion revolves around calculating average velocity for a two-part journey involving walking distances and times. The scenario describes walking 55.0m east to a windmill and then 40.0m west to a bench, with specific times taken for each leg of the journey.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of average velocity using the formula for displacement over time, questioning the addition of time intervals instead of subtraction. Some participants clarify the meaning of the variables in the equation and discuss the implications of defining initial and final times.

Discussion Status

Participants are actively engaging with the problem, offering insights into the correct interpretation of time intervals and displacement. There is a focus on understanding the reasoning behind the calculations, with some guidance provided on how to approach the problem conceptually.

Contextual Notes

There is an emphasis on the distinction between total time for the journey and the individual times for each segment, which is crucial for correctly applying the average velocity formula. The original poster expresses confusion regarding the addition of time intervals, indicating a potential misunderstanding of the problem setup.

stryker105
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Homework Statement


Starting from the front door of your ranch house, you walk 55.0m due east to your windmill, and then you turn around and slowly walk 40.0m west to a bench where you sit and watch the sunrise. It takes you 25.0s to walk from your house to the windmill and then 47.0s to walk from the windmill to the bench.

For the entire trip from your front door to the bench, what is your average velocity?

Homework Equations



[itex]\upsilon[/itex]av-x=x2-x1/t2-t1=Δx/Δt

The Attempt at a Solution


40.0m-55.0m/47.0s+25.0s = -0.208m/s

So I got the answer to this problem by adding the time instead of subtracting it. The program I'm using shows this answer is correct. Though, I'm wondering why adding the time together instead of subtracting it works instead of the other way around. Can someone help explain this to me? Thanks.
 
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stryker105 said:

Homework Statement


Starting from the front door of your ranch house, you walk 55.0m due east to your windmill, and then you turn around and slowly walk 40.0m west to a bench where you sit and watch the sunrise. It takes you 25.0s to walk from your house to the windmill and then 47.0s to walk from the windmill to the bench.

For the entire trip from your front door to the bench, what is your average velocity?

Homework Equations



[itex]\upsilon[/itex]av-x=x2-x1/t2-t1=Δx/Δt

The Attempt at a Solution


40.0m-55.0m/47.0s+25.0s = -0.208m/s

So I got the answer to this problem by adding the time instead of subtracting it. The program I'm using shows this answer is correct. Though, I'm wondering why adding the time together instead of subtracting it works instead of the other way around. Can someone help explain this to me? Thanks.
##t_1## and ##t_2## represent the time (on a clock) at the initial and at the final positions. Their difference ## t_2 - t_1 = \Delta t## is the time it took for the entire motion.
This information is not given in the question. What they give is the time it takes for the two paths, so what they are giving is actually ##\Delta t_{first \, part} ## and ## \Delta t_{second \, part} ##. Since these are time intervals, you must add them to get the total time for the whole motion.
 
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t2 is the time you are at the bench.
t1 is the time you start at ( call it 0).

So Δt=t2-t1=(47.0s+25.0s)-0s
 
stryker105 said:

Homework Equations



[itex]\upsilon[/itex]av-x=x2-x1/t2-t1=Δx/Δt

You are misunderstanding this equation:

x1 is the starting position; and x2 the final position.

t1 is the time at the start; and t2 is the time at the end.

It might help to think of this equation as:

[tex]v_{average} = (x_{final}-x_{start})/(t_{final}-t_{start})[/tex]

Also, for many problems, not all so be careful, you can choose t_start = 0. That's when you start the stopwatch. t_end is when you stop the stopwatch. And ##\Delta t## is what the stopwatch reads.
 
stryker105 said:

Homework Statement


Starting from the front door of your ranch house, you walk 55.0m due east to your windmill, and then you turn around and slowly walk 40.0m west to a bench where you sit and watch the sunrise. It takes you 25.0s to walk from your house to the windmill and then 47.0s to walk from the windmill to the bench.

For the entire trip from your front door to the bench, what is your average velocity?

Homework Equations



[itex]\upsilon[/itex]av-x=x2-x1/t2-t1=Δx/Δt

The Attempt at a Solution


40.0m-55.0m/47.0s+25.0s = -0.208m/s

So I got the answer to this problem by adding the time instead of subtracting it. The program I'm using shows this answer is correct. Though, I'm wondering why adding the time together instead of subtracting it works instead of the other way around. Can someone help explain this to me? Thanks.
The route consist of two stages. You can consider the initial position at the house as xi=0 and the initial time Ti=0. First, you walk due East - your displacement is Δx1= 55 m, and the time is Δt1=25 s. Your position is x1=55 m with respect to the house. Then you turn back moving to west: your displacement is negative, Δx2=-40 m and it took the time Δt2=47 s. Your position is x2= Δx1+Δx2=55-40=15 m measured to East from the house. Your total displacement is Δx=15 m to East, and you walked for 47+25=71 s. The average velocity is Vav=Δx/Δt = (15-0)/71 , and it points to East.

ehild
 
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