Calculating Time on a Moving Sidewalk

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Homework Help Overview

The discussion revolves around calculating the time it takes for a person to walk on a moving sidewalk, given different time measurements for standing still and walking without the sidewalk. The subject area includes concepts of motion, velocity, and time calculations in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various approaches to relate the time taken on the moving sidewalk to the time taken walking without it. Some question the validity of intuitive calculations, while others suggest using equations of motion to derive a solution.

Discussion Status

The discussion is active, with participants sharing equations and reasoning. Some guidance has been provided regarding the use of distance and velocity in the calculations, and there is an ongoing exploration of how to relate the different velocities involved.

Contextual Notes

Participants express uncertainty about the need for specific velocity values and how to properly set up the equations. There is mention of the assumption that distances are equal in different scenarios, which is a key point of discussion.

waldvocm
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Lance takes 167 seconds to get down the gall on the moving sidewalk. While walking (not on the moving sidewalk) it took him 88 seconds. If he walks while on the sidewalk how long will it take him.

I came to the answer to easily so I am not sure if I did it right or if I am missing something

I just took the time it takes him standing on the moving sidewalk and subtracted the time it takes him to walk, not on the moving sidewalk.

My answer is 79 seconds while walking on the moving sidewalk. Is this correct? It seems too easy!
 
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Quite often your intuition works when the equations are linear, but not in this case! You must use the formula for motion at constant speed and figure it out. Don't worry about not knowing the distance - just leave the "d" in and it will cancel out in the end.
 
hmmmm...would this require me to know the velocity as well? I am struggling to figure out which equation you mean.
 
For constant velocity motion, use d = v*t.
If you want an expression for the velocity, just solve the equation for v. The result will have a d in it, but that's okay - d cancels out in the end when you find the time for the guy on the moving sidewalk.

I forgot to suggest you try out your intuition on some cases. For instance, you could say "what if the walker also takes 167 seconds?" In that case your intuitive calculation would suggest a time of zero for the combined motion, which is impossible!
 
I have the equations

d1=v(167) standing on the moving sidewalk
d2=v(88) walking off of the moving sidewalk d1=d2

v(167)=v(88)

I don't know where to go here

d3=v*t3
 
Better make that
d=v1(167) standing on the moving sidewalk
d=v2(88) walking off of the moving sidewalk
d = v3*t3 walking on the moving sidewalk
because the d's are the same and the v's are different.

You know something about that v3, too! Notice that and you'll have one less variable and it will be possible to solve for t3.
 
v3=88v-167v

v3=79v

Ha that brought me back to my original thinking!

I don't know I am really confused
 
v3=88v-167v
says a velocity is equal to a distance - no good!

Say you have a slidewalk moving at 5 km/h and you run at 10 km/h on it. How fast will you be going?
 
15km/h

So, do I add the velocities?

v1+v2=v3

d/167+d/88=d/t3
 
  • #10
Excellent! Cancel those d's and you can calculate your t3.
 

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