Finding the speed of a moving walkway.

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To determine the speed of a moving walkway, the child runs 19 meters in each direction on the walkway, taking a total of 32 seconds for the round trip. The child's running speed is 2.9 m/s relative to the walkway, leading to two equations for the child's speed relative to the ground: (x + 2.9) for the outbound trip and (x - 2.9) for the return. The total distance of 38 meters and the total time of 32 seconds allow for the calculation of average speed, but a more effective approach involves breaking the problem into two separate equations for each direction. This method clarifies the relationship between the walkway's speed and the child's running speed, ultimately leading to a solution. The discussion emphasizes the importance of correctly setting up equations to solve for unknowns in physics problems.
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Homework Statement



While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at a constant speed of 2.9 m/s relative to the surface of the walkway, you decide to try to determine the speed of the walkway itself. You watch the child run on the entire 19-m walkway in one direction, immediately turn around, and run back to his starting point. The entire trip takes a total elapsed time of 32 s. Given this information, what is the speed of the moving walkway relative to the airport terminal?


Homework Equations



v = d/t

The Attempt at a Solution



I know that the child's velocity relative to the ground in one direction is (x+2.9) (where x is the speed of the moving walkway) and in the other direction it is (x-2.9). We did a similar problem to show that average speed is not necessarily the speed in one direction+the speed in the other direction/2.

I know that the whole distance traveled is 19*2 meters, or 38 meters. I also know that it takes the child 32 seconds to do this overall, so the average speed is (38/32) meters/second. With that information, I made this equation:

(x+2.9) * tone way + (x-2.9) * (32-tone way) = 38/32. Time (t) will be the same in both of those, since the time to run the other way is taken care of by subtracting the first t from 32 for the return trip. I need x, but I still have 2 variables. Thanks in advance for helping me out.
 
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Hello maccam912,
maccam912 said:
(x+2.9) * tone way + (x-2.9) * (32-tone way) = 38/32.
Something is definitely not set up right. The left hand side of the above equation has units of distance [meters]. The right hand side has units of velocity [m/s].
I suggest keeping things simple at the start. At least keep things simple when setting up your initial equations.

For example, break up the equations into two, separate equations. Use t1 for the amount of time it takes to go in one direction and use t2 for the other direction. And you also know that t1 + t2 = 32 [sec]. You'll have 3 equations, and 3 unknowns.

(There might be other approaches to this problem, but this is one that works and is easy to set up initially.)

[Edit: Btw, sure, you can modify your original equation to have units of distance on both sides by making the right hand side 38 (instead of 38/32). But you still have two unknowns and only one equation, so you still can't solve it like that. The reason for this is because of the "averaging" idea. By going more general, you've lost the critical piece of information that the the first 19 [m] happens in one direction and the second 19 [m] in the other (as opposed to 18-20 or 17-21, etc). You'll need to break it up into at least two separate equations, so that you can put the 19 [m] back in there somewhere.]
 
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Thank you. Playing around with what you said helped me finally solve this problem. It seems so easy once you know how it's done.
 
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