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A moving walkway at an airport has a speed $v_{1}$ and a length $L$ . A woman stands on the walkway as it moves from one end to the other, while a man in a hurry to reach his flight walks on the walkway with a speed of $v_{2}$ relative to the moving walkway. (a) How long does it take the woman to travel the distance $I Z$ (b) How long does it take the man to travel this distance?
a. $\frac { L } { v _ { 1 } }$
b. $\frac { L } { v _ { 2 } + v _ { 1 } }$
Numerade Educator
Hope College
University of Sheffield
University of Winnipeg
though the time for part of the time required for the woman we can solve immediately this simply be the length divided by be someone visa. But of course, being the woman's constant velocity, so this would be the time required for the woman. Again, we have no acceleration for part B. We can say that the velocity of the man relative to the walkway is equaling the velocity of the man relative to the earth, minus the velocity of the walkway relative to the earth. And so we can say that the absolute the magnitude of the velocity, rather, the magnitude of the lost city of the man relative to the earth is equaling essentially the velocity of the man relative to the world walkway, plus the magnitude of the velocity of the walkway relative to Earth. And so this is equaling visa to plus Visa of one. And so we can say that the time required by the man see, it's gonna be equaling the length divided by the sub two plus visa of one. So it's gonna be a lot less time, because again, he's using the walkway in order to, um, speed up his movement. So this would be for part B. This would be for party. That is the end of the solution. Thank you for watching.