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Relative Velocity of an escalator

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data

    The escalator that leads down into a subway station has a length of 30.0 m and a speed of 1.8 m/s relative to the ground. A student is coming out of the station by running in the wrong direction on this escalator. The local record time for this trick is 11 s. Relative to the escalator, what speed must the student exceed in order to beat the record?

    2. Relevant equations

    Addition/subtraction of relative velocities.

    3. The attempt at a solution

    The velocity of the Student relative to the Ground is equal to the length of the escalator divided by the time it takes to travel the length of the escalator. In this scenario, the time will be the local record time as given.

    We know from the given information that x = +30.0 m and t = 11 s.

    [tex]\vec{v}[/tex]SG = velocity of the Student relative to the Ground = [tex]\frac{x}{t}[/tex] = [tex]\frac{+30.0}{11}[/tex] = +2.727272727 m/s
    [tex]\vec{v}[/tex]SE = velocity of the Student relative to the Escalator
    [tex]\vec{v}[/tex]EG = velocity of the Escalator relative to the Ground = +1.8 m/s

    [tex]\vec{v}[/tex]SG = [tex]\vec{v}[/tex]SE + [tex]\vec{v}[/tex]EG

    [tex]\vec{v}[/tex]SE = [tex]\vec{v}[/tex]SG - [tex]\vec{v}[/tex]EG
    [tex]\vec{v}[/tex]SE = (+2.727272727 m/s) - (+1.8 m/s)
    [tex]\vec{v}[/tex]SE = 0.92727273 m/s

    Relative to the escalator, the student must exceed a speed of 0.93 m/s [to the correct number of significant figures (2)].

    Question

    I feel pretty confident that I've approached this one the right way. Could someone please take the time to look it over and give me some feedback? Thank you... I really appreciate it.
     
  2. jcsd
  3. Feb 1, 2009 #2

    LowlyPion

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    Homework Helper

    That looks ok.
     
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