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Lifeguard swimming to a dock

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data
    A lifeguard who can swim at 1.2 m/s in still water wants to reach a dock positioned perpendicularly directly across a 550 m wide river.

    a. If the current in the river is 0.80 m/s, how long will it take the lifeguard to reach the dock?
    b.If instead she had decided to swim in such a way that will allow her to cross the river in a minimum amount of time, where would she land relative to the dock? (10 marks)


    2. Relevant equations



    3. The attempt at a solution
    I already figured out a with the Pythagorean theorem and so his avg speed is 1.44 m/s and then t= d/Vavg to give me 347,22 sec and so it will take him 345.22 seconds to reach the dock.
    but my question is for b. It has me stumped.
     
  2. jcsd
  3. Mar 13, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi TheronSimon! Welcome to PF! :smile:
    hint: if x is downriver, and y is across the river,

    just concentrate on how fast her y coordinate is increasing :wink:
     
  4. Mar 14, 2012 #3
    but part A. is correct yes?
     
  5. Mar 14, 2012 #4
    and for part b. if i do
    t= 500/1.2
    t= 416.66s
    so she will take 6.94 min to reach the location and then
    d= 0.8*416.66
    d= 332.8m
    so she will land 332.8 m from the dock in 6.9 min
     
  6. Mar 14, 2012 #5

    tiny-tim

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    Hi TheronSimon! :wink:
    Yes, that's fine :smile: (except is it 500/12 or 550/12 ?)
     
  7. Mar 14, 2012 #6
    550 :P darn typos
     
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