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A plane is flying through the air at a velocity 88km/hr

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data

    A plane is flying through the air at a velocity 88km/hr direction N37'E on a day when there is no wind.
    a) After 50 minutes, how far has the plane gone?
    b) How far east is it from it's starting point?
    c) How far north is it from its starting point?

    If there is a wind blowing FROM due south at 17mk/hour, what would be the answers to the question above?

    2. Relevant equations


    time=distance/speed

    3. The attempt at a solution

    50 minutes = distance/ 88 km/h
     
  2. jcsd
  3. Mar 24, 2012 #2

    tiny-tim

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    welcome to pf!

    hi jacknjersey! welcome to pf! :smile:
    ok, now convert minutes to hours …

    (you need everything to be in the same units! :wink:)

    the distance is … ? :smile:

    (time=distance/speed, speed=distance/time, distance=speed*time)
     
  4. Mar 24, 2012 #3
    Re: time=distance/speed

    50 minutes is 0.83 hours

    so now I have distance = speed x time

    distance = 0.83 x 88
    73.04km

    I'm not sure what to use to get b and c - how far north and east? argh.
     
  5. Mar 24, 2012 #4

    tiny-tim

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    well, you know the angle (37° to the right of north) …

    so draw a triangle, and use cos and sin :wink:
     
  6. Mar 24, 2012 #5
    Re: time=distance/speed

    you rock. tx!!
     
  7. Mar 24, 2012 #6
    Re: time=distance/speed

    cos theta = adjacent/ hypotenuse
    cos 37'=x/ 73.04
    x=55.90584235
    x=55.91km

    sin theta = opposite/hypotenuse
    sin 37'= y/73.04
    y=73.04 sin 37
    y=43.96km

    If the wind was blwoing from the south at 17km/hour, the wind is blowing due north.

    therefore:

    after 50 minutes at 17 km p/h, plane has travelled
    distance = 0.83 x 17
    distance = 14.11km

    how far east is it fromit's starting point?


    how far north is it from it's starting point?
     
  8. Mar 24, 2012 #7

    tiny-tim

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    you can regard the wind as being separate from the engine …

    just ad the displacement due to the wind to the displacement due to the engine :wink:
     
  9. Mar 24, 2012 #8
    Re: time=distance/speed

    simply put, 17+88?
     
  10. Mar 24, 2012 #9

    tiny-tim

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    no!!

    add the displacements (the changes in position, the distances)
     
  11. Mar 24, 2012 #10
    Re: time=distance/speed

    the plane went 73.04km. add the wind of 17.
    = 90.04
     
  12. Mar 24, 2012 #11

    tiny-tim

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    haven't you been taught vector addition?

    to add displacements, you have to add the x and y amounts (the components) separately

    (try drawing it)
     
  13. Mar 24, 2012 #12
    Re: time=distance/speed

    do i have all the coordinates?
     
  14. Mar 24, 2012 #13

    tiny-tim

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    yes!!
     
  15. Mar 24, 2012 #14
    Re: time=distance/speed

    Ive done a drawing and to get the resultant vector, i need to draw a tail from the head of the second to the tail of the first and find that length. the length of this vector is the magnitude of the resultant vector.

    Should i paste in my picture?
     
  16. Mar 24, 2012 #15

    tiny-tim

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    hi jacknjersey! :smile:
    that's correct :smile:

    but do b) and c) first (how far east and north), then you can use pythagoras to find the magnitude :wink:
     
  17. Mar 24, 2012 #16
    Re: time=distance/speed

    how far east? 73.04
    how far north from starting point? 17 + 43.96 = 60.96km
    how far has the plan gone? x^2=73.04^2 + 17^2 = 74.99km
     
  18. Mar 25, 2012 #17

    tiny-tim

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    that's right! :smile:
    no 73.04 was the total distance from the engine, you need to use the east distance (and add 0 :wink:)
    no, for the total magnitude, use pythagoras on the total north and total east
     
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