PHYSICS PROBLEM ON ONE DIMENSIONAL MOTION

 Yesterday  i found this interesting problem  .    Have fun in doing this.



On April 15 an airplane takes off at 4 : 40 p.m from Belem. Brazil bound for Villamil, Ecuador (in the Galapagos). The plane lands at 8 : 40 p.m. Villamil local time. The sun sets at 6 : 15 p.m. in Belem (local time), and 7 : 06 p.m. in Villamil (local time). At what time during the flight do the airplane passengers see the sun set? 

Comments

  1. Kantshri BaroniaMay 16, 2012 at 5:47 AM

    what's the solution??

    ReplyDelete
  2. let the distance frm belem to
    vilamil is x km
    nd the distance frm belem to
    the place where the passenger
    sees sunset is y km
    sunsets inbelem at 6.15 nd in vilamil is 7.06
    hence the time difference
    between the two places is 7.06
    - 6.15 = 0.51 hrs= 51 min
    consider time according to
    belem
    the plane reaches vilamil at
    8.40 local time..............
    the time
    in belem when the plane
    reaches vilamil is 8.40 - 51 min =7.49. pm
    therefore time taken for the
    journey is 7.49 - 4.40= 3.09 hrs
    let v (kmph) be the velocity
    then x = v(3.09) km-----------1
    the time taken for the sun set
    frm begining of the journey is
    6.15 - 4.40 =1.75 hrs then y = v (1.75) km-----------2
    for x distance time variation is
    0.51 hrs
    for y distance time variation
    is (T) T =0.51

    dividing eqn 2 by
    eqn 1
    we get y/x = 1.75/3.09 therefore the time variance
    btn the place where the
    passebger sees the sunst nd
    belem is
    T = 0.51 X1.75/3.09=
    0.28 hrs =
    28 min
    therefore the time at the place wer passenger sees
    sunset is =
    sunset at belem
    +time difference btn belem nd
    that place (T)
    = 6.15 hrs+28 min =6.43 p.m
    by prabhat kumar singh

    ReplyDelete

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