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Relative Motion in vectors form

  1. Oct 9, 2004 #1
    The question is as follow:
    A bird is capable of flying at 80km/h. It wishes to fly to its nest which is due East of its present position. Thre is a win blowing from the northwest at 70km/h. Find the direction relative to the air in which the bird must fly to reach its nest.

    I tackled the problem like this:
    1) Velocity of air = (35root(2))i + (-35root(2))j
    Resultant velocity (=velocity of bird) = due east
    Velocity of bird relative to air = 80j

    Is this correct?

    actually I think the method is correct as well, but surely the bird cannot fly just in the north direction (like a helicopter).

    Or should I think like this:
    2) for i direction: 35root(2) = 80cos(theta) - this gives the right angle tho.
    3) for j direction: 35root(2) = 80sin(theta) - I believe this method is correct because you want the resultant direction due east, therefore no j component of the resultant velocity.

    Please help.
    Last edited: Oct 9, 2004
  2. jcsd
  3. Oct 9, 2004 #2

    Doc Al

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    Staff: Mentor

    Let B=bird, A=air, G=ground, and B/A = "bird with respect to ground", etc. The relationship among the relative velocities is: [itex]\vec{V_{B/G}} = \vec{V_{B/A}} + \vec{V_{A/G}}[/itex]. [itex]\vec{V_{A/G}}[/itex] is given, and you know that [itex]\vec{V_{B/G}}[/itex] must have a direction of east. Solve for the components of [itex]\vec{V_{B/A}}[/itex], knowing that it has a magnitude of 80 km/h.
  4. Oct 9, 2004 #3
    I did: v(b/a) = v(b) - v(a)
    v(b/a) = 80i - ((30root2)i - (30root2)j)
    v(b/a) = (80-30root2)i = (30root2)j
    and got an angle of 48.5 degrees. Is this correct? (but the book says it should be 52 degrees)
  5. Oct 9, 2004 #4


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    Staff Emeritus
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    The fact that the bird "is capable of flying at 80km/h" tells you that the speed of the bird (the length of the velocity vector) is 80. It does NOT tell you the direction: that's the whole point of the problem. Draw a picture: the bird wants to fly due east so draw a line in that direction. The wind is 70 km/h from the nw so draw a line like that ending at the tip of the east line. Now draw a line from the beginning of the east line to the beginning of the "nw" line. That's the triangle you need to solve. You know the lengths of two sides (80 and 70) and you can know one angle (45 degrees) so you can solve for the other angles.

    By the way, I'm very puzzled by your remark "surely the bird cannot fly just in the north direction (like a helicopter)."

    ??? are you sure you know which direction north is?! Birds fly north every spring!
  6. Oct 9, 2004 #5
    i tried that but didnt get the right answer. I was being stupid for the north direction thing - thought in two dimensions there are only up and down, and from west to east.
  7. Oct 9, 2004 #6

    Doc Al

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    Staff: Mentor

    Your error is taking the speed of the bird with respect to the ground to be 80. True, it does travel east. But 80 km/h is its speed with respect to the air, not the ground.

    Hint: find the y-component of the bird's velocity with respect to the air.
  8. Oct 9, 2004 #7
    i got:
    v(a) + v(b/a) = v(b)
    (35root2)i - (35root2)j + xi + yj = zi
    but v(b) has no j component: therefore y = 35root2
    x(square) + 35root2(square) = 80(square)
    x = root2950
    tan(theta) = 0.78756153 --> theta = 38.2 degrees
    so the bearing is 52 degrees as required.
    Last edited: Oct 9, 2004
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