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Relative Velocity Question

  1. Oct 18, 2003 #1
    Id be very grateful for any help with this relative velocity question:

    To a man walking North at 4m/s, a constant wind appears to have a speed f 3m/s. The man then changes direction, and when he is walking West at 3m/s the wind appears to have a speed of 4 m/s. There are two possible wind velocities.

    If I and j are unit vectors in the east and north directions, respectively:
    1) Show that -3i + 4j is one possible wind velocity
    2) Find the second possible wind velocity.

    Ok, so this is what I¡¦ve got so far:
    Vm (Velocity of man) = 4m/s = 4j m/s. (When wind appears to have Vw (Velocity of wind) of 3m/s)
    Vm (Velocity of man) =3m/s = 3i m/s (wind appears to have Vw (Velocity of wind) of 4m/s)

    Vm = 4j
    Vwm = xi
    Vw = ai + bj

    Vwm = Vw ¡V Vm
    Xi = ai + bj ¡V 4j
    Xi = ai + bj +0i -4j
    „Ã xi = ai
    „Ã x = a

    „Ã 0 = b ¡V 4
    „Ã b=4

    Vm = -3i
    Vwm = yi ¡V yj
    Vw = ai + bj

    Vwm = Vw ¡V Vm
    Yi ¡V yj = (ai + bj) + 3i
    „Ã yi = ai +3i
    „Ã -y = b
    „Ã y = -4

    Y = a ¡V 3
    -4 = a ¡V 3
    1 = a

    This gives me a velocity of 1i + 4j.

    Have i gone totally wrong, or have i just found the answer to part two instead of part one?
    If so, what do i do next to find the second velocity, or, if its all just totally wrong, how should i actually go about doing the question?

    Thanks in advance for any help you can give.
  2. jcsd
  3. Oct 19, 2003 #2
    Any help is appreciated
  4. Oct 20, 2003 #3
    Anybody at all?
  5. Oct 20, 2003 #4


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    Staff Emeritus
    Science Advisor

    Okay, I saw this the other day, meant to respond and didn't get around to it!

    The man is, originally walking n so, setting up a coordinate system with i pointing east, j pointing north, his velocity vector is, as you say, 4j.
    I'm not at all sure what you mean by
    "Vwm = Vw ¡V Vm
    Xi = ai + bj ¡V 4j
    Xi = ai + bj +0i -4j"

    You defined Vm and Vw but you didn't say what Vwm was except "xi" which isn't a definition since you hadn't define xi! What is "iV"?

    Here's how I would do it: writing the wind's velocity vector as ai+ bj (your "Vw"), the velocity of the wind relative to the man is (ai+ bj)- 4j= ai+ (b-4)j. The speed of the wind relative to the man is √(a2+ (b-4)2). We are told that this 3.

    When the man is walking west at 3 m/s, his velocity vector is -3i (you have 3i: that would be walking east) and the velocity of the wind relative to him is (a+ 3)i+ b. The speed of the wind relative to the man is √((a+ 3)2+ b2). We are told that that is 4.

    Squaring both sides of √(a2+ (b-4)2)= 3, we get (a2+ (b-4)2= 9.
    Squaring both sides of √((a-3)2+ b2) we get (a-3)2+ b2= 16.
    That gives 2 equations to solve for a and b.

    Actually, we don't need to solve to answer the first part. It just asks that we show that -3i+ 4j (that is, a=-3, b= 4) satisfies the conditions. Putting a= -3, b= 4 in the first equation, we have
    (-3)2+ (4-4)2= (-3)2+ 02= 9 which is correct.
    Puting a=-3, b= 4 in the second equation, we have
    (-3+ 3)2+ (4)2= 16 which is, again, correct.
    Yes, a wind speed of -3i+ 4j satisfies both conditions.

    Going back to the equations that must be satisfied,
    (a2+ (b-4)2= 9 and
    (a+3)2+ b2= 16 we can multiply the squares and get:
    a2+ b2- 8b+ 16= 9 and
    a2+ 6a+ 9+ b2 = 16.
    Subtracting the second equation from the first, the "a2" and "b2" terms cancel and we have
    -8b- 6a= -14 or, dividing by -2, 4b+ 3a= 7.
    (Notice that b= 4, a= -3 satisfy this.)

    putting a= 7/3- (4/3)b into the fist equation, we have
    (7/3-(4/3)b)2+ (b-4)2= 9 or
    49/9- (56/9)b+ (16/9)b2+ b2-8b+ 16= 9 which is the quadratic equation (25/9)b2- (128/9)b+ 112/9= 0.

    That is the same as 25b2- 128 b+ 112= 0.

    Knowing that b= 4 is one solution, it shouldn't be hard to factor and find the other possible value for b and then find a.
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