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Difficult vector word problem final step

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data
    The current in a river flows North at 5mph. A boat starts straight across the river at 8mph relative to the water.

    (a) What is the speed of the boat relative tot he land?

    (b) If the river is 2mi wide, how long does it take the boat to cross the river?

    (c) If the boat sets out straight for the opposite side, how far north will it reach the opposite shore?

    (d) If we want to have the boat go straight across the river, at what angle should the boat be headed?


    3. The attempt at a solution
    W = 5
    Wθ = 90
    Wx = 5cos(90) = 0
    Wy = 5sin(90) = 5
    ---
    B = 8
    Bθ = 0
    Bx = 8cos(0) = 8
    By = 8sin(0) = 0

    R = √[64+25] = 9.43
    Rx = 8
    Ry = 5
    Rθ = arctan(5/8) = 32
    ---
    (a) 9.43mph
    ---
    (b) 60/9.43 = 6.36 ∴ it takes 6.36min for the boat to go 1mi. The actual distance the boat travels is given by the hyp of a triangle across the river, the hyp is given by 2/[cos(32)] = 2.36mi ∴ 2.36*6.36mi = 15m to cross the river
    ---
    (c) The distance the boat travels north is give by the opposite side of the triangle across the river, the opp is given by 2.36*sin(32) = 1.25mi North
    ---
    (d) I thought about this for a while, and I can't seem to figure it out... The answer the books gives is 38.7° SE


    Thanks!
     
  2. jcsd
  3. Jan 26, 2013 #2

    haruspex

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    Yes, but there's an easier way. Relative to the water, the how far does the boat have to travel?

    For (d), suppose the boat heads at angle theta South of the straight-across direction. How fast would it be travelling up river?
     
  4. Jan 29, 2013 #3
    I don't understand :S sorry
     
  5. Jan 29, 2013 #4

    haruspex

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    If you take the flowing water as your frame of reference, the boat is heading at 8mph straight across a width of 2 mi, so takes 15 minutes.
    Forget the flow of water for the moment. If the boat were to travel at 8mph but, instead of pointing straight across, it heads at angle theta South of straight across. What will the components of its speed be in the two directions (straight across and South)?
     
  6. Jan 29, 2013 #5
    The x and y components of the boat at the angle theta would be

    Bx = 8cos(θ)
    By = 8cos(θ)
     
  7. Jan 29, 2013 #6

    haruspex

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    They can't both be cos. Which are you calling x and which y?
    When you've sorted that out, what do you get when you add in the effect of the current?
     
  8. Jan 30, 2013 #7
    omg haha sorry. Typing error. The second one was supposed to be sin. I do that sometimes in my homework too :S.

    The x and y comp's of the resultant vector when the boat starts off in a direction theta are

    Rx = 8cos(θ)
    Ry = 8sin(θ) + 5


    wait...I think I understand.

    arctan([8sin(θ)+5]/[8cos(θ)]) = 0

    [8sin(θ)+5]/[8cos(θ)] = tan(0)

    [8sin(θ)+5]/[8cos(θ)] = 0

    8sin(θ)+5 = 0

    sin(θ) = 5/8

    θ = arcsin(5/8) = 38.7°

    Wow thanks dude ^^
     
  9. Jan 30, 2013 #8

    haruspex

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    With the way I defined theta, that should be -5
    If 8sin(θ)+5 = 0 then sin(θ) = -5/8, but given that it should have been -5 in the first place, +5/8 is right. Well done.
     
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