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Math Help - Velocity of boat

  1. #1
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    Velocity of boat

    A straight river is 20 m wide. The velocity of the river at (x, y) is v=[-(3x(20-x))/(100)]j m/min, 0<=x<=20

    A boat leaves the shore at (0, 0) and travels through the water with a constant velocity. It arrives at the opposite shore at (20, 0). The speed of the boat is always radical 20 m/min.

    a. Find the velocity of the boat.
    b. Find the location of the boat at time t.
    c. Sketch the path of the boat.

    I figured the velocity of the boat, since traveling in the positive direction on the x axis, would also be radical 20 m/min. The boat seems to cross the river and end up DIRECTLY across, so I tried to set up a right triangle where the hypotenuse represented the boats velocity (radical 20) and the vertical leg represented the river velocity (-40 m/min? calculated using the integral of the equation from 1 to 20?). I'm not sure what I'm doing and an totally stuck. Thank you.
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  2. #2
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    Quote Originally Posted by jennifer1004 View Post
    A straight river is 20 m wide. The velocity of the river at (x, y) is v=[-(3x(20-x))/(100)]j m/min, 0<=x<=20

    A boat leaves the shore at (0, 0) and travels through the water with a constant velocity. It arrives at the opposite shore at (20, 0). The speed of the boat is always radical 20 m/min.

    a. Find the velocity of the boat.
    b. Find the location of the boat at time t.
    c. Sketch the path of the boat.

    I figured the velocity of the boat, since traveling in the positive direction on the x axis, would also be radical 20 m/min.
    Do you understand the difference between "velocity" and "speed"? The velocity is a vector while speed is a number. So the velocity CANNOT be "radical 20 m/min", that's not a vector. The speed is the magnitude of velocity vector. If the velocity at any point is v_x\vec{i}+ v_y\vec{j}, then the speed is \sqrt{v_x^2+ v_y^2}.

    The boat seems to cross the river and end up DIRECTLY across, so I tried to set up a right triangle where the hypotenuse represented the boats velocity (radical 20) and the vertical leg represented the river velocity (-40 m/min? calculated using the integral of the equation from 1 to 20?). I'm not sure what I'm doing and an totally stuck. Thank you.
    Since the both goes straight across its \vec{j} component of velocity must oppose the velocity of the river (which is NOT -40 m/min. The river speed is dy/dt- integrating (dy/dt)dx doesn't give anything meaningful!), -\frac{3x(20-x)}{100}\vec{j} m/min. So you know the velocity vector is v_x\vec{i}+\frac{3x(20-x)}{100}\vec{j} and you must have \sqrt{v_x^2+ \left(\frac{3x(20-x)}{100}\right)^2}= 20
    Solve that for v_x.
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  3. #3
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    Would v_x=20 if x=0? v_x=4 (rather than -4) if x=20? or just 20-[(60x-3x^2)/100]?
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