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Homework Help: Crossing a River

  1. Sep 13, 2004 #1
    A swimmer wants to cross a river, from point A to point B. The distance d_1 (from A to C) is 200 m, the distance d_2 (from C to B) is 1500 m, and the speed V_r of the current in the river is 5 km/h. Suppose that the swimmer makes an angle of 45 degrees(0.785 radians) with respect to the line from A to C, as indicated in the figure.

    Question #1) To swim directly from A to B, what speed U_s, relative to the water, should the swimmer have? Express the swimmer's speed numerically, to three significant figures, in units of kilometers per hour.

    U_s[/tex] = __________km/h

    ok here's what i think i should do. first i need to find the distance from A to B. so of course a^2 + b^2 = c^2
    1500^2 + 200^2 = C^2
    C= 1513.37m from distance A to B

    ok i know that it makes a 45 degrees angle with respect to the line from A to C. and the inner angle(dont know the math term) or the angle next to it is also 45 degrees(90-45).
    that means i can find the initial velocity of Vx and Vy.

    have to convert meters to km first right? so 1500 would be 1.5
    Vx = 1.5km*cos(45) = 1.06 km/h
    Vy = .2km*sin(45) = .1414 km/h

    ok so i know the distance from A to B, the Vx and Vy initial velocities. im stuck here, dont know what else to do. the picture is attached.

    Attached Files:

    Last edited: Sep 13, 2004
  2. jcsd
  3. Sep 13, 2004 #2
    k changed latex into normal text, i guess latex is not working right now.
  4. Sep 13, 2004 #3
    First, you mentioned the distance between B and C as 150 m, and later said it was 1500.

    A second thing to keep in watch, when you converted m to km, and multiplied it by cos(45) or sin(45), you changed distance to velocity by adding the "/h". Really, 1.5km * cos(45) = 1.06 km

    Keep in mind that the river velocity is 5 km horizontally. The velocity of the swimmer will need to be opposite of that, as if he or she is to swim in a straight line, the overall horizontal movement of the swimmer will need to be 0.

    Now, that you know the horizontal velocity, you can find the overall velocity of the swimmer.
  5. Sep 13, 2004 #4

    i had a typo while changing latex into normal text. it's 1500m from C to B. sorry about that.

    how would i find the overall velocity of the swimmer? do i need to use the final velocity formula(v(t) = V(0) + at)?

    if the horizontal velocity is 5km, wouldnt it help the swimmer to get to points A to B faster? im really trying, but i just dont get it.
  6. Sep 14, 2004 #5
    Has anyone had any luck with this problem?
  7. Sep 14, 2004 #6
    oops, my bad agian, it's actually 150m from C to B lol. sorry...
  8. Sep 14, 2004 #7
    Heres how I did it. I figure this has been unanswered long enough to give more than a hint. I hope im right.

    Well start out in vectors.
    Assume to the left is positive, and up is positive.
    Velocity of Swimmer (relative to standing water) = Vx i + Vy j
    We know that since its at a 45 degree, Vx/Vy = 1, let Vx=Vy=Vs
    Velocity of Swimmer (relative to standing water) = Vs i + Vs j

    The total velocity of the Swimmer as he is crossing the river(relative to the shore) is
    Vs + Vw = Vs i + Vs j - 5000m/hr i = (Vs - 5000) i + Vs j
    Vw is the velocity of the water.

    Now we just use the simple : x=vt

    for x :
    -150 = (Vs-5000)* t /// REMEMBER he moves in the negative direction.
    for y :
    200 = Vs * t
    Solve for t = 200/Vs
    Plug into first eq:
    -150 = (Vs-5000)*(200/Vs)
    Solve for Vs
    Vs = 2857.14 m/hour == 2.86 km/hr
  9. Sep 16, 2004 #8
    Hi everyone. Just figured out the answer to be 4.04km/h.
  10. Sep 16, 2004 #9
    yup it's correct
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