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Relative motion river problem

  1. Mar 17, 2014 #1
    1. The problem statement, all variables and given/known data
    A rower's maximum speed in still water is 6.0 ft/s. He decides to cross a 60 ft. wide river (at his maximum speed), with an apparent heading of 30° north of east. The river's current (relative to the ground) is flowing south at a speed of 1.0 ft/s. Find the rower's speed and direction relative to the ground, as well as the total time it will take him to reach the other side of the river. (Please let me know if you need clarification on this - I am recalling this problem from memory, and do not have the actual problem with me at the moment)


    2. Relevant equations
    vR/G = VW/G + VR/W
    (where R=rower, W=water, G=ground)

    Law of cosines: c2 = a2 + b2 - 2abcos(θ)
    Law of sines: sin(A)/a = sin(B)/b = sin(C)/c

    3. The attempt at a solution

    I set up the vectors in a coordinate plane. I was able to determine the angle between the rower's apparent heading vector and the water's vector to be 60°. Using this, I plugged it into the law of cosines formula to obtain a speed of 5.77 ft/s. I then used the law of sines (with the water's speed and the rower's speed of 5.77 ft/s) to obtain the angle opposite the water's vector, which turned out to be about 8°. Subtracting this from 30°, I got about 22° north of east.

    To find the time it takes to get across, I set up a right triangle, where the base represents the width of the river (60 ft), and the hypotenuse represents the distance the rower actually travels (with the angle between them 22°). Using trig, I found the hypotenuse to be about 64 ft. I then plugged this distance and the rower's speed of 5.77 ft/s into v=d/t to solve for t, and I obtained t= ~11s.
     
  2. jcsd
  3. Mar 17, 2014 #2
    Hey!

    I solved this problem in a bit of a different way and my numbers were different from yours, but not by a large margin.

    You can set it up in a vector notation first, and using the equation for relative motion that you had, I suggest setting this equation up in the fashion of Magnitude * unit vector.

    So, if using this method, you will have an equation with three unknowns, so you know that you must create two more equations using the same variables.

    First off, I get:

    V(xi + yj) = 6(cos30i + sin30j) - 1(j)

    i and j are the standard cartesian unit vectors, V is the magnitude of the rower's relative velocity, and x and y are the coefficients that are also unknown for the direction vector.

    If you wanted to make two equations out of this, try splitting this equation into the x and y components.

    You will still only have two equations, and the third equation can be easy to forget. I won't come out and say it explicitly, but just remember that the magnitude of a unit vector is always 1. After this it is just straight algebra and substitution.

    I can give you the numbers I got if you try this method, or just want to check how close they were. Good luck!
     
  4. Mar 17, 2014 #3
    Hi there! Wow, thank you for the very detailed explanation of your approach. I did not think to use vector notation to solve this problem. I'm going to try to solve this your way as soon as I get home from school!

    By the way, I'm wondering if your numbers were actually 5.57 ft/s and 21 degrees? My numbers were probably slightly off because I was trying to remember what I got when I first solved this problem (so they were rough guesses). After redoing it again, though, those are the numbers I get.

    Also - I was wondering if my way is still a valid approach (especially the last part where I need to find the time it takes to cross the river)?
     
  5. Mar 20, 2014 #4
    The last part is actually exactly how I solved the total time as well, I think I just multiplied the base of the triangle times the tangent of your angle θ.

    But yeah, V = d/t is completely applicable to that scenario, so yes, you're correct.

    I am pretty sure that the approach you used is completely valid. I remember by velocity was something like 5.77 ft/s and my angle was a bit over 21°, and my total time was also ridiculously close to 11 seconds. So, different ways to do the same thing.
     
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