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Boat width problem

  1. Oct 2, 2015 #1
    1. The problem statement, all variables and given/known data
    You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determine the width of the river and the current speed (the speed of the water relative to the earth). You have a small boat with an outboard motor. By measuring the time it takes to cross a pond where the water isn’t flowing, you have calibrated the throttle settings to the speed of the boat in still water. You set the throttle so that the speed of the boat relative to the river is a constant 6.00 m/s . Traveling due north across the river, you reach the opposite bank in 20.1 s . For the return trip, you change the throttle setting so that the speed of the boat relative to the water is 8.20 m/s . You travel due south from one bank to the other and cross the river in 11.2 s .

    How wide is the river?

    2. Relevant equations


    3. The attempt at a solution

    Honestly I don't even know where to start.
    All I did was take the distance travelled by the two trips
    D1=120.6m
    D2=91.84m
     
  2. jcsd
  3. Oct 2, 2015 #2

    DrClaude

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    Staff: Mentor

    Make a sketch of the situation.
     
  4. Oct 2, 2015 #3
    These are the distances you've made through the water, but you want to know the distances over ground. So you need a superposition of the velocity of the boat through water and the velocity of the river over ground. In the end for both directions you must have travelled the same distance (as the width of the river doesn't change).

    Try to find the equations including the velocity of the river over ground (with the unknown vRiver).
     
  5. Oct 12, 2015 #4
    Two equation like D = v ⋅ t

    D ... distance over ground (real distance), must be the same for both cases as the width of the river doesn't change
    t1,2 ... times you needed to cross the river
    v1,2 .. velocities over ground consisting of the superposition of the boat speed through water + the speed of the river over ground

    two equations, two unknowns - try to find the mathematical connection

    hint: you also can solve it graphically using vectors
     
  6. Feb 16, 2016 #5
    Hello!

    I am also trying to figure out this problem and I don't understand what you guys mean about "superposition" or solving it with vectors. How do I go about setting up the equations?
     
  7. Feb 17, 2016 #6
    Think triangles and vectors. The first vector will be pointing at a diagonal north and the second vector will point south.
     
  8. Feb 18, 2016 #7
    Superposition: If you swimming in a river, you move with respect to the water, because you use the water as resistance to push yourself in one direction or the other. If now the water is moving with respect to the ground (which is natural for a river), you have to account for that to. To find your movement over ground (which is the crucial one to calculate how and when you will reach the other side) you have to sum up the your velocity in the water plus the velocity of the river over ground - that's meant with superposition.

    Vectors. Now it is obvious, that if the speed of the river over ground and your speed in the water don't point in the same direction, you can't just add them without taking into account their directions (like scalars). Therefore you can use vectors. If you add the vectors for the two velocities taking into account their correct lengths AND directions, you'll find the final (net) speed of your boat over ground.

    Regarding the equations: What have you already tried? Could you show any possibilty for an attempt?
     
  9. Feb 18, 2016 #8

    SammyS

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    It does have a solution.
     
  10. Feb 18, 2016 #9

    NascentOxygen

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    Staff: Mentor

    That advice holds true for all vector problems. Yet here we see members still attempting this problem without reference to any figure. Why make vector problems difficult for yourself?

    Draw two sketches, one for each crossing.
    Once you have the vector diagram figured out, the problem is usually as good as solved!
     
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