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Fisherman's boat Relative motion

  1. Feb 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A fisherman wishes to cross a river to the shore directly south of his position. The river is 400 m across, and the fisherman's boat is capable of a speed of 2.5m/s. If the river's current has a velocity of 1.5m/s [W], determine

    a) the time it takes the fisherman to cross the river (200 s)
    how far downstream he will be before he reaches the other shore (300 m)

    This is the part I didn't answer.

    If the fisherman wished to land on the shore directly south of his position, how long would it him take to cross the river, and what would be the ground velocity of the boat?
    2. Relevant equations
    d =v*t

    3. The attempt at a solution
    I have calculated the time from question 1, but I'm not sure what is different about time in question 2. I was thinking the ground velocity is 2 m/s since it would be d/t.
     
  2. jcsd
  3. Feb 19, 2015 #2

    jbriggs444

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    How did you calculate the 200 second answer for a)?

    Edit: Scratch that. I see how you got 200 seconds. How did you get 300 meters?
     
  4. Feb 19, 2015 #3
    I made a triangle with 2.5m/s as the hypotenuse, 1.5m/s as one of the sides. and solved for the other side. I did t=d/v (400/2=200).
     
  5. Feb 19, 2015 #4

    jbriggs444

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    Right. Now how did you get 300 meters?
     
  6. Feb 19, 2015 #5
    1.5(200)
     
  7. Feb 19, 2015 #6

    jbriggs444

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    And what is 1.5?
     
  8. Feb 19, 2015 #7
    Velocity of the current.
     
  9. Feb 19, 2015 #8

    jbriggs444

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    Is it the same as the downstream velocity of the boat?
     
  10. Feb 19, 2015 #9
    No.
     
  11. Feb 19, 2015 #10

    jbriggs444

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    The question asks how far downstream the boat goes, not how far downstream the current goes. So multiplying by 1.5 would not be appropriate. But let's back up and re-examine what the question is asking...

    As I read this, there are two parts with two questions each.

    The first part (which you answered) assumes that the fisherman aims his boat due south relative to the moving water. So it drifts downstream at the same rate as the current. The second part (about which you are asking) assumes that the fisherman aims his boat at an angle so that his boat does not drift downstream.

    If this is the case, what is the boat's southward velocity for the first part?
     
  12. Feb 19, 2015 #11
    Yes. I calculated the angle in another part of the question. I got 53 degrees.
     
  13. Feb 19, 2015 #12

    jbriggs444

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    Are you assuming that the boat's ground speed is 2.5 meters per second or that the boat's speed with respect to the water is 2.5 meters per second? The problem assumes the latter.

    53 degrees is not consistent with the intended interpretation of the problem.
     
  14. Feb 19, 2015 #13
    The boat's speed is 2.5m/s
     
  15. Feb 19, 2015 #14

    Nathanael

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    That's not how I understand it. It says, "A fisherman wishes to cross a river to the shore directly south of his position. I understand this to mean he lands (not aims) directly south.

    Did you get this time assuming he aims south? Can you show how you got the time?

    That's one way to do it, since it's constant. I'm not exactly sure what you did to get the time. I found the time from the relative velocity. And this is the time I get for landing south, not aiming south.
     
  16. Feb 19, 2015 #15

    jbriggs444

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    That's how I read it at first as well. But the existence of the second part of the question makes that interpretation untenable.
     
  17. Feb 19, 2015 #16

    Nathanael

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    Or it could be a simple conceptual question not intended to be calculated.
     
  18. Feb 19, 2015 #17

    jbriggs444

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    The numbers are too carefully chosen for that interpretation to be tenable.
     
  19. Feb 19, 2015 #18
    I made a triangle with 2.5m/s as the hypotenuse, 1.5m/s as one of the sides. and solved for the other side. I did t=d/v (400/2=200). That's how I solved time.
     
  20. Feb 19, 2015 #19

    Nathanael

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    Why's that? The speed of the boat is more than that of the river.
     
  21. Feb 19, 2015 #20

    jbriggs444

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    And that's the wrong approach for the first part. 2.5 m/s is not the hypotenuse.
     
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