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[Uni Physics w/ calc] Boat crossing a river

  1. Jan 23, 2015 #1
    The problem is:

    A 230-m-wide river has a uniform flow speed of 1.4 m/s through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 4.6 m/s with respect to the water. There is a clearing on the north bank 63 m upstream from a point directly opposite the clearing on the south bank. (a) At what angle, measured relative to the direction of flow of the river, must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank? (b) How long will the boat take to cross the river and land in the clearing?

    What I have been trying is, multiplying 4.6 by the costheta and sintheta (theta being the angle the boats velocity points toward from west towards north) to get the components of the velocity and then subtracting 1.4 from the x component because of the rivers current. Then I get position information by integrating the velocity equations. I set the x component of the position equal to -63 (I made west positive and east negative) and then solve for t. Then I plug that t into the y component of the position and try to solve for theta. I end up with 230costheta+63sintheta=70. I tried to solve it and quickly realized it is pretty much impossible to find theta from that. So if somebody could figure out what I need to change in how I'm trying to do this problem that would be great, I have been stuck on this problem for 3 days. Last problem in my homework set. Thank you!
     
  2. jcsd
  3. Jan 23, 2015 #2

    haruspex

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    There are two components of position, giving you two equations sharing the same time value.
     
  4. Jan 23, 2015 #3
    The resultant of vector addition should be the path taken by the boat.
     
  5. Jan 23, 2015 #4
    Can someone PLEASE just actually sit down and try to solve this problem and then explain to me the steps they took. I've been looking for help on many website and everyone does the same thing. They read the question, think for what seems to be 3 seconds, then reply something like "add the vectors". That's not helpful. I need someone who really knows their stuff to solve the problem and then explain to me how they went about doing it. I know I'm complaining about the help I have gotten, but it's useless. I went to a TA's office hours and he had 15 people in his office and by the time it would have been my turn the office hours would have ended. I'm desperate.
     
  6. Jan 23, 2015 #5
    Haha. Ok ok. Calm down.

    Let's break it into 2 parts, alright?
    1) First, the river is flowing at 1.4 m/s. This means that if the boat is left in the river on it's own, it will flow at this speed.by its own.
    2) Secondly, the boat by itself has a speed of 4.6 m/s, which means that in still water, it will travel with this speed.

    Now upstream means in the direction opposite to the direction of flow of water. So what you basically have to do is-
    Travel 230 m from South to North.
    Travel 63 m from East to West against the current (against 1.4 m/s).
    Now these two journeys need to take the same time. This will give you your two vectors and the angle between them.

    Got a fair idea?
     

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  7. Jan 23, 2015 #6

    Nathanael

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    My approach for part (a) doesn't involve time. I would write an equation that says "the ratio of (east-west component of boat's velocity minus velocity of the river) to (south-north component of boat's velocity) equals the ratio of east-west distance (63 meters) to south-north distance (230 meters)"
     
  8. Jan 23, 2015 #7
    Hi thank you so much for this response, it's coming together. Although, in the question it wants me to find the angle relative to the flow of the water. In your picture can you point out which angle that is? In your picture is the angle the question is asking for the one that creates a less than 90 degree angle or the one that creates an over 90 degree angle with the desired path (the arrow pointing right and up) of the boat?
     
  9. Jan 23, 2015 #8
    Glad to help :)
    Come on, try it now! Read the question once again and see which angle they are asking for.
     
  10. Jan 23, 2015 #9
    When I tried to do that I end up with 1058costheta-322=289.8sintheta I don't know how to solve for theta from that. Thanks for the response though! I'm thinking maybe I can somehow get the cos or the sin in terms of the other using some other relationship?
     
  11. Jan 23, 2015 #10
    It says the angle relative to the flow of the river does that mean the angle that points toward the flow or against it?
     
  12. Jan 23, 2015 #11

    Nathanael

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    You just need one more constraint. That comes from the nature of sine and cosine; [itex]\sin^2\theta+\cos^2\theta=1[/itex]
     
  13. Jan 23, 2015 #12
    Its the angle BETWEEN the direction of flow of the river and the direction of travel of the boat.
     
  14. Jan 23, 2015 #13

    haruspex

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    That's not how it works. You are supposed to post the detailed steps you took, then others comment. Unfortunately you only posted a description of what you did. That makes it much harder to figure out exactly where you are stuck.
     
  15. Jan 23, 2015 #14
    Okay so I tried some stuff and couldn't get anywhere. Could you please tell me what the constraint is?
     
  16. Jan 23, 2015 #15

    Nathanael

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    Sorry if the way I said it confused you; [itex]\sin^2\theta+\cos^2\theta=1[/itex] is the constraint. You can use it to get a quadratic equation for either cosθ or sinθ.
     
  17. Jan 23, 2015 #16
    So I solved for sin theta and then plugged it into the identity your referencing. Then I used wolfram alpha to find the zeros of the quadratic which were costheta = 0.0305627 and costheta = 0.535652. Now, the issue is I have 2 submissions left on this question before I'm locked out. Does anybody think either of these thetas are correct and if so, which one?
     
  18. Jan 23, 2015 #17

    Nathanael

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    Take the arccos to find the angles. Only one of them is reasonable. Which one?
     
  19. Jan 23, 2015 #18
    Here is a more detailed account of what I did.
    So I have 1058costheta-322=289.8sintheta from earlier. I take this and solve for sintheta. Then, I plug what sintheta equals into sinsquared+cossquared=1. I substitute x for costheta. Then I plug the quadratic into wolfram alpha. It says x=0.535652 and x=0.0305627
     
  20. Jan 23, 2015 #19
    Why is only one of them reasonable? One of them equals 57. something and the other is 88. someth
     
  21. Jan 23, 2015 #20

    Nathanael

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    Well, what if the river was perfectly still. Then what angle would you be traveling?
     
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