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Angle,Velocity and Time

  1. Oct 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Jason is canoeing the Ohio River,Jason forgot his tent on the western shore. Jason is on the Eastern shore directly east of his tent. Jason is going upstream on the river.

    The velocity of the river is Vr +3.311 M/S due North
    The speed of the canoe relative to the water is Vc=5.471 m/s
    The width of the river is W=69702

    A)At what angle relative ti the East-West line must Jason aim his canoe?Theta=
    B)What is Jason's velocity as he crosses the river?V=
    C)How long will it be before Jason gets to his tint?T=

    2. Relevant equations
    sin theta=opp/hyp
    t=d/v
    v=d/t

    3. The attempt at a solution
    A)I'm not sure if I'm correct but to find the angle,I did this:

    sin theta= 3.311 m/s / 5.471 m/s
    theta=36.8 degrees

    b)To find the velocity:
    v=d/t

    69702m/15769s=4.42 m/s

    c)To find the time:
    t=d/v

    69702m/4.42m/s=15769s

    Something is telling me that this problem is not as simple as it seems. Some folks used the pythagorean theorem. Are my answers correct,if not Please help. Thanx
     
    Last edited: Oct 4, 2007
  2. jcsd
  3. Oct 4, 2007 #2

    cristo

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    Staff Emeritus
    Science Advisor

    Your first answer is correct. For the last two, I don't know what you've done: you appear to be using numbers in the second question that you havn't obtained until the third, and where does 4.42 come from?

    I would use pythagoras for the second question.
     
  4. Oct 4, 2007 #3
    Thanx Cristo its truely apreciated,

    Forget the 4.42 stuff thats all wrong I believe.

    But I used the pythagorean for the second and here what i got:

    first to find the adj side

    tan 36.8 degrees= adj/6.9702m

    adj=52127m


    I used the pythagorean:

    I used r=hyp x=adj y=opp
    trying to find the hypotenuse side now

    r=Sqrt x^2+y^2

    r=(52137)^2+(69702)^2
    r=87044m


    5.461m/s-3.11m/s= 2.31 m/s is his velocity

    C)To find the time:T=D/V
    Distance I used the hyp. distance and the velocity I used his speed with respect to the water
    87044m/2.31m/s=37681s
    t=37681s


    B)
    V=D/T
    V=87043.8m/37681s=2.31m/s
    v=2.31m/s
    I think this maybe more like it,let me know if its correct. Thanx again
     
  5. Oct 4, 2007 #4

    learningphysics

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    Homework Helper

    Part b) and c) don't look right to me.

    For part b) use the triangle with one side 3.311 m/s another side 5.471 m/s and angle
    theta=36.8 degrees.

    what is the 3rd side? what does this third side represent?
     
  6. Oct 4, 2007 #5
    learningphysics are you sure because everything works out in this problem for b and c. For B if i just subtract 5.471m/s-3.311m/s=2.31m/s which i think is correct. To find the time it takes i just simply divide the length of that hypotenuse by his velocity with respect to the river of 2.31m/s. DO you still see something wrong? Someone help please...
     
    Last edited: Oct 4, 2007
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