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Relative velocity of a canoe

  1. Jul 10, 2007 #1
    1. The problem statement, all variables and given/known data
    A canoe has a velocity of 0.30m/s northwest relative to the earth. The canoe is on a river that is flowing 0.50m/s west relative to the earth. Find the velocity (magnitude and direction ) of the canoe relative to the river.




    2. Relevant equations
    E = earth
    R = river
    C = canoe
    [tex]V_{C/E} = V_{C/R} + V_{R/E} [/tex]
    [tex]V_{C/E} - V_{R/E} = V_{C/R}[/tex]

    3. The attempt at a solution
    the answer at the back of the book is .36m/s 53.6 degrees east of north

    to get to the book's answer, i used the pythagorean theorem
    [tex]\sqrt{{V_{C/E}}^2 - {V_{R/E}}^2} = V_{C/R}[/tex]
    and i confusingly substituted
    \sqrt{((0.30)cos(45) - .50)^2 - ((.30) sin 45))^2} = V_{C/R}
    then get .36
    and i dont know how the V_{C/E}'s x components got involved ....
     
  2. jcsd
  3. Jul 10, 2007 #2
    ohhh
    ok i got it already:
    since : V_{C/E} - V_{R/E} = V_{C/R}
    x component: -(.21)- (-.50)
    y component: .21 - 0
    (.29)^2 + (.21)^2 = V_resultant^2
    0.36

    [tex]\theta = arctan\frac{.21}{.29} [/tex]
    [tex]\theta = 35.9[/tex]degres north of east
    or 90 - 35.9 = 54.1 e of n
     
  4. Jul 10, 2007 #3

    berkeman

    User Avatar

    Staff: Mentor

    I like these kind of threads! :biggrin:
     
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