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A barge is being towed south at the rate 2 km/hr. A man on the deck walks from west to east at the rate 4ft/s. Find the magnitude and direction of the man's actual velocity.

  • trigonometry -

    1 km = appr 3280.84 feet
    2 km/h = 2(3280.84/3600) ft/s
    = 1.82269 ft/s

    let r be the resultant velocity
    r^2 = 2^2 + 1.82269^2
    r = √7.32219..
    = appr 2.7 ft/s

    tanØ = 1.82269/2 , where Ø is in my right-angled triangle
    Ø = appr 42.34°

    his direction is S 42.34° E

    or by vectors
    vector r = 2(cos270 , sin270) + 1.82269(cos0 , sin0)
    = 2(0, -1) + 1.82269(1,0)
    = (0,-2) + (1.82269,0)
    = (1.82269,-2)

    magnitude = √(1.82269^2 + (-2)^2)
    = 2.7 as above
    tan k = -2/1.82269
    k = 312.34° which is the same as S 42.34 W

  • trigonometry -

    X = 4 Ft/s

    Y = -2km/h = -2000m/3600s * 3.3Ft/m = -1.833 Ft/s

    Tan A = Y/X = -1.833/4 = -0.45825
    A = -24.62o = 24.62o South of East. =
    65.38o East of South = Direction.

    Magnitude = X/Cos A = 4/Cos(-24.62) =
    4.4 Ft/s.

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