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A 200-m-wide river flows due east at a uniform speed of 4.9 m/s. A boat with a speed of 9.0 m/s relative to the water leaves the south bank pointed in a direction 33o west of north. What is the (a) magnitude and (b) direction of the boat's velocity relative to the ground? Give the direction as the angle of the velocity from due north, positive if to the east and negative if to the west. (c) How long does it take for the boat to cross the river?

  • physics -

    (a) Add the water velocity vector to the boat velocity vector with respect to the water. Get the magnitude of the resultant, sqrt (Vx^2 + Vy^2)
    Vx is the downstream component (east), Vy is the cross-stream component (north)
    (b) Get the direction of the resultant. It is the arctangent of the
    Vx/Vy ratio
    (c) Time to cross = (River width)/Vy

  • physics -

    a,b. 33o W. of N.=123o CCW from +x-axis.
    Vr = 9[123o] + 4.9 = -4.9+7.55i + 4.9 = 7.55i = 7.55m/s[90o] = Due North.

    c. V*t = 200m, 7.55t = 200, t = 26.5 s.

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