A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by a current flowing southward at 2.0 meter a second if he ended up exactly across the stream from where he began at what angel to the shore must he swim upstream

1 Answer

  • Answer:

    Angle: [tex]48.19^o[/tex]


    Two-Dimension Motion

    When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

    Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

    To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

    [tex]v_b\ cos\alpha[/tex]

    where [tex]v_b[/tex] is the speed of the boy in still water and [tex]\alpha[/tex] is the angle respect to the shoreline. If the river flows at speed [tex]v_s[/tex], we now set

    [tex]v_b\ cos\alpha=v_s[/tex]

    [tex]\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}[/tex]