Anonymous
Anonymous asked in Science & MathematicsPhysics · 5 months ago

physics help?

A swan is flying at a speed of 17.5 m/s. There is a wind blowing

from the East at 12.5 m/s.

a) If the swan pointed due South, what would be the magnitude and direction of its

velocity relative to the ground.

b) If the swan wishes to travel due South, what would be the magnitude and

direction of its velocity relative to the ground.

c) If the swan travels due South as in part b, what will be its displacement after 8.5

4 Answers

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  • NCS
    Lv 7
    5 months ago
    Favourite answer

    Let's assume that 17.5 m/s is the air speed of the swan (as opposed to the ground speed).

    a) Since the air speed and wind speed vectors are perpendicular, use the Pythagorean theorem:

    mag v = √(17.5² + 12.5²) m/s = 21.5 m/s

    and the direction is

    Θ = arctan(12.5/17.5) = 35.5º W of South

    or 54.5º S of W

    which is 234º ccw from "east"

    b) If the swan wants to travel south, it must counter the wind. Measuring the angle from south,

    17.5m/s * sinΘ = 12.5 m/s

    Θ = arcsin(12.5/17.5) = 45.6º E of South

    which is 44.4º S of E, or

    314º ccw from "east"

    The wind having been effectively canceled, the ground speed is

    v = 17.5m/s * cos45.6º = 12.2 m/s

    c) After 8.5 what?

    After 8.5 s, d = 104 m

    After 8.5 min, 6250 m

    After 8.5 h, 375 km

    Hope this helps!

  • 5 months ago

    see diagrams for solutions

    a

    (from east means toward west)

    When you get a good response,

    please consider giving a best answer.

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  • oubaas
    Lv 7
    5 months ago

    a)

    Vx = -12.5 m/sec 

    Vy = -17.5 m/sec

    module of V = √ 12,5^2+17,5^2 = 21,51 m/sec

    heading = arctan (-12.5 / -17.5) = 35.54 ° WoS

    b)

    module of V' = 17.5 m/sec

    V'x = 12.5 m/sec 

    V'y = √ -(12,5^2)+17,5^2 = -12.25 m/sec 

    heading = arctan (V'y/V'x) = arctan(-12.25/12.5) = -44.4° (44.4° SoE)

    c)

    d = V'y*t = 12.25*8.5 = 104.1 m 

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