Answer: The speed is 22.2 mi/hr and woman is 8° north relative to the surface of the water.
Step-by-step explanation:
Since we have given that
Speed at which woman walks due west = 3 mi/hr
Speed at which ship moving due north = 22 mi/hr
Now it forms "Pythagorus theorem":
So, it becomes,
[tex]H^2=B^2+P^2\\\\H^2=3^2+22^2\\\\H^2=9+484\\\\H^2=493\\\\H=\sqrt{493}\\H=22.20\ mi/hr[/tex]
And the direction of the woman relative to the surface of the water is given by
[tex]\theta=\sin^{-1}x(\dfrac{3}{22.2})=7.76\approx 8^\circ[/tex]
Hence, the speed is 22.2 mi/hr and woman is 8° north relative to the surface of the water.
The speed of the woman is 22.2 mi\h and the direction is [tex]8^{\circ}[/tex] in the northwest relative to the surface of the water.
Let the vector AB represents a woman walking due west on the deck of a ship at 3 mi/h.
And the vector BC represents the ship moving north at a speed of 22 mi/h
Then the magnitude of the sum of vector AC represents the speed for the women relative to the surface of the water with an angle [tex]\theta[/tex]
Then the perpendicular sides of the triangle ABC (diagram is attached below) are,
[tex]|AB|=3\\|BC|=22[/tex]
So, the magnitude of AC is,
[tex]|AC|=\sqrt{3^2+22^2}\\=22.2[/tex]
And,
[tex]\theta=tan^{-1}(\frac{3}{22})\\=7.765\\=8^{\circ}[/tex]
Therefore, the speed of the woman is 22.2 mi\h and the direction is [tex]8^{\circ}[/tex] in the northwest relative to the surface of the water.
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