A ferry is crossing a river. If the ferry is headed due north with a speed of 3.2 m/s relative to the water and the river's velocity is 4.3 m/s to the east, What will the ferry's velocity relative to the shore be?
(Choose the nearest answer)
To determine the ferry's velocity relative to the shore, we need to combine the velocity of the ferry relative to the water with the velocity of the river.
To do this, we can use vector addition. The velocity of the ferry relative to the shore (Vfs) will be the vector sum of the velocity of the ferry relative to the water (Vfw) and the velocity of the river (Vwr).
Vfs = Vfw + Vwr
Given:
Vfw = 3.2 m/s to the north
Vwr = 4.3 m/s to the east
To add these vectors, we need to calculate their components along the north-south (y) and east-west (x) axes.
The x-component of Vfw is 0 m/s since it is purely in the north direction.
The y-component of Vfw is +3.2 m/s since it is pointing north.
The x-component of Vwr is +4.3 m/s to the east.
The y-component of Vwr is 0 m/s since it is purely in the east direction.
Now, let's add the components together:
Vfsx = Vfwx + Vwrx = 0 + 4.3 = 4.3 m/s to the east
Vfsy = Vfwy + Vwry = 3.2 + 0 = 3.2 m/s to the north
Finally, we can calculate the magnitude and direction of the ferry's velocity relative to the shore using the Pythagorean theorem and trigonometry.
Magnitude (Vfs) = √(Vfsx^2 + Vfsy^2) = √((4.3)^2 + (3.2)^2) = √(18.49 + 10.24) = √28.73 ≈ 5.35 m/s
Direction (θ) = tan^(-1)(Vfsy / Vfsx) = tan^(-1)(3.2 / 4.3) ≈ 36.4° north of east
Therefore, the ferry's velocity relative to the shore is approximately 5.35 m/s, directed approximately 36.4° north of east.