A passenger walks from one side of a ferry to the other as it approaches a dock. If the passenger's velocity is 1.70 m/s due north relative to the ferry, and 4.5 m/s at an angle of 34.0° west of north relative to the water, what are the direction and magnitude of the ferry's velocity relative to the water?
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To determine the direction and magnitude of the ferry's velocity relative to the water, we can use vector addition.
First, let's define the given velocities:
1. The passenger's velocity relative to the ferry is 1.70 m/s due north.
2. The passenger's velocity relative to the water is 4.5 m/s at an angle of 34.0° west of north.
To find the ferry's velocity relative to the water, we need to add the passenger's velocity relative to the water (which includes the passenger's velocity relative to the ferry) to the ferry's velocity relative to the passenger.
1. Start by converting the passenger's velocity relative to the water into its horizontal (x) and vertical (y) components.
Horizontal Component:
V_x = 4.5 m/s * cos(34.0°)
Vertical Component:
V_y = 4.5 m/s * sin(34.0°)
2. Now, add the passenger's velocities relative to the water and the ferry's velocity relative to the passenger.
Horizontal Component:
V_ferry_x = V_x + (ferry's velocity relative to the passenger in the x-direction)
Vertical Component:
V_ferry_y = V_y + (ferry's velocity relative to the passenger in the y-direction)
Since the passenger's velocity relative to the ferry is due north, we know that the ferry's velocity relative to the passenger is due south. Therefore:
V_ferry_x = V_x
V_ferry_y = V_y - 1.70 m/s
3. Finally, calculate the magnitude and direction of the ferry's velocity relative to the water using the Pythagorean theorem and inverse tangent.
Magnitude:
V_ferry = √(V_ferry_x² + V_ferry_y²)
Direction:
θ = tan^(-1)(V_ferry_y / V_ferry_x)
Plug in the values and solve to find the final answer.