What Is the Ferry's Velocity Relative to the Water?

[planke]

Homework Statement

A passenger walks from one side of a ferry to the other as it approaches a dock. If the passenger's velocity is 1.90 m/s due north relative to the ferry, and 4.5 m/s at an angle of 33.0° west of north relative to the water, what are the direction and magnitude of the ferry's velocity relative to the water?

_____ degrees west of north
_____ m/s

Homework Equations

Vfw = Vfp + Vwp
theta (angle) = tan-1 (Ay/Ax)
A (magnitude) = sqroot (Ax^2 + Ay^2)

The Attempt at a Solution

Vfw = 1.9 ^y + (-4.5) cos (33) ^x

theta = tan-1 (1.9 / -4.5 cost (33)) = -26.7 degrees
magnitude = sqroot (1.9^2 + (-4.5 cos (33))^2) = 4.2 m/s

I don't understand what I did wrong.

 

[J_21]

I think you just found a few equations and are jamming in the numbers you found in the question in hope of getting the right answer. I'm not sure where you made your mistake because the proper way to do the question into understand what is really happening. I'm not sure what level of physics you are doing right now but this is how I would solve the problem:

Break up into x and y components (I would suggest x being the east and west direction and y being the north and south direction).

y components: (note that whether to use cos and sin for this part can most easily be found by drawing out the vectors)

Vpw = Vfw + Vpf
4.5 cos33 = Vfw + 1.9
=> Vfw = -1.87

x components:

Vpw = Vfw + Vpf
4.5sin33 = Vfw + 0
=> Vfw = 2.45

The magnetude and direction of the vector can be found using the formulas that you posted

 

[Doc Al]

Vfw = Vfp + Vwp

That should be:

Vfw + Vwp = Vfp

Vfw = 1.9 ^y + (-4.5) cos (33) ^x

Don't forget that Vwp has both x and y components. Note that the angle is given with respect to north (the y-axis, not the x-axis), so you'll need to redo the x-component.