Question

Can you walk me through this problem? Also, what does "relative to the water" mean?---

A passenger walks from one side of a ferry to the other as it approaches a dock. The passenger's v=1.55 m/s North relative to the ferry and 4.50 m/s at an angle of 30 degrees West of North relative to the water.

A) what is the magnitude of the ferry's velocity relative to the water?

B) what is the direction of the ferry's velocity relative to the water?

Answer 1

This is a problem of vector subtraction. Draw a triangle.

Point north, you have the vector with magnitude 1.55 m/s.
30 degrees to the west of that, you have a vector of 4.5 m/s.
You must find the difference between these using geometry - or you can take the components of each vector and add.

I used the Law of Cosines:
c^2 = a^2 + b^2 - 2*a*b*cos(C)
The vector you are solving for has a length of 'c',
you have a=1.55 and b=4.5, and you have C=30 degrees.
c^2 = 2.4025+20.25-12.081

c^2 = 10.5715
c = 3.25
That's the magnitude.

Using trig, you can find the vertical component of the angled vector is 4.5cos(30) = 3.9. Subtracting the north vector, you have the vertical component of the resultant.
3.9-1.55 = 2.35
Taking 4.5sin(30) = 2.25 gives you the horizontal vector.

Inverse tangent of (2.35/2.25) gives you the angle.
Angle = 46.25 degrees north of west. To find degrees west of north, take 90-46.25= 43.75 degrees.
It really helps to draw a picture.

Pythag can confirm your magnitude. c = sqrt(2.35^2 + 2.25^2) = 3.25